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How would you find an approximation of the Inverse Mellin transform of $\Gamma(s)$ $\zeta(s/2)$ near $x=0$?

sigma
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1 Answers1

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$\mathcal{M}^{-1}\left\{\Gamma(s)\zeta\left(\dfrac{s}{2}\right)\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{\Gamma(s)}{\Gamma\left(\dfrac{s}{2}\right)}\int_0^\infty\dfrac{x^{\frac{s}{2}-1}}{e^x-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^{s-1}\Gamma\left(\dfrac{s}{2}\right)\Gamma\left(\dfrac{s+1}{2}\right)}{\sqrt\pi~\Gamma\left(\dfrac{s}{2}\right)}\int_0^\infty\dfrac{x^{s-2}}{e^{x^2}-1}~d(x^2)\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^s}{\sqrt\pi}\int_0^\infty x^\frac{s-1}{2}e^{-x}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^s}{\sqrt\pi}\int_0^\infty x^{s-1}e^{-x^2}~d(x^2)\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^{s+1}}{\sqrt\pi}\int_0^\infty x^{s-1}e^{-x^2}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2^{s+1}}{\sqrt\pi}\int_0^\infty\dfrac{x^{s-1}}{2^{s-1}}e^{-\frac{x^2}{4}}~d\left(\dfrac{x}{2}\right)\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\mathcal{M}^{-1}\left\{\dfrac{2}{\sqrt\pi}\int_0^\infty x^se^{-\frac{x^2}{4}}~dx\int_0^\infty\dfrac{x^{s-1}}{e^{x^2}-1}~dx\right\}$

$=\dfrac{2}{\sqrt\pi}\int_0^\infty\dfrac{xe^{-\frac{x^2}{4t^2}}}{t^2(e^{t^2}-1)}~dt$ (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 $a,b>0$)

$=2\sum\limits_{n=1}^\infty e^{-x\sqrt n}$

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doraemonpaul
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    all you need is $n^{-s} \Gamma(s/2) = \int_0^\infty x^{s/2-1} e^{-n^2 x} dx$ (change of variable $y = n^2 x$) so that $\Gamma(s/2) \zeta(s) = \sum_{n=1}^\infty n^{-s} \Gamma(s/2) = \sum_{n=1}^\infty \int_0^\infty x^{s/2-1} e^{-n^2 x} dx$ $ = \int_0^\infty x^{s/2-1} \sum_{n=1}^\infty e^{-n^2 x} dx$ $ = 2\int_0^\infty y^{s-1} \sum_{n=1}^\infty e^{-n^2 y^2} dy^2 $ ($x= y^2$) (well-defined for $Re(s) > 1$) so that $M^{-1}[\Gamma(s/2) \zeta(s)] = 2 \sum_{n=1}^\infty e^{-n^2 x^2}$ – reuns Sep 11 '16 at 18:00
  • and it seems obvious you didn't understand the inverse Mellin transform. Do you know the Laplace/Fourier transform ? It is the same with $x = e^t$ – reuns Sep 11 '16 at 18:04
  • @user1952009 This question is about $\mathcal{M}^{-1}\left{\Gamma(s)\zeta\left(\dfrac{s}{2}\right)\right}$ rather than $\mathcal{M}^{-1}\left{\Gamma\left(\dfrac{s}{2}\right)\zeta(s)\right}$ . – doraemonpaul Sep 11 '16 at 18:46
  • right, sorry then, but integrating $\int_0^\infty e^{-(ax^2+b/x^2)} dx$ is ridiculous, use the same method I wrote – reuns Sep 11 '16 at 18:55