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However, its direct utilisation for problems with infinite lattices is problematic due to slowly converging sums over the lattice. Here I develop a way to compute the problematic sums in the reciprocal space, making the \begin_inset Formula $T$ \end_inset -matrix method very suitable for infinite periodic systems as well. \end_layout \begin_layout Section Formulation of the problem \end_layout \begin_layout Standard Assume a system of compact EM scatterers in otherwise homogeneous and isotropic medium, and assume that the system, i.e. both the medium and the scatterers, have linear response. A scattering problem in such system can be written as \begin_inset Formula \[ A_{α}=T_{α}P_{α}=T_{α}(\sum_{β}S_{α\leftarrowβ}A_{β}+P_{0α}) \] \end_inset where \begin_inset Formula $T_{α}$ \end_inset is the \begin_inset Formula $T$ \end_inset -matrix for scatterer α, \begin_inset Formula $A_{α}$ \end_inset is its vector of the scattered wave expansion coefficient (the multipole indices are not explicitely indicated here) and \begin_inset Formula $P_{α}$ \end_inset is the local expansion of the incoming sources. \begin_inset Formula $S_{α\leftarrowβ}$ \end_inset is ... and ... is ... \end_layout \begin_layout Standard ... \end_layout \begin_layout Standard \begin_inset Formula \[ \sum_{β}(\delta_{αβ}-T_{α}S_{α\leftarrowβ})A_{β}=T_{α}P_{0α}. \] \end_inset \end_layout \begin_layout Standard Now suppose that the scatterers constitute an infinite lattice \end_layout \begin_layout Standard \begin_inset Formula \[ \sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{\vect aα}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=T_{\vect aα}P_{0\vect aα}. \] \end_inset Due to the periodicity, we can write \begin_inset Formula $S_{\vect aα\leftarrow\vect bβ}=S_{α\leftarrowβ}(\vect b-\vect a)$ \end_inset and \begin_inset Formula $T_{\vect aα}=T_{\alpha}$ \end_inset . In order to find lattice modes, we search for solutions with zero RHS \begin_inset Formula \[ \sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{α}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=0 \] \end_inset and we assume periodic solution \begin_inset Formula $A_{\vect b\beta}(\vect k)=A_{\vect a\beta}e^{i\vect k\cdot\vect r_{\vect b-\vect a}}$ \end_inset , yielding \begin_inset Formula \begin{eqnarray*} \sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{α}S_{\vect aα\leftarrow\vect bβ})A_{\vect a\beta}\left(\vect k\right)e^{i\vect k\cdot\vect r_{\vect b-\vect a}} & = & 0,\\ \sum_{\vect bβ}(\delta_{\vect{0b}}\delta_{αβ}-T_{α}S_{\vect 0α\leftarrow\vect bβ})A_{\vect 0\beta}\left(\vect k\right)e^{i\vect k\cdot\vect r_{\vect b}} & = & 0,\\ \sum_{β}(\delta_{αβ}-T_{α}\underbrace{\sum_{\vect b}S_{\vect 0α\leftarrow\vect bβ}e^{i\vect k\cdot\vect r_{\vect b}}}_{W_{\alpha\beta}(\vect k)})A_{\vect 0\beta}\left(\vect k\right) & = & 0,\\ A_{\vect 0\alpha}\left(\vect k\right)-T_{α}\sum_{\beta}W_{\alpha\beta}\left(\vect k\right)A_{\vect 0\beta}\left(\vect k\right) & = & 0. \end{eqnarray*} \end_inset Therefore, in order to solve the modes, we need to compute the \begin_inset Quotes eld \end_inset lattice Fourier transform \begin_inset Quotes erd \end_inset of the translation operator, \begin_inset Formula \begin{equation} W_{\alpha\beta}(\vect k)\equiv\sum_{\vect b}S_{\vect 0α\leftarrow\vect bβ}e^{i\vect k\cdot\vect r_{\vect b}}.\label{eq:W definition} \end{equation} \end_inset \end_layout \begin_layout Section Computing the Fourier sum of the translation operator \end_layout \begin_layout Standard The problem evaluating \begin_inset CommandInset ref LatexCommand eqref reference "eq:W definition" \end_inset is the asymptotic behaviour of the translation operator, \begin_inset Formula $S_{\vect 0α\leftarrow\vect bβ}\sim\left|\vect r_{\vect b}\right|^{-1}e^{ik_{0}\left|\vect r_{\vect b}\right|}$ \end_inset that makes the convergence of the sum quite problematic for any \begin_inset Formula $d>1$ \end_inset -dimensional lattice. \begin_inset Foot status open \begin_layout Plain Layout Note that \begin_inset Formula $d$ \end_inset here is dimensionality of the lattice, not the space it lies in, which I for certain reasons assume to be three. (TODO few notes on integration and reciprocal lattices in some appendix) \end_layout \end_inset In electrostatics, one can solve this problem with Ewald summation. Its basic idea is that if what asymptoticaly decays poorly in the direct space, will perhaps decay fast in the Fourier space. I use the same idea here, but everything will be somehow harder than in electrostatics. \end_layout \begin_layout Standard Let us re-express the sum in \begin_inset CommandInset ref LatexCommand eqref reference "eq:W definition" \end_inset in terms of integral with a delta comb \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} W_{\alpha\beta}(\vect k)=\int\ud^{d}\vect r\dc{\basis u}(\vect r)S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})e^{i\vect k\cdot\vect r}.\label{eq:W integral} \end{equation} \end_inset The translation operator \begin_inset Formula $S$ \end_inset is now a function defined in the whole 3d space; \begin_inset Formula $\vect r_{\alpha},\vect r_{\beta}$ \end_inset are the displacements of scatterers \begin_inset Formula $\alpha$ \end_inset and \begin_inset Formula $\beta$ \end_inset in a unit cell. The arrow notation \begin_inset Formula $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})$ \end_inset means \begin_inset Quotes eld \end_inset translation operator for spherical waves originating in \begin_inset Formula $\vect r+\vect r_{\beta}$ \end_inset evaluated in \begin_inset Formula $\vect r_{\alpha}$ \end_inset \begin_inset Quotes erd \end_inset and obviously \begin_inset Formula $S$ \end_inset is in fact a function of a single 3d argument, \begin_inset Formula $S(\vect r_{\alpha}\leftarrow\vect r+\vect r_{\beta})=S(\vect 0\leftarrow\vect r+\vect r_{\beta}-\vect r_{\alpha})=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)=S(-\vect r-\vect r_{\beta}+\vect r_{\alpha})$ \end_inset . Expression \begin_inset CommandInset ref LatexCommand eqref reference "eq:W integral" \end_inset can be rewritten as \begin_inset Formula \[ W_{\alpha\beta}(\vect k)=\left(2\pi\right)^{\frac{d}{2}}\uaft{(\dc{\basis u}S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0))\left(\vect k\right)} \] \end_inset where changed the sign of \begin_inset Formula $\vect r/\vect{\bullet}$ \end_inset has been swapped under integration, utilising evenness of \begin_inset Formula $\dc{\basis u}$ \end_inset . Fourier transform of product is convolution of Fourier transforms, so (using formula \begin_inset CommandInset ref LatexCommand eqref reference "eq:Dirac comb uaFt" \end_inset for the Fourier transform of Dirac comb) \begin_inset Formula \begin{eqnarray} W_{\alpha\beta}(\vect k) & = & \left(\left(\uaft{\dc{\basis u}}\right)\ast\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\right)(\vect k)\nonumber \\ & = & \frac{\left|\det\recb{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\left(\dc{\recb{\basis u}}^{(d)}\ast\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\right)\left(\vect k\right)\nonumber \\ & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}\left(\uaft{S(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\left(\vect k-\vect K\right)\label{eq:W sum in reciprocal space}\\ & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}e^{i\left(\vect k-\vect K\right)\cdot\left(-\vect r_{\beta}+\vect r_{\alpha}\right)}\left(\uaft{S(\vect{\bullet}\leftarrow\vect 0)}\right)\left(\vect k-\vect K\right)\nonumber \end{eqnarray} \end_inset \begin_inset Note Note status open \begin_layout Plain Layout Factor \begin_inset Formula $\left(2\pi\right)^{\frac{d}{2}}$ \end_inset cancels out with the \begin_inset Formula $\left(2\pi\right)^{-\frac{d}{2}}$ \end_inset factor appearing in the convolution/product formula in the unitary angular momentum convention. \end_layout \end_inset As such, this is not extremely helpful because the the \emph on whole \emph default translation operator \begin_inset Formula $S$ \end_inset has singularities in origin, hence its Fourier transform \begin_inset Formula $\uaft S$ \end_inset will decay poorly. \end_layout \begin_layout Standard However, Fourier transform is linear, so we can in principle separate \begin_inset Formula $S$ \end_inset in two parts, \begin_inset Formula $S=S^{\textup{L}}+S^{\textup{S}}$ \end_inset . \begin_inset Formula $S^{\textup{S}}$ \end_inset is a short-range part that decays sufficiently fast with distance so that its direct-space lattice sum converges well; \begin_inset Formula $S^{\textup{S}}$ \end_inset must as well contain all the singularities of \begin_inset Formula $S$ \end_inset in the origin. The other part, \begin_inset Formula $S^{\textup{L}}$ \end_inset , will retain all the slowly decaying terms of \begin_inset Formula $S$ \end_inset but it also has to be smooth enough in the origin, so that its Fourier transform \begin_inset Formula $\uaft{S^{\textup{L}}}$ \end_inset decays fast enough. (The same idea lies behind the Ewald summation in electrostatics.) Using the linearity of Fourier transform and formulae \begin_inset CommandInset ref LatexCommand eqref reference "eq:W definition" \end_inset and \begin_inset CommandInset ref LatexCommand eqref reference "eq:W sum in reciprocal space" \end_inset , the operator \begin_inset Formula $W_{\alpha\beta}$ \end_inset can then be re-expressed as \begin_inset Formula \begin{eqnarray} W_{\alpha\beta}\left(\vect k\right) & = & W_{\alpha\beta}^{\textup{S}}\left(\vect k\right)+W_{\alpha\beta}^{\textup{L}}\left(\vect k\right)\nonumber \\ W_{\alpha\beta}^{\textup{S}}\left(\vect k\right) & = & \sum_{\vect R\in\basis u\ints^{d}}S^{\textup{S}}(\vect 0\leftarrow\vect R+\vect r_{\beta}-\vect r_{\alpha})e^{i\vect k\cdot\vect R}\label{eq:W Short definition}\\ W_{\alpha\beta}^{\textup{L}}\left(\vect k\right) & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\sum_{\vect K\in\recb{\basis u}\ints^{d}}\left(\uaft{S^{\textup{L}}(\vect{\bullet}-\vect r_{\beta}+\vect r_{\alpha}\leftarrow\vect 0)}\right)\left(\vect k-\vect K\right)\label{eq:W Long definition} \end{eqnarray} \end_inset where both sums should converge nicely. \end_layout \begin_layout Section Finding a good decomposition \end_layout \begin_layout Standard The remaining challenge is therefore finding a suitable decomposition \begin_inset Formula $S^{\textup{L}}+S^{\textup{S}}$ \end_inset such that both \begin_inset Formula $S^{\textup{S}}$ \end_inset and \begin_inset Formula $\uaft{S^{\textup{L}}}$ \end_inset decay fast enough with distance and are expressable analytically. With these requirements, I do not expect to find gaussian asymptotics as in the electrostatic Ewald formula—having \begin_inset Formula $\sim x^{-t}$ \end_inset , \begin_inset Formula $t>d$ \end_inset asymptotics would be nice, making the sums in \begin_inset CommandInset ref LatexCommand eqref reference "eq:W Short definition" \end_inset , \begin_inset CommandInset ref LatexCommand eqref reference "eq:W Long definition" \end_inset absolutely convergent. \end_layout \begin_layout Standard The translation operator \begin_inset Formula $S$ \end_inset for compact scatterers in 3d can be expressed as \begin_inset Formula \[ S_{l',m',t'\leftarrow l,m,t}\left(\vect r\leftarrow\vect 0\right)=\sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\theta_{\vect r},\phi_{\vect r}\right)z_{p}^{(J)}\left(k_{0}\left|\vect r\right|\right) \] \end_inset where \begin_inset Formula $Y_{l,m}\left(\theta,\phi\right)$ \end_inset are the spherical harmonics, \begin_inset Formula $z_{p}^{(J)}\left(r\right)$ \end_inset some of the Bessel or Hankel functions (probably \begin_inset Formula $h_{p}^{(1)}$ \end_inset in all meaningful cases; TODO) and \begin_inset Formula $c_{p}^{l,m,t\leftarrow l',m',t'}$ \end_inset are some ugly but known coefficients (REF Xu 1996, eqs. 76,77). \end_layout \begin_layout Standard The spherical Hankel functions can be expressed analytically as (REF DLMF 10.49.6, 10.49.1) \begin_inset Formula \begin{equation} h_{n}^{(1)}(r)=e^{ir}\sum_{k=0}^{n}\frac{i^{k-n-1}}{r^{k+1}}\frac{\left(n+k\right)!}{2^{k}k!\left(n-k\right)!},\label{eq:spherical Hankel function series} \end{equation} \end_inset so if we find a way to deal with the radial functions \begin_inset Formula $s_{k_{0},q}(r)=e^{ik_{0}r}\left(k_{0}r\right)^{-q}$ \end_inset , \begin_inset Formula $q=1,2$ \end_inset in 2d case or \begin_inset Formula $q=1,2,3$ \end_inset in 3d case, we get absolutely convergent summations in the direct space. \end_layout \begin_layout Subsection 2d \end_layout \begin_layout Standard Assume that all scatterers are placed in the plane \begin_inset Formula $\vect z=0$ \end_inset , so that the 2d Fourier transform of the long-range part of the translation operator in terms of Hankel transforms, according to \begin_inset CommandInset ref LatexCommand eqref reference "eq:Fourier v. Hankel tf 2d" \end_inset , reads \end_layout \begin_layout Standard \begin_inset Formula \begin{multline*} \uaft{S_{l',m',t'\leftarrow l,m,t}^{\textup{L}}\left(\vect{\bullet}\leftarrow\vect 0\right)}(\vect k)=\\ \sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\frac{\pi}{2},0\right)e^{i(m'-m)\phi}i^{m'-m}\pht{m'-m}{h_{p}^{(1)\textup{L}}\left(k_{0}\vect{\bullet}\right)}\left(\left|\vect k\right|\right) \end{multline*} \end_inset Here \begin_inset Formula $h_{p}^{(1)\textup{L}}=h_{p}^{(1)}-h_{p}^{(1)\textup{S}}$ \end_inset is a long range part of a given spherical Hankel function which has to be found and which contains all the terms with far-field ( \begin_inset Formula $r\to\infty$ \end_inset ) asymptotics proportional to \begin_inset Formula $\sim e^{ik_{0}r}\left(k_{0}r\right)^{-q}$ \end_inset , \begin_inset Formula $q\le Q$ \end_inset where \begin_inset Formula $Q$ \end_inset is at least two in order to achieve absolute convergence of the direct-space sum, but might be higher in order to speed the convergence up. \end_layout \begin_layout Standard Obviously, all the terms \begin_inset Formula $\propto s_{k_{0},q}(r)=e^{ik_{0}r}\left(k_{0}r\right)^{-q}$ \end_inset , \begin_inset Formula $q>Q$ \end_inset of the spherical Hankel function \begin_inset CommandInset ref LatexCommand eqref reference "eq:spherical Hankel function series" \end_inset can be kept untouched as part of \begin_inset Formula $h_{p}^{(1)\textup{S}}$ \end_inset , as they decay fast enough. \end_layout \begin_layout Standard The remaining task is therefore to find a suitable decomposition of \begin_inset Formula $s_{k_{0},q}(r)=e^{ik_{0}r}\left(k_{0}r\right)^{-q}$ \end_inset , \begin_inset Formula $q\le Q$ \end_inset into short-range and long-range parts, \begin_inset Formula $s_{k_{0},q}(r)=s_{k_{0},q}^{\textup{S}}(r)+s_{k_{0},q}^{\textup{L}}(r)$ \end_inset , such that \begin_inset Formula $s_{k_{0},q}^{\textup{L}}(r)$ \end_inset contains all the slowly decaying asymptotics and its Hankel transforms decay desirably fast as well, \begin_inset Formula $\pht n{s_{k_{0},q}^{\textup{L}}}\left(k\right)=o(z^{-Q})$ \end_inset , \begin_inset Formula $z\to\infty$ \end_inset . The latter requirement calls for suitable regularisation functions— \begin_inset Formula $s_{q}^{\textup{L}}$ \end_inset must be sufficiently smooth in the origin, so that \begin_inset Formula \begin{equation} \pht n{s_{k_{0},q}^{\textup{L}}}\left(k\right)=\int_{0}^{\infty}s_{k_{0},q}^{\textup{L}}\left(r\right)rJ_{n}\left(kr\right)\ud r=\int_{0}^{\infty}s_{k_{0},q}\left(r\right)\rho\left(r\right)rJ_{n}\left(kr\right)\ud r\label{eq:2d long range regularisation problem statement} \end{equation} \end_inset exists and decays fast enough. \begin_inset Formula $J_{\nu}(r)\sim\left(r/2\right)^{\nu}/\Gamma\left(\nu+1\right)$ \end_inset (REF DLMF 10.7.3) near the origin, so the regularisation function should be \begin_inset Formula $\rho(r)=o(r^{q-n-1})$ \end_inset only to make \begin_inset Formula $\pht n{s_{q}^{\textup{L}}}$ \end_inset converge. The additional decay speed requirement calls for at least \begin_inset Formula $\rho(r)=o(r^{q-n+Q-1})$ \end_inset , I guess. At the same time, \begin_inset Formula $\rho(r)$ \end_inset must converge fast enough to one for \begin_inset Formula $r\to\infty$ \end_inset . \end_layout \begin_layout Standard The electrostatic Ewald summation uses regularisation with \begin_inset Formula $1-e^{-cr^{2}}$ \end_inset . However, such choice does not seem to lead to an analytical solution (really? could not something be dug out of DLMF 10.22.54?) for the current problem \begin_inset CommandInset ref LatexCommand eqref reference "eq:2d long range regularisation problem statement" \end_inset . But it turns out that the family of functions \begin_inset Formula \begin{equation} \rho_{\kappa,c}(r)\equiv\left(1-e^{-cr}\right)^{\text{\kappa}},\quad c>0,\kappa\in\nats\label{eq:binom regularisation function} \end{equation} \end_inset might lead to satisfactory results; see below. \end_layout \begin_layout Standard \begin_inset Note Note status open \begin_layout Plain Layout In natural/dimensionless units; \begin_inset Formula $x=k_{0}r$ \end_inset , \begin_inset Formula $\tilde{k}=k/k_{0}$ \end_inset , \begin_inset Formula $č=c/k_{0}$ \end_inset \begin_inset Formula \[ s_{q}(x)\equiv e^{ix}x^{-q} \] \end_inset \begin_inset Formula \[ \tilde{\rho}_{\kappa,č}(x)\equiv\left(1-e^{-čx}\right)^{\text{\kappa}}=\left(1-e^{-\frac{c}{k_{0}}k_{0}r}\right)^{\kappa}=\left(1-e^{-cr}\right)^{\kappa}=\rho_{\kappa,c}(r) \] \end_inset \begin_inset Formula \[ s_{q}^{\textup{L}}\left(x\right)\equiv s_{q}(x)\tilde{\rho}_{\kappa,č}(x)=e^{ix}x^{-q}\left(1-e^{-čx}\right)^{\text{\kappa}} \] \end_inset \begin_inset Formula \begin{eqnarray*} \pht n{s_{q}^{\textup{L}}}\left(\tilde{k}\right) & = & \int_{0}^{\infty}s_{q}^{\textup{L}}\left(x\right)xJ_{n}\left(\tilde{k}x\right)\ud x=\int_{0}^{\infty}s_{q}\left(x\right)\tilde{\rho}_{\kappa,č}(x)xJ_{n}\left(\tilde{k}x\right)\ud x\\ & = & \int_{0}^{\infty}s_{k_{0},q}\left(r\right)\rho_{\kappa,c}(r)\left(k_{0}r\right)J_{n}\left(kr\right)\ud\left(k_{0}r\right)\\ & = & k_{0}^{2}\int_{0}^{\infty}s_{k_{0},q}\left(r\right)\rho_{\kappa,c}(r)rJ_{n}\left(kr\right)\ud r\\ & = & k_{0}^{2}\pht n{s_{k_{0},q}^{\textup{L}}}\left(k\right) \end{eqnarray*} \end_inset \end_layout \end_inset \end_layout \begin_layout Standard \begin_inset Note Note status open \begin_layout Plain Layout Another analytically feasible possibility could be \begin_inset Formula \begin{equation} \rho_{p}^{\textup{ig.}}\equiv e^{-p/x^{2}}.\label{eq:inverse gaussian regularisation function} \end{equation} \end_inset \end_layout \begin_layout Plain Layout Nope, propably did not work. \end_layout \end_inset \end_layout \begin_layout Subsubsection Hankel transforms of the long-range parts, „binomial“ regularisation \begin_inset CommandInset label LatexCommand label name "sub:Hankel-transforms-binom-reg" \end_inset \end_layout \begin_layout Standard Let \begin_inset Note Note status open \begin_layout Plain Layout \begin_inset Formula $\rho_{\kappa,c}$ \end_inset from \begin_inset CommandInset ref LatexCommand eqref reference "eq:binom regularisation function" \end_inset serve as the regularisation fuction and \end_layout \end_inset \end_layout \begin_layout Standard \begin_inset Formula \begin{eqnarray} \pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & \equiv & \int_{0}^{\infty}\frac{e^{ik_{0}r}}{\left(k_{0}r\right)^{q}}J_{n}\left(kr\right)\left(1-e^{-cr}\right)^{\kappa}r\,\ud r\nonumber \\ & = & k_{0}^{-q}\int_{0}^{\infty}r^{1-q}J_{n}\left(kr\right)\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}e^{-(\sigma c-ik_{0})r}\ud r\nonumber \\ & \underset{\equiv}{\textup{form.}} & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\pht n{s_{q,k_{0}}^{\textup{L}1,\sigma c}}\left(k\right).\label{eq:2D Hankel transform of regularized outgoing wave, decomposition} \end{eqnarray} \end_inset From [REF DLMF 10.22.49] one digs \begin_inset Note Note status open \begin_layout Plain Layout \begin_inset Formula \begin{eqnarray*} \mu & \leftarrow & 2-q\\ \nu & \leftarrow & n\\ b & \leftarrow & k\\ a & \leftarrow & c-ik_{0} \end{eqnarray*} \end_inset \end_layout \end_inset \begin_inset Formula \begin{multline} \pht n{s_{q,k_{0}}^{\textup{L}1,c}}\left(k\right)=\frac{k^{n}Γ\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(c-ik_{0}\right)^{2-q+n}}\hgfr\left(\frac{2-q+n}{2},\frac{3-q+n}{2};1+n;\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right),\\ \Re\left(2-q+n\right)>0,\Re(c-ik_{0}\pm k)\ge0,\label{eq:2D Hankel transform of exponentially suppressed outgoing wave as 2F1} \end{multline} \end_inset and using [REF DLMF 15.9.17] and \begin_inset Note Note status open \begin_layout Plain Layout \begin_inset Formula $P_{\nu}^{\mu}=P_{-\nu-1}^{\mu}$ \end_inset \end_layout \end_inset [REF DLMF 14.9.5] \end_layout \begin_layout Standard \begin_inset Note Note status collapsed \begin_layout Plain Layout \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\hgfr\left(\frac{2-q+n}{2},\frac{3-q+n}{2};1+n;\frac{-k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)\\ \mbox{(D15.2.2)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}Γ\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\sum_{s=0}^{\infty}\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{3-q+n}{2}\right)_{s}}{Γ(1+n+s)s!}\left(\frac{-k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{s},\quad\left|\frac{-k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right|<1\\ \end{eqnarray*} \end_inset \end_layout \begin_layout Plain Layout \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\hgfr\left(\frac{2-q+n}{2},\frac{3-q+n}{2};1+n;\frac{-k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)\\ \mbox{(D15.8.2)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}(\\ & & \pi\frac{\left(\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{2-q+n}{2}}}{Γ\left(\frac{3-q+n}{2}\right)\text{Γ}\left(1+n-\frac{2-q+n}{2}\right)}\hgfr\left(\begin{array}{c} \frac{2-q+n}{2},\frac{2-q+n}{2}-\left(1+n\right)+1\\ 1/2 \end{array};-\frac{\left(\sigma c-ik_{0}\right)^{2}}{k^{2}}\right)\\ & - & \pi\frac{\left(\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{3-q+n}{2}}}{\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(1+n-\frac{3-q+n}{2}\right)}\hgfr\left(\begin{array}{c} \frac{3-q+n}{2},\frac{3-q+n}{2}-\left(1+n\right)+1\\ 3/2 \end{array};-\frac{\left(\sigma c-ik_{0}\right)^{2}}{k^{2}}\right))\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\pi(\\ & & \frac{\left(\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{2-q+n}{2}}}{\text{Γ}\left(\frac{3-q+n}{2}\right)\text{Γ}\left(\frac{q+n}{2}\right)}\hgfr\left(\begin{array}{c} \frac{2-q+n}{2},\frac{2-q-n}{2}\\ 1/2 \end{array};-\frac{\left(\sigma c-ik_{0}\right)^{2}}{k^{2}}\right)\\ & - & \frac{\left(\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{3-q+n}{2}}}{\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(\frac{q+n-1}{2}\right)}\hgfr\left(\begin{array}{c} \frac{3-q+n}{2},\frac{3-q-n}{2}\\ 3/2 \end{array};-\frac{\left(\sigma c-ik_{0}\right)^{2}}{k^{2}}\right))\\ \mbox{(D15.2.2)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\pi\sum_{s=0}^{\infty}(\\ & & \frac{\left(\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{2-q+n}{2}}}{\text{Γ}\left(\frac{3-q+n}{2}\right)\text{Γ}\left(\frac{q+n}{2}\right)}\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{1}{2}+s\right)s!}\left(-\frac{\left(\sigma c-ik_{0}\right)^{2}}{k^{2}}\right)^{s}\\ & - & \frac{\left(\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{3-q+n}{2}}}{\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(\frac{q+n-1}{2}\right)}\frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{3}{2}+s\right)s!}\left(-\frac{\left(\sigma c-ik_{0}\right)^{2}}{k^{2}}\right)^{s})\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\kor{k^{n}}\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}\kor{\left(\sigma c-ik_{0}\right)^{2-q+n}}}\pi\sum_{s=0}^{\infty}\left(-1\right)^{s}(\\ & & \frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{3-q+n}{2}\right)\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}k^{-2+q\kor{-n}-2s}\left(\sigma c-ik_{0}\right)^{\kor{2-q+n}+2s}\\ & - & \frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}k^{-3+q\kor{-n}-2s}\left(\sigma c-ik_{0}\right)^{\kor{3-q+n}+2s})\\ \mbox{} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}}\pi\sum_{s=0}^{\infty}\left(-1\right)^{s}(\\ & & \frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{3-q+n}{2}\right)\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\kor{k^{-2+q-2s}}\kor{\left(\sigma c-ik_{0}\right)^{2s}}\\ & - & \frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\kor{k^{-3+q-2s}}\kor{\left(\sigma c-ik_{0}\right)^{1+2s}})\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}}\pi\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\\ & & \times\left(\underbrace{\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{3-q+n}{2}\right)\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}}_{\equiv c_{q,n,s}}-\underbrace{\frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}}_{č_{q,n,s}}\frac{\left(\sigma c-ik_{0}\right)}{k}\right)\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}}\pi\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\kor{\left(\sigma c-ik_{0}\right)^{2s}}c_{q,n,s}-\frac{\left(\sigma c-ik_{0}\right)^{2s+1}}{k}č_{q,n,s}\right)\\ \mbox{(binom.)} & = & \kor{\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}}\frac{\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}}\pi\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(c_{q,n,s}\sum_{t=0}^{2s}\binom{2s}{t}\left(\kor{\sigma}c\right)^{t}\left(-ik_{0}\right)^{2s-t}-č_{q,n,s}\sum_{t=0}^{2s+1}\binom{2s+1}{t}\left(\kor{\sigma}c\right)^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)\\ \mbox{(conds?)} & = & \frac{\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}}\pi\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(c_{q,n,s}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-č_{q,n,s}\sum_{t=0}^{2s+1}\binom{2s+1}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right) \end{eqnarray*} \end_inset now the Stirling number of the 2nd kind \begin_inset Formula $\begin{Bmatrix}t\\ \kappa \end{Bmatrix}=0$ \end_inset if \begin_inset Formula $\kappa>t$ \end_inset . \end_layout \begin_layout Plain Layout What about the gamma fn on the left? Using DLMF 5.5.5, which says \begin_inset Formula $Γ(2z)=\pi^{-1/2}2^{2z-1}\text{Γ}(z)\text{Γ}(z+\frac{1}{2})$ \end_inset we have \begin_inset Formula \[ \text{Γ}\left(2-q+n\right)=\frac{2^{1-q+n}}{\sqrt{\pi}}\text{Γ}\left(\frac{2-q+n}{2}\right)\text{Γ}\left(\frac{3-q+n}{2}\right), \] \end_inset so \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \frac{\kor{\text{Γ}\left(2-q+n\right)}}{\kor{2^{n}}k_{0}^{q}}\kor{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)}\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)}\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)\\ & = & \frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)\\ \mbox{(D5.2.5)} & = & \frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}+s\right)\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\text{Γ}\left(\frac{3-q+n}{2}+s\right)\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s+1}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right) \end{eqnarray*} \end_inset The two terms have to be treated fifferently depending on whether q \begin_inset Formula $q+n$ \end_inset is even or odd. \end_layout \begin_layout Plain Layout First, assume that \begin_inset Formula $q+n$ \end_inset is even, so the left term has gamma functions and pochhammer symbols with integer arguments, while the right one has half-integer arguments. As \begin_inset Formula $n$ \end_inset is non-negative and \begin_inset Formula $q$ \end_inset is positive, \begin_inset Formula $\frac{q+n}{2}$ \end_inset is positive, and the Pochhammer symbol \begin_inset Formula $\left(\frac{2-q-n}{2}\right)_{s}=0$ \end_inset if \begin_inset Formula $s\ge\frac{q+n}{2}$ \end_inset , which transforms the sum over \begin_inset Formula $s$ \end_inset to a finite sum for the left term. However, there still remain divergent terms if \begin_inset Formula $\frac{2-q+n}{2}+s\le0$ \end_inset (let's handle this later; maybe D15.8.6–7 may be then be useful)! Now we need to perform some transformations of variables to make the other sum finite as well \end_layout \begin_layout Plain Layout Pár kroků zpět: \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\kor{\text{Γ}\left(2-q+n\right)}}{\kor{2^{n}}k_{0}^{q}}\kor{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\times\left(\underbrace{\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{3-q+n}{2}\right)}\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}}_{\equiv c_{q,n,s}}-\underbrace{\frac{\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)}\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}}_{č_{q,n,s}}\frac{\left(\sigma c-ik_{0}\right)}{k}\right)\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\times\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}-\frac{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\frac{\left(\sigma c-ik_{0}\right)}{k}\right) \end{eqnarray*} \end_inset \end_layout \begin_layout Plain Layout If \begin_inset Formula $q+n$ \end_inset is even and \begin_inset Formula $2-q+n\le0$ \end_inset \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\kor{\hgfr}\left(\frac{2-q+n}{2},\frac{3-q+n}{2};1+n;\frac{-k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)\\ \mbox{(D15.1.2)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)\koru{\text{Γ}(1+n)}}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\koru{\hgf}\left(\frac{2-q+n}{2},\kor{\frac{3-q+n}{2};1+n;\frac{-k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}}\right)\\ \mbox{(D15.8.6)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\kor{k^{n}}\text{Γ}\left(2-q+n\right)\text{Γ}(1+n)}{2^{n}k_{0}^{q}\kor{\left(\sigma c-ik_{0}\right)^{2-q+n}}}\koru{\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\kor{\left(\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{2-q+n}{2}}}}\hgf\left(\begin{array}{c} \frac{2-q+n}{2},\koru{\kor{1-\left(1+n\right)+\frac{2-q+n}{2}}}\\ \koru{\kor{1-\frac{3-q+n}{2}+\frac{2-q+n}{2}}} \end{array};\koru{\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}}\right)\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\koru{k^{q-2}}\text{Γ}\left(2-q+n\right)\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\koru{\frac{3}{2}\left(2-q+n\right)}}}\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\kor{\hgf\left(\begin{array}{c} \frac{2-q+n}{2},\koru{\frac{2-q-n}{2}}\\ \koru{1/2} \end{array};\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)}\\ \mbox{(D15.2.1)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\kor{\text{Γ}\left(2-q+n\right)}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\koru{\sum_{s=0}^{\infty}\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\left(\frac{1}{2}\right)_{s}s!}\left(\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)^{s}}\\ \mbox{(D5.5.5)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{\kor{2^{n}}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\koru{\frac{2^{1-q\kor{+n}}}{\sqrt{\pi}}\kor{\text{Γ}\left(\frac{2-q+n}{2}\right)}\text{Γ}\left(\frac{3-q+n}{2}\right)}\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\sum_{s=0}^{\infty}\frac{\kor{\left(\frac{2-q+n}{2}\right)_{s}}\left(\frac{2-q-n}{2}\right)_{s}}{\left(\frac{1}{2}\right)_{s}s!}\left(\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)^{s}\\ \mbox{(D5.2.5)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\koru{2^{1-q}}}{\sqrt{\pi}}\text{Γ}\left(\frac{3-q+n}{2}\right)\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\sum_{s=0}^{\infty}\frac{\koru{\text{Γ}\left(\frac{2-q+n}{2}+s\right)}\left(\frac{2-q-n}{2}\right)_{s}}{\left(\frac{1}{2}\right)_{s}s!}\left(\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)^{s}\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{2^{1-q}}{\sqrt{\pi}}\text{Γ}\left(\frac{3-q+n}{2}\right)\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\sum_{s=0}^{\frac{q+n}{2}}\frac{\text{Γ}\left(\frac{2-q+n}{2}+s\right)\left(\frac{2-q-n}{2}\right)_{s}}{\left(\frac{1}{2}\right)_{s}s!}\left(\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)^{s}\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{2^{1-q}}{\sqrt{\pi}}\text{Γ}\left(\frac{3-q+n}{2}\right)\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\sum_{s=0}^{\frac{q+n}{2}}\frac{\text{Γ}\left(\frac{2-q+n}{2}+s\right)\left(\frac{2-q-n}{2}\right)_{s}}{\left(\frac{1}{2}\right)_{s}s!}\left(\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)^{s} \end{eqnarray*} \end_inset now \begin_inset Formula $\left(\frac{2-q-n}{2}\right)_{s}=0$ \end_inset whenever \begin_inset Formula $s\ge\frac{q+n}{2}$ \end_inset and \begin_inset Formula $\text{Γ}\left(\frac{2-q+n}{2}+s\right)$ \end_inset is singular whenever \begin_inset Formula $s\le-\frac{2-q+n}{2}$ \end_inset , so we are no less fucked than before. Maybe let's try the other variable transformation. Or what about (D15.8.27)? \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}\left(2-q+n\right)\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\kor{\hgf\left(\begin{array}{c} \frac{2-q+n}{2},\frac{2-q-n}{2}\\ 1/2 \end{array};\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)}\\ \mbox{(D15.8.27)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}\left(2-q+n\right)\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\kor{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\koru{\frac{\kor{Γ\left(\frac{3-q+n}{2}\right)}Γ\left(\frac{3-q-n}{2}\right)}{2Γ\left(\frac{1}{2}\right)Γ\left(2-q+\frac{1}{2}\right)}\left(\hgf\left(\begin{array}{c} 2-q+n,2-q-n\\ 2-q+\frac{1}{2} \end{array};\frac{1}{2}-\frac{\sigma c-ik_{0}}{ik}\right)+\hgf\left(\begin{array}{c} 2-q+n,2-q-n\\ 2-q+\frac{1}{2} \end{array};\frac{1}{2}+\frac{\sigma c-ik_{0}}{ik}\right)\right)}\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}\left(2-q+n\right)\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\kor{\text{Γ}\koru{\left(\frac{3-q+n}{2}-\frac{2-q+n}{2}\right)}}\text{Γ}\left(\frac{3-q-n}{2}\right)}{\left(1+n\right)_{-\frac{2-q+n}{2}}2\kor{\text{Γ}\left(\frac{1}{2}\right)}\text{Γ}\left(2-q+\frac{1}{2}\right)}\left(\hgf\left(\begin{array}{c} 2-q+n,2-q-n\\ 2-q+\frac{1}{2} \end{array};\frac{1}{2}-\frac{\sigma c-ik_{0}}{ik}\right)+\hgf\left(\begin{array}{c} 2-q+n,2-q-n\\ 2-q+\frac{1}{2} \end{array};\frac{1}{2}+\frac{\sigma c-ik_{0}}{ik}\right)\right)\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}\left(2-q+n\right)\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\text{Γ}\left(\frac{3-q-n}{2}\right)}{\left(1+n\right)_{-\frac{2-q+n}{2}}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\kor{\left(\hgf\left(\begin{array}{c} 2-q+n,2-q-n\\ 2-q+\frac{1}{2} \end{array};\frac{1}{2}-\frac{\sigma c-ik_{0}}{ik}\right)+\hgf\left(\begin{array}{c} 2-q+n,2-q-n\\ 2-q+\frac{1}{2} \end{array};\frac{1}{2}+\frac{\sigma c-ik_{0}}{ik}\right)\right)}\\ \mbox{(D15.2.1)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}\left(2-q+n\right)\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\text{Γ}\left(\frac{3-q-n}{2}\right)}{\left(1+n\right)_{-\frac{2-q+n}{2}}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\koru{\sum_{s=0}^{\infty}\left(\frac{\left(2-q+n\right)_{s}\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\kor{\left(\left(\frac{1}{2}-\frac{\sigma c-ik_{0}}{ik}\right)^{s}+\left(\frac{1}{2}+\frac{\sigma c-ik_{0}}{ik}\right)^{s}\right)}\right)}\\ \mbox{(binom)} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\text{Γ}\left(\frac{3-q-n}{2}\right)}{\left(1+n\right)_{-\frac{2-q+n}{2}}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\kor{\left(2-q+n\right)_{s}}\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\koru{\sum_{r=0}^{s}\binom{s}{r}\left(\frac{\sigma c-ik_{0}}{ik}\right)^{r}2^{r-s}\left(\left(-1\right)^{r}+1\right)}\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\text{Γ}\left(\frac{3-q-n}{2}\right)}{\kor{\left(1+n\right)_{-\frac{2-q+n}{2}}}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\koru{\text{Γ}\left(2-q+n+s\right)}\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=0}^{s}\binom{s}{r}\left(\frac{\sigma c-ik_{0}}{ik}\right)^{r}2^{r-s}\left(\left(-1\right)^{r}+1\right)\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\kor{\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}}\frac{\koru{\text{Γ}\left(1+n\right)}\text{Γ}\left(\frac{3-q-n}{2}\right)}{\koru{\text{Γ}\left(\frac{q+n}{2}\right)}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=0}^{s}\binom{s}{r}\kor{\left(\frac{\sigma c-ik_{0}}{ik}\right)^{r}}2^{r-s}\left(\left(-1\right)^{r}+1\right)\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}}\frac{\text{Γ}\left(1+n\right)\text{Γ}\left(\frac{3-q-n}{2}\right)}{Γ\left(\frac{q+n}{2}\right)2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=0}^{s}\binom{s}{r}\koru{\left(ik\right)^{-r}}\koru{\kor{\left(\sigma c-ik_{0}\right)^{r-\frac{3}{2}\left(2-q+n\right)}}}2^{r-s}\left(\left(-1\right)^{r}+1\right)\\ (bionm) & = & \kor{\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}}\frac{\text{Γ}\left(1+n\right)\text{Γ}\left(\frac{3-q-n}{2}\right)}{\text{Γ}\left(\frac{q+n}{2}\right)2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=0}^{s}\binom{s}{r}\left(ik\right)^{-r}\koru{\sum_{w=0}^{\infty|r-\frac{3}{2}\left(2-q+n\right)}\binom{r-\frac{3}{2}\left(2-q+n\right)}{w}\kor{\sigma^{w}}c^{w}\left(-ik_{0}^{r-\frac{3}{2}\left(2-q+n\right)-w}\right)}2^{r-s}\left(\left(-1\right)^{r}+1\right)\\ & = & \koru{\kappa!\left(-1\right)^{\kappa}}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}}\frac{\text{Γ}\left(1+n\right)\text{Γ}\left(\frac{3-q-n}{2}\right)}{\text{Γ}\left(\frac{q+n}{2}\right)2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=\kor 0}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=\kor 0}^{s}\binom{\kor s}{\kor r}\left(ik\right)^{-r}\sum_{w=\kor 0}^{\infty|r-\frac{3}{2}\left(2-q+n\right)}\binom{r-\frac{3}{2}\left(2-q+n\right)}{\kor w}\koru{\kor{\begin{Bmatrix}w\\ \kappa \end{Bmatrix}}}c^{w}\left(-ik_{0}^{r-\frac{3}{2}\left(2-q+n\right)-w}\right)2^{r-s}\left(\left(-1\right)^{r}+1\right)\\ & = & \kappa!\left(-1\right)^{\kappa}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}}\frac{\text{Γ}\left(1+n\right)\text{Γ}\left(\frac{3-q-n}{2}\right)}{\text{Γ}\left(\frac{q+n}{2}\right)2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=\koru{\kappa}}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=\koru{\kappa}}^{s}\binom{s}{r}\left(ik\right)^{-r}\sum_{w=\koru{\kappa}}^{\infty|r-\frac{3}{2}\left(2-q+n\right)}\binom{r-\frac{3}{2}\left(2-q+n\right)}{w}\begin{Bmatrix}w\\ \kappa \end{Bmatrix}c^{w}\left(-ik_{0}^{r-\frac{3}{2}\left(2-q+n\right)-w}\right)2^{r-s}\left(\left(-1\right)^{r}+1\right)\\ & = & \kappa!\left(-1\right)^{\kappa}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}}\frac{\text{Γ}\left(1+n\right)\text{Γ}\left(\frac{3-q-n}{2}\right)}{\text{Γ}\left(\frac{q+n}{2}\right)2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=\kappa}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=\kappa}^{s}\binom{s}{r}\left(ik\right)^{-r}\sum_{w=\kappa}^{\infty|r-\frac{3}{2}\left(2-q+n\right)}\binom{r-\frac{3}{2}\left(2-q+n\right)}{w}\begin{Bmatrix}w\\ \kappa \end{Bmatrix}c^{w}\left(-ik_{0}^{r-\frac{3}{2}\left(2-q+n\right)-w}\right)2^{r-s}\left(\left(-1\right)^{r}+1\right) \end{eqnarray*} \end_inset \end_layout \begin_layout Plain Layout The previous things are valid only if \begin_inset Formula $q$ \end_inset has a small non-integer part, \begin_inset Formula $q=q'+\varepsilon$ \end_inset . They might still play a role in the series (especially in the infinite ones) when taking the limit \begin_inset Formula $\varepsilon\to0$ \end_inset . However, we got rid of the singularities in \begin_inset Formula $\text{Γ}\left(2-q+n+s\right)$ \end_inset if \begin_inset Formula $\kappa$ \end_inset is large enough. \end_layout \begin_layout Plain Layout and we get same shit as before due to the singular \begin_inset Formula $\text{Γ}\left(2-q+n+s\right)$ \end_inset . However, \begin_inset Formula \begin{eqnarray*} (...) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\text{Γ}\left(\frac{3-q-n}{2}\right)}{\left(1+n\right)_{-\frac{2-q+n}{2}}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=0}^{s}\binom{s}{r}\left(\frac{\sigma c-ik_{0}}{ik}\right)^{r}2^{r-s}\kor{\left(\left(-1\right)^{r}+1\right)}\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\text{Γ}\left(\frac{3-q-n}{2}\right)}{\left(1+n\right)_{-\frac{2-q+n}{2}}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=0}^{\koru{floor(s/2)}}\binom{s}{\koru{2r}}\left(\frac{\sigma c-ik_{0}}{ik}\right)^{\koru{2r}}2^{\koru{2r}-s}\left(\left(-1\right)^{\koru{2r}}+1\right) \end{eqnarray*} \end_inset \begin_inset Formula \begin{eqnarray*} (...) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\text{Γ}\left(\frac{3-q-n}{2}\right)}{\left(1+n\right)_{-\frac{2-q+n}{2}}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=0}^{s}\binom{s}{r}\kor{\left(\frac{\sigma c-ik_{0}}{ik}\right)^{r}}2^{r-s}\left(\left(-1\right)^{r}+1\right)\\ binom & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\text{Γ}\left(\frac{3-q-n}{2}\right)}{\left(1+n\right)_{-\frac{2-q+n}{2}}2\text{Γ}\left(2-q+\frac{1}{2}\right)}\sum_{s=0}^{\infty}\frac{\text{Γ}\left(2-q+n+s\right)\left(2-q-n\right)_{s}}{\left(2-q+\frac{1}{2}\right)_{s}s!}\sum_{r=0}^{s}\binom{s}{r}\koru{\left(ik\right)^{-r}\sum_{b=0}^{r}\binom{r}{b}\sigma^{b}c^{b}\left(-ik_{0}\right)^{r-b}}2^{r-s}\left(\left(-1\right)^{r}+1\right)\\ & = \end{eqnarray*} \end_inset \end_layout \begin_layout Plain Layout aaah. Let's assume that \begin_inset Formula $q$ \end_inset is not exactly \begin_inset Formula \begin{eqnarray*} & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\kor{\text{Γ}\left(2-q+n\right)}\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\sum_{s=0}^{\infty}\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\left(\frac{1}{2}\right)_{s}s!}\left(\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)^{s}\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{q-2}\text{Γ}\left(2-q+n\right)\text{Γ}(1+n)}{2^{n}k_{0}^{q}\left(\sigma c-ik_{0}\right)^{\frac{3}{2}\left(2-q+n\right)}}\frac{\left(\frac{3-q+n}{2}\right)_{-\frac{2-q+n}{2}}}{\left(1+n\right)_{-\frac{2-q+n}{2}}}\sum_{s=0}^{\infty}k^{-2s}\frac{\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\left(\frac{1}{2}\right)_{s}s!}\left(\frac{\left(\sigma c-ik_{0}\right)^{2}}{-k^{2}}\right)^{s} \end{eqnarray*} \end_inset zpět \end_layout \begin_layout Plain Layout \begin_inset Formula \begin{eqnarray*} & = & \frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\sum_{t=0}^{2s+1}\binom{2s+1}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right)\\ & = & \frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\kappa!\left(-1\right)^{\kappa}\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)\text{Γ}\left(1+s\right)}\sum_{t=0}^{2s}\binom{2s}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s-t}-\frac{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)\text{Γ}\left(1+s\right)}\sum_{t=0}^{2s+1}\binom{2s+1}{t}\begin{Bmatrix}t\\ \kappa \end{Bmatrix}c^{t}\left(-ik_{0}\right)^{2s+1-t}k^{-1}\right) \end{eqnarray*} \end_inset \end_layout \end_inset \begin_inset Note Note status collapsed \begin_layout Plain Layout \begin_inset Formula \begin{eqnarray*} a & \leftarrow & \frac{2-q+n}{2}\\ c & \leftarrow & 1+n\\ z & \leftarrow & \frac{-k^{2}}{\left(c-ik_{0}\right)^{2}} \end{eqnarray*} \end_inset \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}1,c}}\left(k\right) & = & \frac{k^{n}Γ\left(2-q+n\right)}{2^{n}k_{0}^{q}\left(c-ik_{0}\right)^{2-q+n}}2^{n}\left(\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{-\frac{n}{2}}\left(1-\left(\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right)\right)^{-\frac{2-q+n}{2}+\frac{n}{2}}P_{2-q+n-(1+n)}^{1-(1+n)}\left(\frac{1}{\sqrt{1-\left(\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right)}}\right)\\ & = & \frac{k^{n}Γ\left(2-q+n\right)}{k_{0}^{q}\left(c-ik_{0}\right)^{2-q+n}}\left(\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{-\frac{n}{2}}\left(1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{\frac{q}{2}-1}P_{1-q}^{-n}\left(\frac{1}{\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}}\right) \end{eqnarray*} \end_inset \begin_inset Formula \[ \left|\ph\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right|<\pi,\quad\left|\ph\left(1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}\right)\right|<\pi \] \end_inset in other words, neither \begin_inset Formula $-k^{2}/\left(c-ik_{0}\right)^{2}$ \end_inset nor \begin_inset Formula $1+k^{2}/\left(c-ik_{0}\right)^{2}$ \end_inset can be non-positive real number. For assumed positive \begin_inset Formula $k_{0}$ \end_inset and non-negative \begin_inset Formula $c$ \end_inset and \begin_inset Formula $k$ \end_inset , the former case can happen only if \begin_inset Formula $k=0$ \end_inset and the latter only if \begin_inset Formula $c=0\wedge k_{0}=k$ \end_inset . \begin_inset Formula \begin{eqnarray*} \left|\ph\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right|<\pi & \Leftrightarrow & \left|\ph\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right|\neq\pi\\ \varphi & \equiv & \ph\left(c-ik_{0}\right)<0,\\ \ph k & \equiv & 0\\ \ph\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}} & = & 2\varphi\\ \rightsquigarrow\left|\varphi\right| & \neq & \pi/2\\ \rightsquigarrow c & \neq & k_{0}\\ \left|\ph\left(1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}\right)\right| & = & \left|-2\varphi+\ph\left(\left(c-ik_{0}\right)^{2}+k^{2}\right)\right| \end{eqnarray*} \end_inset Finally, swapping the first two arguments of \begin_inset Formula $\hgfr$ \end_inset in the hypergeometric represenation [REF DLMF 14.3.6] (note [REF DLMF §14.21(iii)] that this also holds for complex arguments) of Legendre functions gives \begin_inset Formula $P_{\nu}^{\mu}=P_{-\nu-1}^{\mu}$ \end_inset , so the above result can be written \begin_inset Formula \[ \pht n{s_{q,k_{0}}^{\textup{L}1,c}}\left(k\right)=\frac{k^{n}\text{Γ}\left(2-q+n\right)}{k_{0}^{q}\left(c-ik_{0}\right)^{2-q+n}}\left(\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{-\frac{n}{2}}\left(1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{\frac{q}{2}-1}P_{q}^{-n}\left(\frac{1}{\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}}\right). \] \end_inset Let's polish it a bit more \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}1,c}}\left(k\right) & = & \frac{Γ\left(2-q+n\right)}{k_{0}^{q}\left(c-ik_{0}\right)^{2-q}}\left(-1\right)^{-\frac{n}{2}}\left(1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{\frac{q}{2}-1}P_{q}^{-n}\left(\frac{1}{\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}}\right)\\ & = & \frac{\text{Γ}\left(2-q+n\right)}{k_{0}^{q}}\left(-1\right)^{-\frac{n}{2}}\left(\left(c-ik_{0}\right)^{2}+k^{2}\right)^{\frac{q}{2}-1}P_{q}^{-n}\left(\frac{1}{\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}}\right). \end{eqnarray*} \end_inset \end_layout \end_inset \size footnotesize \begin_inset Formula \begin{multline} \pht n{s_{q,k_{0}}^{\textup{L}1,c}}\left(k\right)=\frac{k^{n}Γ\left(2-q+n\right)}{k_{0}^{q}\left(c-ik_{0}\right)^{2-q+n}}\left(\frac{-k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{-\frac{n}{2}}\left(1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}\right)^{\frac{q}{2}-1}P_{q}^{-n}\left(\frac{1}{\sqrt{1+\frac{k^{2}}{\left(c-ik_{0}\right)^{2}}}}\right),\\ k>0\wedge k_{0}>0\wedge c\ge0\wedge\lnot\left(c=0\wedge k_{0}=k\right)\label{eq:2D Hankel transform of exponentially suppressed outgoing wave expanded} \end{multline} \end_inset \size default with principal branches of the hypergeometric functions, associated Legendre functions, and fractional powers. The conditions from \begin_inset CommandInset ref LatexCommand eqref reference "eq:2D Hankel transform of exponentially suppressed outgoing wave as 2F1" \end_inset should hold, but we will use \begin_inset CommandInset ref LatexCommand eqref reference "eq:2D Hankel transform of exponentially suppressed outgoing wave expanded" \end_inset formally even if they are violated, with the hope that the divergences eventually cancel in \begin_inset CommandInset ref LatexCommand eqref reference "eq:2D Hankel transform of regularized outgoing wave, decomposition" \end_inset . \end_layout \begin_layout Standard \begin_inset Note Note status collapsed \begin_layout Plain Layout Let's do it. \begin_inset Formula \begin{eqnarray*} \pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right) & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)}{k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\left(\frac{-k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{n}{2}}\left(1+\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{\frac{q}{2}-1}P_{q}^{-n}\left(\frac{1}{\sqrt{1+\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}}}\right)\\ & = & \sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k^{n}\text{Γ}\left(2-q+n\right)}{k_{0}^{q}\left(\sigma c-ik_{0}\right)^{2-q+n}}\left(\frac{-k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{-\frac{n}{2}}\left(1+\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}\right)^{\frac{q}{2}-1}P_{q}^{-n}\left(\frac{1}{\sqrt{1+\frac{k^{2}}{\left(\sigma c-ik_{0}\right)^{2}}}}\right) \end{eqnarray*} \end_inset \end_layout \end_inset One problematic element here is the gamma function \begin_inset Formula $\text{Γ}\left(2-q+n\right)$ \end_inset which is singular if the argument is zero or negative integer, i.e. if \begin_inset Formula $q-n\ge2$ \end_inset ; which is painful especially because of the case \begin_inset Formula $q=2,n=0$ \end_inset . The associated Legendre function can be expressed as a finite \begin_inset Quotes eld \end_inset polynomial \begin_inset Quotes erd \end_inset if \begin_inset Formula $q\ge n$ \end_inset . In other cases, different expressions can be obtained from \begin_inset CommandInset ref LatexCommand ref reference "eq:2D Hankel transform of exponentially suppressed outgoing wave as 2F1" \end_inset using various transformation formulae from either DLMF or \begin_inset ERT status open \begin_layout Plain Layout \backslash begin{russian} \end_layout \end_inset Прудников \begin_inset ERT status open \begin_layout Plain Layout \backslash end{russian} \end_layout \end_inset . \end_layout \begin_layout Standard In fact, Mathematica is usually able to calculate the transforms for specific values of \begin_inset Formula $\kappa,q,n$ \end_inset , but it did not find any general formula for me. The resulting expressions are finite sums of algebraic functions, Table \begin_inset CommandInset ref LatexCommand ref reference "tab:Asymptotical-behaviour-Mathematica" \end_inset shows how fast they decay with growing \begin_inset Formula $k$ \end_inset for some parameters. One particular case where Mathematica did not help at all is \begin_inset Formula $q=2,n=0$ \end_inset , which is unfortunately important. \end_layout \begin_layout Standard \begin_inset Float table wide false sideways false status open \begin_layout Plain Layout \align center \size footnotesize \begin_inset Tabular \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $\kappa=0$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $n$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 0 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $q$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize x \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize w \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 0 \end_layout \end_inset \end_inset \begin_inset space \hspace*{\fill} \end_inset \begin_inset Tabular \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $\kappa=1$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $n$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 0 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 3 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 4 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $q$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize w \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 3 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize x \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize w \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \end_inset \begin_inset space \hspace*{\fill} \end_inset \begin_inset Tabular \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $\kappa=2$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $n$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 0 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 3 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 4 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize \begin_inset Formula $q$ \end_inset \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 0/w \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 3 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 4 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 3 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 3 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize x \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 3 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 2 \end_layout \end_inset \begin_inset Text \begin_layout Plain Layout \size footnotesize 1 \end_layout \end_inset \end_inset \end_layout \begin_layout Plain Layout \begin_inset Caption Standard \begin_layout Plain Layout Asymptotical behaviour of some \begin_inset CommandInset ref LatexCommand eqref reference "eq:2D Hankel transform of regularized outgoing wave, decomposition" \end_inset obtained by Mathematica for \begin_inset Formula $k\to\infty$ \end_inset . The table entries are the \begin_inset Formula $N$ \end_inset of \begin_inset Formula $\pht n{s_{q,k_{0}}^{\textup{L}\kappa,c}}\left(k\right)=o\left(1/k^{N}\right)$ \end_inset . The special entry \begin_inset Quotes eld \end_inset x \begin_inset Quotes erd \end_inset means that Mathematica was not able to calculate the integral, and \begin_inset Quotes eld \end_inset w \begin_inset Quotes erd \end_inset denotes that the first returned term was not simply of the kind \begin_inset Formula $(\ldots)k^{-N-1}$ \end_inset . \begin_inset CommandInset label LatexCommand label name "tab:Asymptotical-behaviour-Mathematica" \end_inset \end_layout \end_inset \end_layout \end_inset \begin_inset Note Note status open \begin_layout Plain Layout \begin_inset ERT status open \begin_layout Plain Layout \backslash begin{russian} \end_layout \end_inset Градштейн и Рыжик \begin_inset ERT status open \begin_layout Plain Layout \backslash end{russian} \end_layout \end_inset 6.512.1 has expression for \begin_inset Formula $\int_{0}^{\infty}J_{\mu}\left(ax\right)J_{\nu}\left(bx\right)\ud x$ \end_inset , \begin_inset Formula $\Re\left(\mu+\nu\right)>-1$ \end_inset in terms of hypergeometric function. Unfortunately, no corresponding and general enough expression for \begin_inset Formula $\int_{0}^{\infty}J_{\mu}\left(ax\right)Y_{\nu}\left(bx\right)\ud x$ \end_inset . \end_layout \end_inset \end_layout \begin_layout Paragraph Case \begin_inset Formula $n=0,q=2$ \end_inset \end_layout \begin_layout Standard As shown in a separate note, \end_layout \begin_layout Standard \begin_inset Formula \[ \pht 0{s_{2,k_{0}}^{\textup{L}\kappa,c}}\left(k\right)=-\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{1}{k_{0}^{2}}\sinh^{-1}\left(\frac{\sigma c-ik_{0}}{k}\right) \] \end_inset for \begin_inset Formula $\kappa\ge?$ \end_inset , \begin_inset Formula $k>k_{0}?$ \end_inset \end_layout \begin_layout Paragraph Case \begin_inset Formula $n=1,q=3$ \end_inset \end_layout \begin_layout Standard As shown in separate note (check whether copied correctly) \begin_inset Formula \[ \pht 1{s_{3,k_{0}}^{\textup{L}\kappa>3,c}}\left(k\right)=-\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\left(-ik_{0}+c\sigma\right)\sqrt{1-\left(\frac{k_{0}+ic\sigma}{k}\right)^{2}}-ik\sin^{-1}\left(\frac{k_{0}+ic\sigma}{k}\right)}{2k_{0}^{3}} \] \end_inset for \begin_inset Formula $\kappa\ge3$ \end_inset , \begin_inset Formula $k>k_{0}?$ \end_inset \end_layout \begin_layout Paragraph Case \begin_inset Formula $n=0,q=3$ \end_inset \end_layout \begin_layout Standard As shown in separate note (check whether copied correctly) \lang finnish \begin_inset Note Note status collapsed \begin_layout Plain Layout \lang finnish Sum[((-1)^(1 + sig)*(k*Sqrt[(k^2 - (k0 + I*c*sig)^2)/k^2] + (k0 + I*c*sig)*ArcSi n[(k0 + I*c*sig)/k])*Binomial[kap, sig])/k0^3, {sig, 0, kap}] \end_layout \end_inset \begin_inset Formula \begin{eqnarray*} \pht 0{s_{3,k_{0}}^{\textup{L}\kappa>3,c}}\left(k\right) & = & -\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k\sqrt{1-\left(\frac{k_{0}+ic\sigma}{k}\right)^{2}}+\left(k_{0}+ic\sigma\right)\sin^{-1}\left(\frac{k_{0}+ic\sigma}{k}\right)}{k_{0}^{3}} \end{eqnarray*} \end_inset \lang english for \begin_inset Formula $\kappa\ge2$ \end_inset , \begin_inset Formula $k>k_{0}?$ \end_inset \end_layout \begin_layout Standard \begin_inset Note Note status open \begin_layout Plain Layout From Wikipedia page on binomial coefficient, eq. (10) and around: \end_layout \begin_layout Plain Layout When \begin_inset Formula $P(x)$ \end_inset is of degree less than or equal to \begin_inset Formula $n$ \end_inset , \begin_inset Formula \[ \sum_{j=0}^{n}(-1)^{j}\binom{n}{j}P(n-j)=n!a_{n} \] \end_inset where \begin_inset Formula $a_{n}$ \end_inset is the coefficient of degree \begin_inset Formula $n$ \end_inset in \begin_inset Formula $P(x)$ \end_inset . \end_layout \begin_layout Plain Layout More generally, \begin_inset Formula \[ \sum_{j=0}^{n}(-1)^{j}\binom{n}{j}P(m+(n-j)d)=d^{n}n!a_{n} \] \end_inset where \begin_inset Formula $m$ \end_inset and \begin_inset Formula $d$ \end_inset are complex numbers. \end_layout \end_inset \begin_inset Note Note status open \begin_layout Subsubsection Hankel transforms of the long-range parts, alternative regularisation with \begin_inset Formula $e^{-p/x^{2}}$ \end_inset \begin_inset CommandInset label LatexCommand label name "sub:Hankel-transforms-ig-reg" \end_inset \end_layout \begin_layout Plain Layout From [REF \begin_inset ERT status open \begin_layout Plain Layout \backslash begin{russian} \end_layout \end_inset Прудников, том 2 \begin_inset ERT status open \begin_layout Plain Layout \backslash end{russian} \end_layout \end_inset , 2.12.9.14] \begin_inset Formula \begin{multline} \int_{0}^{\infty}x^{\alpha-1}e^{-p/x^{2}}J_{\nu}\left(cx\right)\,\ud x=\frac{2^{\alpha-1}}{c^{\alpha}}Γ\begin{bmatrix}\left(\alpha+\nu\right)/2\\ 1+\left(\nu-\alpha\right)/2 \end{bmatrix}{}_{0}F_{2}\left(1-\frac{\nu+\alpha}{2},1+\frac{\nu-\alpha}{2};\frac{c^{2}p}{4}\right)\\ +\frac{c^{\nu}p^{\left(\alpha+\nu\right)/2}}{2^{\nu+1}}\text{Γ}\begin{bmatrix}\left(\alpha+\nu\right)/2\\ \nu+1 \end{bmatrix}{}_{0}F_{2}\left(1+\frac{\nu+\alpha}{2},\nu+1;\frac{c^{2}p}{4}\right),\qquad[c,\Re p>0;\Re\alpha<3/2].\label{eq:prudnikov2 eq 2.12.9.14} \end{multline} \end_inset Let now \begin_inset Formula $\rho_{p}^{\textup{ig.}}$ \end_inset from \begin_inset CommandInset ref LatexCommand eqref reference "eq:inverse gaussian regularisation function" \end_inset serve as the regularisation fuction and \begin_inset Formula \[ \pht n{s_{q,k_{0}}^{\textup{L}'p}}\left(k\right)\equiv\int_{0}^{\infty}\frac{e^{ik_{0}r}}{\left(k_{0}r\right)^{q}}J_{n}\left(kr\right)e^{-p/r^{2}}r\,\ud r. \] \end_inset And it seems that this is a dead-end, because \begin_inset CommandInset ref LatexCommand eqref reference "eq:prudnikov2 eq 2.12.9.14" \end_inset cannot deal with the \begin_inset Formula $e^{ik_{0}r}$ \end_inset part. Damn. \end_layout \end_inset \end_layout \begin_layout Subsection 3d (TODO) \end_layout \begin_layout Standard \begin_inset Formula \begin{multline*} \uaft{S_{l',m',t'\leftarrow l,m,t}\left(\vect{\bullet}\leftarrow\vect 0\right)}(\vect k)=\\ \sum_{p}c_{p}^{l',m',t'\leftarrow l,m,t}\ush p{m'-m}\left(\theta_{\vect k},\phi_{\vect k}\right)\left(-i\right)^{p}\usht p{z_{p}^{(J)}}\left(\left|\vect k\right|\right) \end{multline*} \end_inset \end_layout \begin_layout Section Exponentially converging decompositions \end_layout \begin_layout Standard (As in Linton, Thompson, Journal of Computational Physics 228 (2009) 1815–1829 [LT].) \end_layout \begin_layout Standard [LT] offers a better decomposition than above. Let \begin_inset Formula \begin{eqnarray*} \sigma_{n}^{m}\left(\vect{\beta}\right) & = & \sum_{\vect R\in\Lambda}^{'}e^{i\vect{\beta}\cdot\vect R}\swv_{n}^{m}\left(\vect R\right),\\ \swv_{n}^{m}\left(\vect r\right) & = & Y_{n}^{m}\left(\hat{\vect r}\right)h_{n}\left(\left|\vect r\right|\right). \end{eqnarray*} \end_inset Then, we have a decomposition \begin_inset Formula $\sigma_{n}^{m}=\sigma_{n}^{m(0)}+\sigma_{n}^{m(1)}+\sigma_{n}^{m(2)}$ \end_inset . The real-space sum part \begin_inset Formula $\sigma_{n}^{m(2)}$ \end_inset is already \begin_inset Quotes eld \end_inset convention independent \begin_inset Quotes erd \end_inset in [LT(4.5)] (i.e. the result is also expressed in terms of \begin_inset Formula $Y_{n}^{m}$ \end_inset , so it is valid regardless of normalisation or CS-phase convention used inside \begin_inset Formula $Y_{n}^{m}$ \end_inset ): \begin_inset Formula \[ \sigma_{n}^{m(2)}=-\frac{2^{n+1}i}{k^{n+1}\sqrt{\pi}}\sum_{\vect R\in\Lambda}^{'}\left|\vect R\right|^{n}e^{i\vect{\beta}\cdot\vect R}Y_{n}^{m}\left(\vect R\right)\int_{\eta}^{\infty}e^{-\left|\vect R\right|^{2}\xi^{2}}e^{-k/4\xi^{2}}\xi^{2n}\ud\xi. \] \end_inset However the other parts in [LT] are convention dependend, so let me fix it here. [LT] use the convention [LT(A.7)] \begin_inset Formula \begin{eqnarray*} P_{n}^{m}\left(0\right) & = & \frac{\left(-1\right)^{\left(n-m\right)/2}\left(n+m\right)!}{2^{n}\left(\left(n-m\right)/2\right)!\left(\left(n+m\right)/2\right)!}\qquad n+m\mbox{ even,}\\ Y_{n}^{m}\left(\theta,\phi\right) & = & \left(-1\right)^{m}\sqrt{\frac{\left(2n+1\right)\left(n-m\right)!}{4\pi\left(n+m\right)!}}P_{n}^{m}\left(\cos\theta\right)e^{im\phi}, \end{eqnarray*} \end_inset noting that the former formula is valid also for negative \begin_inset Formula $m$ \end_inset (as can be checked by substituting [LT(A.4)]). Therefore \begin_inset Formula \begin{eqnarray*} Y_{n}^{m}\left(\frac{\pi}{2},\phi\right) & = & \left(-1\right)^{m}\sqrt{\frac{\left(2n+1\right)\left(n-m\right)!}{4\pi\left(n+m\right)!}}\frac{\left(-1\right)^{\left(n-m\right)/2}\left(n+m\right)!}{2^{n}\left(\left(n-m\right)/2\right)!\left(\left(n+m\right)/2\right)!}e^{im\phi}\\ & = & \frac{\left(-1\right)^{\left(n+m\right)/2}\sqrt{\left(2n+1\right)\left(n-m\right)!\left(n+m\right)!}}{\sqrt{\pi}2^{n+1}\left(\left(n-m\right)/2\right)!\left(\left(n+m\right)/2\right)!}e^{im\phi} \end{eqnarray*} \end_inset Let us substitute this into [LT(4.5)] \begin_inset Formula \begin{eqnarray*} \sigma_{n}^{m(1)} & = & -\frac{i^{n+1}}{2k^{2}\mathscr{A}}\left(-1\right)^{\left(n+m\right)/2}\sqrt{\left(2n+1\right)\left(n-m\right)!\left(n+m\right)!}\sum_{\vect K_{pq}\in\Lambda^{*}}^{'}\sum_{j=0}^{\left[\left(n-\left|m\right|/2\right)\right]}\frac{\left(-1\right)^{j}\left(\beta_{pq}/2k\right)^{n-2j}e^{im\phi_{\vect{\beta}_{pq}}}\Gamma_{j,pq}}{j!\left(\frac{1}{2}\left(n-m\right)-j\right)!\left(\frac{1}{2}\left(n+m\right)-j\right)!}\left(\frac{\gamma_{pq}}{2}\right)^{2j-1}\\ & = & -\frac{i^{n+1}}{2k^{2}\mathscr{A}}\sqrt{\pi}2^{n+1}\left(\left(n-m\right)/2\right)!\left(\left(n+m\right)/2\right)!Y_{n}^{m}\left(0,\phi_{\vect{\beta}_{pq}}\right)\sum_{\vect K_{pq}\in\Lambda^{*}}^{'}\sum_{j=0}^{\left[\left(n-\left|m\right|/2\right)\right]}\frac{\left(-1\right)^{j}\left(\beta_{pq}/2k\right)^{n-2j}\Gamma_{j,pq}}{j!\left(\frac{1}{2}\left(n-m\right)-j\right)!\left(\frac{1}{2}\left(n+m\right)-j\right)!}\left(\frac{\gamma_{pq}}{2}\right)^{2j-1}\\ & = & -\frac{i^{n+1}}{k\mathscr{A}}\sqrt{\pi}2\left(\left(n-m\right)/2\right)!\left(\left(n+m\right)/2\right)!Y_{n}^{m}\left(0,\phi_{\vect{\beta}_{pq}}\right)\sum_{\vect K_{pq}\in\Lambda^{*}}^{'}\sum_{j=0}^{\left[\left(n-\left|m\right|/2\right)\right]}\frac{\left(-1\right)^{j}\left(\beta_{pq}/k\right)^{n-2j}\Gamma_{j,pq}}{j!\left(\frac{1}{2}\left(n-m\right)-j\right)!\left(\frac{1}{2}\left(n+m\right)-j\right)!}\left(\gamma_{pq}\right)^{2j-1}, \end{eqnarray*} \end_inset which basically replaces an ugly prefactor with another, similarly ugly one. See [LT] for the meanings of the \begin_inset Formula $pq$ \end_inset -indexed symbols. Note that the latter version does not depend on the sign of \begin_inset Formula $m$ \end_inset (except for that which is already included in \begin_inset Formula $Y_{n}^{m}$ \end_inset ). \end_layout \begin_layout Standard To have it complete, \begin_inset Formula \[ \sigma_{n}^{m(0)}=\frac{\delta_{n0}\delta_{m0}}{4\pi}\Gamma\left(-\frac{1}{2},-\frac{k}{4\eta^{2}}\right)=\frac{\delta_{n0}\delta_{m0}}{\sqrt{4\pi}}\Gamma\left(-\frac{1}{2},-\frac{k}{4\eta^{2}}\right)Y_{n}^{m}. \] \end_inset \end_layout \begin_layout Section Major TODOs and open questions \end_layout \begin_layout Itemize Check if \begin_inset CommandInset ref LatexCommand eqref reference "eq:2D Hankel transform of exponentially suppressed outgoing wave expanded" \end_inset gives a satisfactory result for the case \begin_inset Formula $q=2,n=0$ \end_inset . \end_layout \begin_layout Itemize Analyse the behaviour \begin_inset Formula $k\to k_{0}$ \end_inset . \end_layout \begin_layout Itemize Find a general algorithm for generating the expressions of the Hankel transforms. \end_layout \begin_layout Itemize Three-dimensional case. \end_layout \begin_layout Section (Appendix) Fourier vs. Hankel transform \end_layout \begin_layout Subsection Three dimensions \end_layout \begin_layout Standard Given a nice enough function \begin_inset Formula $f$ \end_inset of a real 3d variable, assume its factorisation into radial and angular parts \begin_inset Formula \[ f(\vect r)=\sum_{l,m}f_{l,m}(\left|\vect r\right|)\ush lm\left(\theta_{\vect r},\phi_{\vect r}\right). \] \end_inset Acording to (REF Baddour 2010, eqs. 13, 16), its Fourier transform can then be expressed in terms of Hankel transforms (CHECK normalisation of \begin_inset Formula $j_{n}$ \end_inset , REF Baddour (1)) \begin_inset Formula \[ \uaft f(\vect k)=\frac{4\pi}{\left(2\pi\right)^{\frac{3}{2}}}\sum_{l,m}\left(-i\right)^{l}\left(\bsht{f_{l,m}}{}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right) \] \end_inset where the spherical Hankel transform \begin_inset Formula $\bsht l{}$ \end_inset of degree \begin_inset Formula $l$ \end_inset is defined as (REF Baddour eq. 2) \begin_inset Formula \[ \bsht lg(k)\equiv\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right). \] \end_inset Using this convention, the inverse spherical Hankel transform is given by (REF Baddour eq. 3) \begin_inset Formula \[ g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\,k^{2}\bsht lg(k)j_{l}(k), \] \end_inset so it is not unitary. \end_layout \begin_layout Standard An unitary convention would look like this: \begin_inset Formula \begin{equation} \usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition} \end{equation} \end_inset Then \begin_inset Formula $\usht l{}^{-1}=\usht l{}$ \end_inset and the unitary, angular-momentum Fourier transform reads \begin_inset Formula \begin{eqnarray} \uaft f(\vect k) & = & \frac{4\pi}{\left(2\pi\right)^{\frac{3}{2}}}\sqrt{\frac{\pi}{2}}\sum_{l,m}\left(-i\right)^{l}\left(\usht l{f_{l,m}}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right)\nonumber \\ & = & \sum_{l,m}\left(-i\right)^{l}\left(\usht l{f_{l,m}}\right)\left(\left|\vect k\right|\right)\ush lm\left(\theta_{\vect k},\phi_{\vect k}\right).\label{eq:Fourier v. Hankel tf 3d} \end{eqnarray} \end_inset Cool. \end_layout \begin_layout Subsection Two dimensions \end_layout \begin_layout Standard Similarly in 2d, let the expansion of \begin_inset Formula $f$ \end_inset be \begin_inset Formula \[ f\left(\vect r\right)=\sum_{m}f_{m}\left(\left|\vect r\right|\right)e^{im\phi_{\vect r}}, \] \end_inset its Fourier transform is then (CHECK this, it is taken from the Wikipedia article on Hankel transform) \begin_inset Formula \begin{equation} \uaft f\left(\vect k\right)=\sum_{m}i^{m}e^{im\phi_{\vect k}}\pht mf_{m}\left(\left|\vect k\right|\right)\label{eq:Fourier v. Hankel tf 2d} \end{equation} \end_inset where the Hankel transform of order \begin_inset Formula $m$ \end_inset is defined as \begin_inset Formula \begin{eqnarray} \pht mg\left(k\right) & = & \int_{0}^{\infty}\ud r\,g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}\\ & = & \left(-1\right)^{m}\int_{0}^{\infty}\ud r\,g(r)J_{-m}(kr)r \end{eqnarray} \end_inset which is already self-inverse, \begin_inset Formula $\pht m{}^{-1}=\pht m{}$ \end_inset (hence also unitary). \end_layout \begin_layout Section (Appendix) Multidimensional Dirac comb \end_layout \begin_layout Subsection 1D \end_layout \begin_layout Standard This is all from Wikipedia \end_layout \begin_layout Subsubsection Definitions \end_layout \begin_layout Standard \begin_inset Formula \begin{eqnarray*} Ш(t) & \equiv & \sum_{k=-\infty}^{\infty}\delta(t-k)\\ Ш_{T}(t) & \equiv & \sum_{k=-\infty}^{\infty}\delta(t-kT)=\frac{1}{T}Ш\left(\frac{t}{T}\right) \end{eqnarray*} \end_inset \end_layout \begin_layout Subsubsection Fourier series representation \end_layout \begin_layout Standard \begin_inset Formula \begin{equation} Ш_{T}(t)=\sum_{n=-\infty}^{\infty}e^{2\pi int/T}\label{eq:1D Dirac comb Fourier series} \end{equation} \end_inset \end_layout \begin_layout Subsubsection Fourier transform \end_layout \begin_layout Standard With unitary ordinary frequency Ft., i.e. \end_layout \begin_layout Standard \begin_inset Formula \[ \uoft f(\vect{\xi})\equiv\int_{\mathbb{R}^{n}}f(\vect x)e^{-2\pi i\vect x\cdot\vect{\xi}}\ud^{n}\vect x \] \end_inset we have \begin_inset Formula \begin{equation} \uoft{Ш_{T}}(f)=\frac{1}{T}Ш_{\frac{1}{T}}(f)=\sum_{n=-\infty}^{\infty}e^{-i2\pi fnT}\label{eq:1D Dirac comb Ft ordinary freq} \end{equation} \end_inset and with unitary angular frequency Ft., i.e. \begin_inset Formula \begin{equation} \uaft f(\vect k)\equiv\frac{1}{\left(2\pi\right)^{n/2}}\int_{\mathbb{R}^{n}}f(\vect x)e^{-i\vect x\cdot\vect k}\ud^{n}\vect x\label{eq:Ft unitary angular frequency} \end{equation} \end_inset we have (CHECK) \begin_inset Formula \[ \uaft{Ш_{T}}(\omega)=\frac{\sqrt{2\pi}}{T}Ш_{\frac{2\pi}{T}}(\omega)=\frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty}e^{-i\omega nT} \] \end_inset \end_layout \begin_layout Subsection Dirac comb for multidimensional lattices \end_layout \begin_layout Subsubsection Definitions \end_layout \begin_layout Standard Let \begin_inset Formula $d$ \end_inset be the dimensionality of the real vector space in question, and let \begin_inset Formula $\basis u\equiv\left\{ \vect u_{i}\right\} _{i=1}^{d}$ \end_inset denote a basis for some lattice in that space. Let the corresponding lattice delta comb be \begin_inset Formula \[ \dc{\basis u}\left(\vect x\right)\equiv\sum_{n_{1}=-\infty}^{\infty}\ldots\sum_{n_{d}=-\infty}^{\infty}\delta\left(\vect x-\sum_{i=1}^{d}n_{i}\vect u_{i}\right). \] \end_inset \end_layout \begin_layout Standard Furthemore, let \begin_inset Formula $\rec{\basis u}\equiv\left\{ \rec{\vect u}_{i}\right\} _{i=1}^{d}$ \end_inset be the reciprocal lattice basis, that is the basis satisfying \begin_inset Formula $\vect u_{i}\cdot\rec{\vect u_{j}}=\delta_{ij}$ \end_inset . This slightly differs from the usual definition of a reciprocal basis, here denoted \begin_inset Formula $\recb{\basis u}\equiv\left\{ \recb{\vect u_{i}}\right\} _{i=1}^{d}$ \end_inset , which satisfies \begin_inset Formula $\vect u_{i}\cdot\recb{\vect u_{j}}=2\pi\delta_{ij}$ \end_inset instead. \end_layout \begin_layout Subsubsection Factorisation of a multidimensional lattice delta comb \end_layout \begin_layout Standard By simple drawing, it can be seen that \begin_inset Formula \[ \dc{\basis u}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right) \] \end_inset where \begin_inset Formula $c_{\basis u}$ \end_inset is some numerical volume factor. In order to determine \begin_inset Formula $c_{\basis u}$ \end_inset , let us consider only the \begin_inset Quotes eld \end_inset zero tooth \begin_inset Quotes erd \end_inset of the comb, leading to \begin_inset Formula \[ \delta^{d}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\delta\left(\vect x\cdot\rec{\vect u_{i}}\right). \] \end_inset From the scaling property of delta function, \begin_inset Formula $\delta(ax)=\left|a\right|^{-1}\delta(x)$ \end_inset , we get \begin_inset Formula \[ \delta^{d}(\vect x)=c_{\basis u}\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert ^{-1}\delta\left(\vect x\cdot\frac{\rec{\vect u_{i}}}{\left\Vert \rec{\vect u_{i}}\right\Vert }\right). \] \end_inset \end_layout \begin_layout Standard From the Osgood's book (p. 375): \end_layout \begin_layout Standard \begin_inset Formula \[ \dc A(\vect x)=\frac{1}{\left|\det A\right|}\dc{}^{(d)}\left(A^{-1}\vect x\right) \] \end_inset \begin_inset Note Note status open \begin_layout Plain Layout Applying both sides to a test function that is one at the origin, we get \begin_inset Formula $c_{\basis u}=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert $ \end_inset SRSLY?, and hence \begin_inset Formula \begin{equation} \dc{\basis u}(\vect x)=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right).\label{eq:Dirac comb factorisation} \end{equation} \end_inset \end_layout \end_inset \end_layout \begin_layout Subsubsection Fourier series representation \end_layout \begin_layout Standard \begin_inset Note Note status open \begin_layout Plain Layout Utilising the Fourier series for 1D Dirac comb \begin_inset CommandInset ref LatexCommand eqref reference "eq:1D Dirac comb Fourier series" \end_inset and the factorisation \begin_inset CommandInset ref LatexCommand eqref reference "eq:Dirac comb factorisation" \end_inset , we get \begin_inset Formula \begin{eqnarray*} \dc{\basis u}(\vect x) & = & \prod_{j=1}^{d}\left\Vert \rec{\vect u_{j}}\right\Vert \sum_{n_{j}=-\infty}^{\infty}e^{2\pi in_{i}\vect x\cdot\rec{\vect u_{i}}}\\ & = & \left(\prod_{j=1}^{d}\left\Vert \rec{\vect u_{j}}\right\Vert \right)\sum_{\vect n\in\mathbb{Z}^{d}}e^{2\pi i\vect x\cdot\sum_{k=1}^{d}n_{k}\rec{\vect u_{k}}}. \end{eqnarray*} \end_inset \end_layout \end_inset \end_layout \begin_layout Subsubsection Fourier transform (OK) \end_layout \begin_layout Standard From the Osgood's book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf, p. 379 \end_layout \begin_layout Standard \begin_inset Formula \[ \uoft{\dc{\basis u}}\left(\vect{\xi}\right)=\left|\det\rec{\basis u}\right|\dc{\rec{\basis u}}^{(d)}\left(\vect{\xi}\right). \] \end_inset And consequently, for unitary/angular frequency it is \end_layout \begin_layout Standard \begin_inset Formula \begin{eqnarray} \uaft{\dc{\basis u}}\left(\vect k\right) & = & \frac{1}{\left(2\pi\right)^{\frac{d}{2}}}\uoft{\dc{\basis u}}\left(\frac{\vect k}{2\pi}\right)\nonumber \\ & = & \frac{\left|\det\rec{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{\rec{\basis u}}^{(d)}\left(\frac{\vect k}{2\pi}\right)\nonumber \\ & = & \left(2\pi\right)^{\frac{d}{2}}\left|\det\rec{\basis u}\right|\dc{\recb{\basis u}}\left(\vect k\right)\nonumber \\ & = & \frac{\left|\det\recb{\basis u}\right|}{\left(2\pi\right)^{\frac{d}{2}}}\dc{\recb{\basis u}}\left(\vect k\right).\label{eq:Dirac comb uaFt} \end{eqnarray} \end_inset \end_layout \begin_layout Standard \begin_inset Note Note status open \begin_layout Plain Layout On the third line, we used the stretch theorem, getting \begin_inset Formula \[ \dc{\recb{\basis u}}\left(\vect k\right)=\dc{2\pi\rec{\basis u}}\left(\vect k\right)=\left(2\pi\right)^{-d}\dc{\rec{\basis u}}\left(\frac{\vect k}{2\pi}\right) \] \end_inset \end_layout \end_inset \end_layout \begin_layout Subsubsection Convolution \end_layout \begin_layout Standard \begin_inset Formula \[ \left(f\ast\dc{\basis u}\right)(\vect x)=\sum_{\vect t\in\basis u\ints^{d}}f(\vect x-\vect t) \] \end_inset \end_layout \begin_layout Standard \begin_inset Note Note status open \begin_layout Plain Layout So, from the stretch theorem \begin_inset Formula $\uoft{(f(A\vect x))}=\frac{1}{\left|\det A\right|}\uoft{f\left(A^{-T}\vect{\xi}\right)}=\left|\det A^{-T}\right|\uoft{f\left(A^{-T}\vect{\xi}\right)}$ \end_inset \end_layout \begin_layout Plain Layout From \begin_inset CommandInset ref LatexCommand eqref reference "eq:Dirac comb factorisation" \end_inset and \begin_inset CommandInset ref LatexCommand eqref reference "eq:1D Dirac comb Ft ordinary freq" \end_inset \begin_inset Formula \[ \uoft{\dc{\basis u}}(\vect{\xi})=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right). \] \end_inset \end_layout \end_inset \end_layout \end_body \end_document