3
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.