518 lines
15 KiB
Plaintext
518 lines
15 KiB
Plaintext
#LyX 2.4 created this file. For more info see https://www.lyx.org/
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\end_header
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\begin_body
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\begin_layout Title
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1D in 3D Ewald sum
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\end_layout
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\begin_layout Standard
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\begin_inset FormulaMacro
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\newcommand{\ud}{\mathrm{d}}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\abs}[1]{\left|#1\right|}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\vect}[1]{\mathbf{#1}}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\uvec}[1]{\hat{\mathbf{#1}}}
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\end_inset
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\lang english
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\begin_inset FormulaMacro
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\newcommand{\ush}[2]{Y_{#1}^{#2}}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\ushD}[2]{Y'_{#1}^{#2}}
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset FormulaMacro
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\newcommand{\vsh}{\vect A}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\vshD}{\vect{A'}}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\wfkc}{\vect y}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\wfkcout}{\vect u}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\wfkcreg}{\vect v}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\wckcreg}{a}
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\end_inset
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\begin_inset FormulaMacro
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\newcommand{\wckcout}{f}
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\end_inset
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\end_layout
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\begin_layout Standard
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[Linton, (2.24)] with slightly modified notation and setting
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\begin_inset Formula $d_{c}=2$
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\end_inset
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:
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\begin_inset Formula
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\[
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G_{\Lambda}^{(1;\kappa)}\left(\vect r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect r}\int_{1/\eta}^{\infty e^{i\pi/4}}e^{-\kappa^{2}\gamma^{2}t^{2}/4}e^{-\left|\vect r^{\bot}\right|^{2}/t^{2}}t^{-1}\ud t
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\]
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\end_inset
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or, evaluated at point
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\begin_inset Formula $\vect s+\vect r$
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\end_inset
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instead
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\begin_inset Formula
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\[
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G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{1/\eta}^{\infty e^{i\pi/4}}e^{-\kappa^{2}\gamma^{2}t^{2}/4}e^{-\left|\vect s^{\bot}+\vect r^{\bot}\right|^{2}/t^{2}}t^{-1}\ud t
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\]
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\end_inset
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The integral can be by substitutions taken into the form
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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\lang english
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\begin_inset Formula
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\[
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G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{2\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\int_{1/\eta}^{\infty\exp\left(i\pi/4\right)}e^{-\kappa^{2}\gamma_{m}^{2}\zeta^{2}/4}e^{-\left|\vect r_{\bot}\right|^{2}/\zeta^{2}}\zeta^{1-d_{c}}\ud\zeta
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\]
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\end_inset
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Try substitution
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\begin_inset Formula $t=\zeta^{2}$
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\end_inset
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: then
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\begin_inset Formula $\ud t=2\zeta\,\ud\zeta$
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\end_inset
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(
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\begin_inset Formula $\ud\zeta=\ud t/2t^{1/2}$
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\end_inset
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) and
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\begin_inset Formula
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\[
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G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\int_{1/\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\kappa^{2}\gamma_{m}^{2}t/4}e^{-\left|\vect r_{\bot}\right|^{2}/t}t^{\frac{-d_{c}}{2}}\ud t
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\]
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\end_inset
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Try subst.
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\begin_inset Formula $\tau=k^{2}\gamma_{m}^{2}/4$
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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\lang english
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\begin_inset Formula
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\[
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G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}e^{i\vect K_{m}\cdot\vect r}\left(\frac{\kappa\gamma_{m}}{2}\right)^{d_{c}}\int_{\kappa^{2}\gamma_{m}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{m}^{2}/4\tau}\tau^{\frac{-d_{c}}{2}}\ud\tau
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\]
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\end_inset
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\end_layout
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\end_inset
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\begin_inset Formula
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\[
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G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{\kappa^{2}\gamma_{m}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{m}^{2}/4\tau}\tau^{-1}\ud\tau
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\]
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\end_inset
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\end_layout
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\begin_layout Standard
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\begin_inset Foot
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status open
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\begin_layout Plain Layout
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[Linton, (2.25)] with slightly modified notation:
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\begin_inset Formula
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\[
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G_{\Lambda}^{(1;\kappa)}\left(\vect r\right)=-\frac{1}{\sqrt{4\pi}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect r}\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}\left|\vect r^{\bot}\right|^{2j}}{j!}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2j-1}\Gamma_{j\vect K}
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\]
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\end_inset
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We want to express an expansion in a shifted point, so let's substitute
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\begin_inset Formula $\vect r\to\vect s+\vect r$
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\end_inset
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\begin_inset Formula
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\[
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G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)=-\frac{1}{\sqrt{4\pi}\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\sum_{j=0}^{\infty}\frac{\left(-1\right)^{j}\left|\vect s^{\bot}+\vect r^{\bot}\right|^{2j}}{j!}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2j-1}\Gamma_{j\vect K}
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\]
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\end_inset
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\end_layout
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\end_inset
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Let's do the integration to get
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\begin_inset Formula $\tau_{l}^{m}\left(\vect s,\vect k\right)$
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\end_inset
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\begin_inset Formula
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\[
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\left(\vect s+\vect r\right)}\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1}\ud\tau
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\]
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\end_inset
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The
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\begin_inset Formula $\vect r$
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\end_inset
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-dependent plane wave factor can be also written as
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\begin_inset Formula
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\begin{align*}
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e^{i\vect K\cdot\vect r} & =e^{i\left|\vect K\right|\vect r\cdot\uvec K}=4\pi\sum_{lm}i^{l}\mathcal{J}'_{l}^{m}\left(\left|\vect K\right|\vect r\right)\ush lm\left(\uvec K\right)\\
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& =4\pi\sum_{lm}i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD lm\left(\uvec{\vect r}\right)\ush lm\left(\uvec K\right)
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\end{align*}
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\end_inset
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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or the other way around
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\begin_inset Formula
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\[
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e^{i\vect K\cdot\vect r}=4\pi\sum_{lm}i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec{\vect r}\right)\ushD lm\left(\uvec K\right)
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\]
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\end_inset
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\end_layout
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\end_inset
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so
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\begin_inset Formula
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\[
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD lm\left(\uvec r\right)\ush lm\left(\uvec K\right)\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1}\ud\tau
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\]
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\end_inset
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\end_layout
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\begin_layout Standard
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Now we set the conventions: let the lattice lie on the
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\begin_inset Formula $z$
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\end_inset
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axis, so that
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\begin_inset Formula $\vect s_{\bot},\vect r_{\bot}$
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\end_inset
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lie in the
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\begin_inset Formula $xy$
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\end_inset
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-plane, (TODO check the meaning of
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\begin_inset Formula $\vect k$
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\end_inset
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and possible additional phase factor.) If we write
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\begin_inset Formula $\vect s_{\bot}=\uvec xs_{\bot}\cos\Phi+\uvec ys_{\bot}\sin\Phi$
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\end_inset
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,
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\begin_inset Formula $\vect r_{\bot}=\uvec xr_{\bot}\cos\phi+\uvec yr_{\bot}\sin\phi$
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\end_inset
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, we have
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\begin_inset Formula
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\[
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\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}=s_{\bot}^{2}+r_{\bot}^{2}+2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right).
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\]
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\end_inset
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Also, in this convention
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\begin_inset Formula $\ush lm\left(\uvec K\right)=0$
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\end_inset
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for
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\begin_inset Formula $m\ne0$
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\end_inset
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, so
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\begin_inset Formula
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\[
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{l}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD l0\left(\uvec r\right)\ush l0\left(\uvec K\right)\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left(s_{\bot}^{2}+r_{\bot}^{2}+2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right)\right)^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1}\ud\tau
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\]
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\end_inset
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\end_layout
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\begin_layout Standard
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Let's also fix the spherical harmonics for now,
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\begin_inset Formula
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\[
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\ushD lm\left(\uvec r\right)=\lambda'_{lm}e^{-im\phi}P_{l}^{-m}\left(\cos\theta\right)
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\]
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\end_inset
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\end_layout
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\begin_layout Standard
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The angular integral (assuming it can be separated from the rest like this)
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is
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\begin_inset Formula
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\[
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I_{l}^{m}\equiv\int\ud\Omega_{\vect r}\,\ushD lm\left(\uvec r\right)e^{-\left(r_{\bot}^{2}+2s_{\bot}r_{\bot}\cos\left(\phi-\Phi\right)\right)^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}
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\]
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\end_inset
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\end_layout
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\begin_layout Standard
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Let's further extract the azimuthal part
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\begin_inset Formula $\left(w\equiv2r_{\bot}s_{\bot}\kappa^{2}\gamma_{\vect K}^{2}/4\tau\right)$
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\end_inset
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\begin_inset Formula
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\[
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e^{-im\Phi}A_{l}^{m}\equiv\int_{0}^{2\pi}e^{-im\phi}e^{-w\cos\left(\phi-\Phi\right)}\ud\phi=e^{-im\Phi}\int_{0}^{2\pi}e^{-im\varphi}e^{-w\cos\varphi}\ud\varphi
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\]
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\end_inset
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\begin_inset Formula
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\begin{align*}
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A_{l}^{m} & =\int_{0}^{2\pi}e^{-im\varphi}\sum_{n=0}^{\infty}\frac{\left(-w\cos\varphi\right)^{n}}{n!}\ud\varphi\\
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& =\int_{0}^{2\pi}e^{-im\varphi}\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\left(e^{i\varphi}+e^{-i\varphi}\right)^{n}\ud\varphi\\
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& =\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\int_{0}^{2\pi}e^{-im\varphi}\sum_{k=0}^{n}\binom{n}{k}e^{ik\varphi}e^{-i\left(n-k\right)\varphi}\ud\varphi\\
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& =\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\int_{0}^{2\pi}e^{i\left(2k-n-m\right)\varphi}\ud\varphi\\
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& =2\pi\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\delta_{2k-n-m=0}\\
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& =2\pi\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\delta_{2k-n-m=0}\\
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& =2\pi\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\binom{n}{k}\delta_{2k-n-m=0}\\
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& =2\pi\sum_{k=0}^{\infty}\frac{\left(-w\right)^{2k-m}}{2^{2k-m}\left(2k-m\right)!}\binom{2k-m}{k}\delta_{2k-m\ge k}\\
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& =2\pi\sum_{k=0}^{\infty}\frac{\left(-w\right)^{2k-m}}{2^{2k-m}}\frac{1}{k!\left(k-m\right)!}\delta_{k-m\ge0}\\
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& =2\pi\sum_{k=\max\left(m,0\right)}^{\infty}\left(-\frac{w}{2}\right)^{2k-m}\frac{1}{k!\left(k-m\right)!}
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\end{align*}
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\end_inset
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\begin_inset Note Note
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status collapsed
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\begin_layout Plain Layout
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\begin_inset Formula
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\begin{align*}
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A_{l}^{m} & =\int_{0}^{2\pi}e^{-im\varphi}\sum_{n=0}^{\infty}\frac{\left(-w\cos\varphi\right)^{n}}{n!}\ud\varphi\\
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& =\int_{0}^{2\pi}e^{-im\varphi}\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\left(e^{i\varphi}+e^{-i\varphi}\right)^{n}\ud\varphi\\
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& =\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\int_{0}^{2\pi}e^{-im\varphi}\sum_{k=0}^{n}\binom{n}{k}e^{i\left(n-k\right)\varphi}e^{-ik\varphi}\ud\varphi\\
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& =\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\int_{0}^{2\pi}e^{i\left(-2k+n-m\right)\varphi}\ud\varphi\\
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& =2\pi\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\delta_{-2k+n-m=0}\\
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& =2\pi\sum_{n=0}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\sum_{k=0}^{n}\binom{n}{k}\delta_{-2k+n-m=0}\\
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& =2\pi\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{\left(-w\right)^{n}}{2^{n}n!}\binom{n}{k}\delta_{-2k+n-m=0}\\
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& =2\pi\sum_{k=0}^{\infty}\frac{\left(-w\right)^{2k+m}}{2^{2k+m}\left(2k+m\right)!}\binom{2k+m}{k}\delta_{2k+m\ge k}\\
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& =2\pi\sum_{k=0}^{\infty}\frac{\left(-w\right)^{2k+m}}{2^{2k+m}}\frac{1}{k!\left(k+m\right)!}\delta_{k+m\ge0}\\
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& =2\pi\sum_{k=\max\left(-m,0\right)}^{\infty}\frac{\left(-w\right)^{2k+m}}{2^{2k+m}}\frac{1}{k!\left(k+m\right)!}
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\end{align*}
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\end_inset
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\end_layout
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\end_inset
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Althought it's not superobvious, this sum is symmetric w.r.t.
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sign change in
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\begin_inset Formula $m$
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\end_inset
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.
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\end_layout
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\begin_layout Standard
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Let's do the polar integration next:
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\begin_inset Formula $r_{\bot}=r\sin\theta$
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\end_inset
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\begin_inset Formula
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\[
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B_{l}^{m}\equiv\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l}^{-m}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)e^{-\left(\sin\theta\right)^{2}r^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\left(-\sin\theta\,rs_{\bot}\kappa^{2}\gamma_{\vect K}^{2}/4\tau\right)^{2k-m}
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\]
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\end_inset
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Label
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\begin_inset Formula $u\equiv r^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau,v\equiv rs_{\bot}\kappa^{2}\gamma_{\vect K}^{2}/4\tau$
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\end_inset
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; then
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\begin_inset Formula
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\begin{align*}
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B_{l}^{m} & =\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l}^{-m}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)e^{-u\left(\sin\theta\right)^{2}}\left(-v\sin\theta\right)^{2k-m}\\
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& =\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l}^{-m}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)\left(-v\sin\theta\right)^{2k-m}\sum_{a=0}^{\infty}\frac{\left(-u\right)^{a}}{a!}\left(\sin\theta\right)^{2a}\\
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& =\left(-v\right)^{2k-m}\sum_{a=0}^{\infty}\frac{\left(-u\right)^{a}}{a!}\int_{0}^{\pi}\sin\theta\ud\theta\,P_{l}^{-m}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)\left(\sin\theta\right)^{2a+2k-m}
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\end{align*}
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\end_inset
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\end_layout
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\end_body
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\end_document
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