Notes on evaluating Δ_n factor in the lattice sums.

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Marek Nečada 2020-05-26 16:00:20 +03:00
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#LyX 2.4 created this file. For more info see https://www.lyx.org/
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\begin_body
\begin_layout Standard
\begin_inset FormulaMacro
\newcommand{\ud}{\mathrm{d}}
\end_inset
\begin_inset Formula
\begin{equation}
\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}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
Integration per partes:
\end_layout
\begin_layout Standard
\begin_inset Formula
\[
\int t^{-\frac{1}{2}-n}\ud t=\frac{t^{\frac{1}{2}-n}}{\frac{1}{2}-n};
\]
\end_inset
\begin_inset Formula
\[
\frac{\ud}{\ud t}e^{-t+\frac{z^{2}}{4t}}=\left(-1-\frac{z^{2}}{4t^{2}}\right)e^{-t+\frac{z^{2}}{4t}}
\]
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{align*}
\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\\
& =-x^{\frac{1}{2}-n}e^{-x+\frac{z^{2}}{4x}}+\Delta_{n-1}+\frac{z^{2}}{4}\Delta_{n+1},
\end{align*}
\end_inset
\begin_inset Formula
\begin{equation}
\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}
\end{equation}
\end_inset
There are obviously wrong signs in Kambe II, (A 3.3).
\end_layout
\begin_layout Standard
Eq.
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Delta recurrence"
plural "false"
caps "false"
noprefix "false"
\end_inset
is obviously unsuitable for numerical computation when
\begin_inset Formula $z$
\end_inset
approaches 0.
However, the definition
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Delta definition"
plural "false"
caps "false"
noprefix "false"
\end_inset
suggests that the function should be analytical around
\begin_inset Formula $z=0$
\end_inset
.
If
\begin_inset Formula $z=0$
\end_inset
, one has (by definition of incomplete Г function)
\begin_inset Formula
\begin{equation}
\Delta_{n}(x,0)=\Gamma\left(\frac{1}{2}-n,x\right).\label{eq:Delta:z = 0}
\end{equation}
\end_inset
For convenience, label
\begin_inset Formula $w=z^{2}/4$
\end_inset
and
\begin_inset Formula
\[
\Delta'_{n}\left(x,w\right)\equiv\int_{x}^{\infty}t^{-\frac{1}{2}-n}e^{-t+\frac{w}{t}}\ud t.
\]
\end_inset
Differentiating by parameter
\begin_inset Formula $w$
\end_inset
(which should be fine as long as the integration contour does not go through
zero) gives
\begin_inset Formula
\[
\frac{\partial\Delta'_{n}\left(x,w\right)}{\partial w}=\Delta'_{n+1}\left(x,w\right),
\]
\end_inset
so by recurrence
\begin_inset Formula
\[
\frac{\partial^{k}}{\partial w^{k}}\Delta'_{n}\left(x,w\right)=\Delta'_{n+k}\left(x,w\right).
\]
\end_inset
Together with
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:Delta:z = 0"
plural "false"
caps "false"
noprefix "false"
\end_inset
, this gives an expansion around
\begin_inset Formula $w=0$
\end_inset
:
\begin_inset Formula
\[
\Delta_{n}'\left(x,w\right)=\sum_{k=0}^{\infty}\Gamma\left(\frac{1}{2}-n-k,x\right)\frac{w^{k}}{k!},
\]
\end_inset
\begin_inset Formula
\[
\Delta_{n}\left(x,z\right)=\sum_{k=0}^{\infty}\Gamma\left(\frac{1}{2}-n-k,x\right)\frac{\left(z/2\right)^{2k}}{k!}.
\]
\end_inset
The big negative first arguments in incomplete
\begin_inset Formula $\Gamma$
\end_inset
functions should be good (at least I think so, CHECKME), as well as the
\begin_inset Formula $1/k!$
\end_inset
factor (of course).
I am not sure what the convergence radius is, but for
\begin_inset Formula $\left|z\right|<2$
\end_inset
there seems to be absolutely no problem in using this formula.
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