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I have been strugling with finding the inverse Mellin tranform of the following function:

\begin{align} F(s) &= b^{\frac{2s-2\alpha}{2\alpha-1}}\sin\left[\frac{\pi}{2}\left(\frac{2s-2\alpha}{2\alpha-1}\right)\right]\Gamma(s)\Gamma\left(\frac{2\alpha-2s}{2\alpha-1}\right)\Gamma\left(1-\frac{2\alpha-2s}{2\alpha-1}\right), \end{align} where $b > 0, \alpha > \frac{1}{2}$ and $\frac{1}{2} < s < \alpha$. Such that the inverse Mellin transform is given by \begin{align} f(t) = \frac{1}{2\pi i} \int_{a-i \infty}^{a + i \infty} ds F(s)t^{-s}, \end{align} with $\frac{1}{2} < a < \alpha$. I have tried using Euler's reflection formula together with double angle formula for the sine to simplify $F(s)$ and it simplifies a lot, however since Euler's reflection formula is not valid for negative integers it turns out to be wrong since the pole structure of the two integrands are not the same. I have also tried to directly evaluate the integral by closing the contour and applying the residue theorem but I am not sure how to argue that the 'closing the contour' is allowed (how those extra contour parts go to zero). The last thing that I tried was to use mathematica (InverseMellinTransform) but it just returns my input. Any suggestions? In particular, I am interested in the behavior of $f(t)$ for $t\to 0$ and $t \to \infty$.

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  • $\begingroup$ I believe $F(s)$ simplifies to $F(s)=-\frac{1}{2} \pi\, b^{\frac{2 (s-\alpha)}{2 \alpha-1}} \sec\left(\frac{\pi (s-\alpha)}{2 \alpha-1}\right)\, \Gamma(s)$, but I'm not sure if its useful. $\endgroup$ Apr 29, 2023 at 18:18
  • $\begingroup$ This is indeed what I also had. However, I'm guessing you used Euler's reflection formula right? If this is the case, it is not valid for some values. The concequence of this is that your form has a different pole structure of the original one, which makes it not useful right? (I could be wrong however). $\endgroup$
    – Audrique
    Apr 29, 2023 at 19:43

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