Notes
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@ -241,6 +241,64 @@ For
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\end_inset
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\end_layout
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\begin_layout Section
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\lang english
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Ewald long range integral
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\end_layout
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\begin_layout Standard
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\lang english
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Linton has (2.24):
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\begin_inset Formula
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\[
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G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{2\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}\int_{1/\eta}^{\infty\exp\left(i\pi/4\right)}e^{-\kappa^{2}\gamma_{m}^{2}\zeta^{2}/4}e^{-\left|\vect r_{\bot}\right|^{2}/\zeta^{2}}\zeta^{1-d_{c}}\ud\zeta
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\]
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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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\[
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G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}\int_{1/\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\kappa^{2}\gamma_{m}^{2}t/4}e^{-\left|\vect r_{\bot}\right|^{2}/t}t^{\frac{-d_{c}}{2}}\ud t
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\]
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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 Standard
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\lang english
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\begin_inset Formula
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\[
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G_{\Lambda}^{\left(1\right)}\left(\vect r\right)=\frac{\pi^{-d_{c}/2}}{4\mathcal{A}}\sum_{\vect K_{m}\in\Lambda^{*}}\left(\frac{\kappa\gamma_{m}}{2}\right)^{d_{c}}\int_{\kappa^{2}\gamma_{m}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{m}^{2}/4\tau}\tau^{\frac{-d_{c}}{2}}\ud\tau
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\]
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\end_inset
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\end_layout
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\end_body
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@ -371,5 +371,16 @@ Let's look at the
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Ewald parameter (integration limits)
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\[
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\Linton{\eta}=\sqrt{\frac{1}{2\Kambe{\omega}}}
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\]
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\end_inset
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(Based on comparison of some function arguments, not checked.)
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\end_layout
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\end_body
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\end_document
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