Some 1D lattice sums
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@ -3399,6 +3399,39 @@ The only diverging factor here is apparently
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\end_inset
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\end_layout
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\begin_layout Subsection
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1D
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\end_layout
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\begin_layout Standard
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One-dimensional lattice sums are provided in [REF LT, sect.
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3].
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However, these are the
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\begin_inset Quotes eld
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\end_inset
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non-shifted
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\begin_inset Quotes erd
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\end_inset
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sums,
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\begin_inset Formula
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\begin{eqnarray*}
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\ell_{n}\left(\beta\right) & = & \sum_{j\in\ints}^{'}e^{i\beta aj}\mathcal{H}_{n}^{0}\left(aj\hat{\vect z}\right)\\
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& = & \sum_{j\in\ints}^{'}e^{i\beta aj}h_{n}\left(\left|aj\right|\right)Y_{n}^{0}\\
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& = & \sqrt{\frac{2n+1}{4\pi}}\sum_{j\in\ints}^{'}P_{n}^{0}\left(\sgn j\right)h_{n}\left(\left|aj\right|\right)e^{i\beta aj}\\
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& = & \sqrt{\frac{2n+1}{4\pi}}\sum_{j\in\ints}^{'}\left(\sgn j\right)^{n}h_{n}\left(\left|aj\right|\right)e^{i\beta aj},
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\end{eqnarray*}
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\end_inset
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where we used
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\begin_inset Formula $P_{n}^{m}\left(\pm1\right)=\left(\pm1\right)^{n}\delta_{m0}$
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\end_inset
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\end_layout
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\begin_layout Section
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@ -3485,7 +3518,7 @@ where the spherical Hankel transform
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2)
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\begin_inset Formula
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\[
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\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).
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\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).
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\]
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\end_inset
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@ -3495,7 +3528,7 @@ Using this convention, the inverse spherical Hankel transform is given by
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3)
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\begin_inset Formula
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\[
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g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\,k^{2}\bsht lg(k)j_{l}(k),
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g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\, k^{2}\bsht lg(k)j_{l}(k),
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\]
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\end_inset
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@ -3508,7 +3541,7 @@ so it is not unitary.
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An unitary convention would look like this:
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\begin_inset Formula
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\begin{equation}
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\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}
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\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}
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\end{equation}
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\end_inset
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@ -3562,8 +3595,8 @@ where the Hankel transform of order
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is defined as
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\begin_inset Formula
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\begin{eqnarray}
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\pht mg\left(k\right) & = & \int_{0}^{\infty}\ud r\,g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}\\
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& = & \left(-1\right)^{m}\int_{0}^{\infty}\ud r\,g(r)J_{-m}(kr)r
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\pht mg\left(k\right) & = & \int_{0}^{\infty}\ud r\, g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}\\
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& = & \left(-1\right)^{m}\int_{0}^{\infty}\ud r\, g(r)J_{-m}(kr)r
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\end{eqnarray}
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\end_inset
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