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

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

  • $\begingroup$ For the Mellin transformed function, you have to provide the strip on which it is defined; the reverse transform depends on this strip ($c$ has to within the strip). $\endgroup$ – Fabian Dec 16 '12 at 17:36
  • $\begingroup$ in this case , can we find a function so $$ \frac{\zeta (1-s)}{\zeta (s)}= \int_{0}^{\infty}dt f(t)t^{s-1} $$ $\endgroup$ – Jose Garcia Dec 16 '12 at 21:46
  • $\begingroup$ again, for which values of $\text{Re} s$ this relation should hold? $\endgroup$ – Fabian Dec 16 '12 at 22:14
  • $\begingroup$ Hint: $\int_0^\infty x^s\sin ax~dx=a^{-s-1}\Gamma(s+1)\cos\dfrac{\pi s}{2}$ , according to eqworld.ipmnet.ru/en/auxiliary/inttrans/FourSin2.pdf $\endgroup$ – doraemonpaul Jul 5 '15 at 0:24

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 $$ \mathcal{M}^{-1}\left[\frac{1}{2^s \pi^{s-1} \sin(\frac{\pi s}{2})\Gamma(1-s)}\right] = 2 \cos(2 \pi x) $$

  • $\begingroup$ We can use the residue theorem and I'm not convinced it will give your first claim. How did you obtain that ? $\endgroup$ – reuns Nov 11 '17 at 12:51
  • $\begingroup$ @reuns It was output by Mathematica's InverseMellinTransform routine. However, they might have a bug and I trust your wisdom more. What do you think the answer should be? (Or what makes you doubt that is the answer?) $\endgroup$ – Benedict W. J. Irwin Nov 11 '17 at 13:24
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    $\begingroup$ Apply the residue theorem to $\frac{1}{2\pi i}\int_{1/2-i\infty}^{1/2+i\infty} \frac{1}{\cos( \frac{\pi s}{2})\Gamma (s)}x^{-s}ds$ $\endgroup$ – reuns Nov 11 '17 at 13:27
  • $\begingroup$ I'm still getting the same answer now that I have done it fully. Sum of the residues is $$ \frac{2}{\pi}\sum_{n=0}^\infty \frac{(-1)^{n+1} a^{2n+1}}{(2n)! x^{2n+1}} = \frac{2 a \cos(a/x)}{\pi x} $$ $\endgroup$ – Benedict W. J. Irwin Nov 12 '17 at 10:05

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