Notes on evaluating Δ_n factor in the lattice sums.
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Marek Nečada
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#LyX 2.4 created this file. For more info see https://www.lyx.org/
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\newcommand{\ud}{\mathrm{d}}
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\end_inset
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\begin_inset Formula
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\begin{equation}
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\Delta_{n}(x,z)\equiv\int_{x}^{\infty}t^{-\frac{1}{2}-n}e^{-t+\frac{z^{2}}{4t}}\ud t\label{eq:Delta definition}
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\end{equation}
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\begin_layout Standard
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Integration per partes:
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\int t^{-\frac{1}{2}-n}\ud t=\frac{t^{\frac{1}{2}-n}}{\frac{1}{2}-n};
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\]
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\end_inset
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\begin_inset Formula
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\[
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\frac{\ud}{\ud t}e^{-t+\frac{z^{2}}{4t}}=\left(-1-\frac{z^{2}}{4t^{2}}\right)e^{-t+\frac{z^{2}}{4t}}
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\]
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{align*}
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\left(\frac{1}{2}-n\right)\Delta_{n} & =-x^{\frac{1}{2}-n}e^{-x+\frac{z^{2}}{4x}}+\int_{x}^{\infty}t^{\frac{1}{2}-n}e^{-t+\frac{z^{2}}{4t}}\ud t+\frac{z^{2}}{4}\int_{x}^{\infty}t^{\frac{-3}{2}-n}e^{-t+\frac{z^{2}}{4t}}\ud t\\
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& =-x^{\frac{1}{2}-n}e^{-x+\frac{z^{2}}{4x}}+\Delta_{n-1}+\frac{z^{2}}{4}\Delta_{n+1},
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\end{align*}
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\end_inset
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\begin_inset Formula
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\begin{equation}
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\Delta_{n+1}=\frac{4}{z^{2}}\left(\left(\frac{1}{2}-n\right)\Delta_{n}-\Delta_{n-1}+x^{\frac{1}{2}-n}e^{-x+\frac{z^{2}}{4x}}\right).\label{eq:Delta recurrence}
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\end{equation}
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\end_inset
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There are obviously wrong signs in Kambe II, (A 3.3).
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\begin_layout Standard
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Eq.
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:Delta recurrence"
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plural "false"
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is obviously unsuitable for numerical computation when
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\begin_inset Formula $z$
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\end_inset
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approaches 0.
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However, the definition
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:Delta definition"
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plural "false"
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caps "false"
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noprefix "false"
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\end_inset
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suggests that the function should be analytical around
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\begin_inset Formula $z=0$
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\end_inset
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.
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If
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\begin_inset Formula $z=0$
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\end_inset
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, one has (by definition of incomplete Г function)
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\begin_inset Formula
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\begin{equation}
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\Delta_{n}(x,0)=\Gamma\left(\frac{1}{2}-n,x\right).\label{eq:Delta:z = 0}
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\end{equation}
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\end_inset
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For convenience, label
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\begin_inset Formula $w=z^{2}/4$
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\end_inset
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and
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\begin_inset Formula
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\[
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\Delta'_{n}\left(x,w\right)\equiv\int_{x}^{\infty}t^{-\frac{1}{2}-n}e^{-t+\frac{w}{t}}\ud t.
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\]
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\end_inset
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Differentiating by parameter
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\begin_inset Formula $w$
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\end_inset
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(which should be fine as long as the integration contour does not go through
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zero) gives
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\begin_inset Formula
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\[
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\frac{\partial\Delta'_{n}\left(x,w\right)}{\partial w}=\Delta'_{n+1}\left(x,w\right),
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\]
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\end_inset
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so by recurrence
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\begin_inset Formula
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\[
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\frac{\partial^{k}}{\partial w^{k}}\Delta'_{n}\left(x,w\right)=\Delta'_{n+k}\left(x,w\right).
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\]
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\end_inset
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Together with
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:Delta:z = 0"
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plural "false"
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caps "false"
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noprefix "false"
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\end_inset
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, this gives an expansion around
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\begin_inset Formula $w=0$
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\end_inset
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:
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\begin_inset Formula
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\[
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\Delta_{n}'\left(x,w\right)=\sum_{k=0}^{\infty}\Gamma\left(\frac{1}{2}-n-k,x\right)\frac{w^{k}}{k!},
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\]
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\end_inset
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\begin_inset Formula
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\[
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\Delta_{n}\left(x,z\right)=\sum_{k=0}^{\infty}\Gamma\left(\frac{1}{2}-n-k,x\right)\frac{\left(z/2\right)^{2k}}{k!}.
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\]
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\end_inset
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The big negative first arguments in incomplete
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\begin_inset Formula $\Gamma$
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\end_inset
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functions should be good (at least I think so, CHECKME), as well as the
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\begin_inset Formula $1/k!$
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\end_inset
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factor (of course).
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I am not sure what the convergence radius is, but for
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\begin_inset Formula $\left|z\right|<2$
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\end_inset
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there seems to be absolutely no problem in using this formula.
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\end_document
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