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Is it possible to compute the inverse transform of

$ \frac{1}{a^{-s}\cos( \frac{\pi s}{2})\Gamma (s)} $

or similarly is it possible to compute the Inverse Mellin transform ??

$ \frac{ \zeta (1-s)}{\zeta (s)} $

$ \frac{ \zeta (s)}{\zeta (1-s)} $


The Mellin inverse is given by

$ \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}dsF(s)x^{-s} $

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    Hint: $\int_0^\infty x^s\sin ax~dx=a^{-s-1}\Gamma(s+1)\cos\dfrac{\pi s}{2}$ , according to http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourSin2.pdf2015-07-05

1 Answers 1

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For the first one $ \mathcal{M}^{-1}\left[ \frac{1}{a^{-s}\cos( \frac{\pi s}{2})\Gamma (s)} \right] = \frac{2 a \cos\left(\frac{a}{x}\right)}{\pi x} $

For the $\zeta$ expressions we can use the Riemann functional equation $ \zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s) $ Then find the inverse Mellin transforms $ \mathcal{M}^{-1}\left[2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right)\Gamma(1-s)\right] = \frac{2 \cos(\frac{2\pi}{x})}{x} $ and $ \mathcal{M}^{-1}\left[\frac{1}{2^s \pi^{s-1} \sin(\frac{\pi s}{2})\Gamma(1-s)}\right] = 2 \cos(2 \pi x) $

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    I'm still getting the same answer now that I have done it fully. Sum of the residues is $ \frac{2}{\pi}\sum_{n=0}^\infty \frac{(-1)^{n+1} a^{2n+1}}{(2n)! x^{2n+1}} = \frac{2 a \cos(a/x)}{\pi x} $2017-11-12