Inverse Mellin transform of $\Gamma(s)$ $\zeta(s/2)$ How would you find an approximation of the Inverse Mellin transform of $\Gamma(s)$ $\zeta(s/2)$ near $x=0$?
 A: $\mathcal{M}^{-1}\left\{\Gamma(s)\zeta\left(\dfrac{s}{2}\right)\right\}$
$=\mathcal{M}^{-1}\left\{\dfrac{\Gamma(s)}{\Gamma\left(\dfrac{s}{2}\right)}\int_0^\infty\dfrac{x^{\frac{s}{2}-1}}{e^x-1}~dx\right\}$
$=\mathcal{M}^{-1}\left\{\dfrac{2^{s-1}\Gamma\left(\dfrac{s}{2}\right)\Gamma\left(\dfrac{s+1}{2}\right)}{\sqrt\pi~\Gamma\left(\dfrac{s}{2}\right)}\int_0^\infty\dfrac{x^{s-2}}{e^{x^2}-1}~d(x^2)\right\}$
$=\mathcal{M}^{-1}\left\{\dfrac{2^s}{\sqrt\pi}\int_0^\infty x^\frac{s-1}{2}e^{-x}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$
$=\mathcal{M}^{-1}\left\{\dfrac{2^s}{\sqrt\pi}\int_0^\infty x^{s-1}e^{-x^2}~d(x^2)\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$
$=\mathcal{M}^{-1}\left\{\dfrac{2^{s+1}}{\sqrt\pi}\int_0^\infty x^{s-1}e^{-x^2}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$
$=\mathcal{M}^{-1}\left\{\dfrac{2^{s+1}}{\sqrt\pi}\int_0^\infty\dfrac{x^{s-1}}{2^{s-1}}e^{-\frac{x^2}{4}}~d\left(\dfrac{x}{2}\right)\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$
$=\mathcal{M}^{-1}\left\{\dfrac{2}{\sqrt\pi}\int_0^\infty x^se^{-\frac{x^2}{4}}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$
$=\dfrac{2}{\sqrt\pi}\int_0^\infty\dfrac{xe^{-\frac{x^2}{4t^2}}}{t^2(e^{t^2}-1)}~dt$ (according to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/mellin.pdf)
$=-\dfrac{2x}{\sqrt\pi}\int_0^\infty\dfrac{e^{-\frac{x^2}{4t^2}}}{e^{t^2}-1}~d\left(\dfrac{1}{t}\right)$
$=\dfrac{2x}{\sqrt\pi}\int_0^\infty\dfrac{e^{-\frac{x^2t^2}{4}}}{e^\frac{1}{t^2}-1}~dt$
$=\dfrac{2x}{\sqrt\pi}\int_0^\infty\dfrac{e^{-\frac{x^2t^2}{4}-\frac{1}{t^2}}}{1-e^{-\frac{1}{t^2}}}~dt$
$=\dfrac{2x}{\sqrt\pi}\int_0^\infty\sum\limits_{n=0}^\infty e^{-\frac{x^2t^2}{4}-\frac{n+1}{t^2}}~dt$
$=2\sum\limits_{n=0}^\infty e^{-x\sqrt{n+1}}$ (according to How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$)
$=2\sum\limits_{n=1}^\infty e^{-x\sqrt n}$
