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 $t \to \infty$.