Evaluate Complex Integral with $\frac{\Gamma(\frac{s}{2})} {\Gamma\big({\beta +1\over 2} - {s\over 2}\big)}$ I am proving this integral:
$$ \int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(\,x^{-1}\sigma\beta^{1 \over 2}\,\right)^{s}\ \Gamma\left(\,s \over 2\,\right) \Gamma\left(\,{\beta +1 \over 2} - {s \over 2}\,\right)\,{\rm d}s = \Gamma\left(\,{\beta +1 \over 2}\,\right)\left(\,1+{1 \over \beta}\left(x \over \sigma\right)^2\,\right)^{-{\beta +1 \over 2}}$$
where $\beta>0$, $\sigma>0$ and $x$ is real number
The clue I have is that Cauchy's residue theorem is applicable for the evaluation but I cant figure out how can the simplification be made
 A: The equation should actually be 
$$\frac{1}{4\pi j} \int_{c - j\infty}^{c + j\infty} (x^{-1}\sigma \beta^{\frac{1}{2}})^s\, \Gamma\left(\frac{s}{2}\right) \Gamma\left(\frac{\beta + 1}{2} - \frac{s}{2}\right)\, \mathrm{d}s = \Gamma\left(\frac{\beta + 1}{2}\right) \left[1 + \frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\right]^{-\frac{\beta + 1}{2}}$$
for $0 < c <\beta + 1$. Let $F(x;\sigma,\beta)$ be the expression on the left-hand side. Then 
\begin{align}F(x;\sigma,\beta) &= \frac{1}{2\pi j}\int_{\frac{c}{2} - j\infty}^{\frac{c}{2} + j\infty} (x^{-1}\sigma\beta^{\frac{1}{2}})^{2s}\, \Gamma(s) \Gamma\left(\frac{\beta + 1}{2} - s\right)\, \mathrm{d}s\\
&= \frac{\Gamma\left(\frac{\beta+1}{2}\right)}{2\pi j}\int_{\frac{c}{2} - j\infty}^{\frac{c}{2} + j\infty} \left[\frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\right]^{-s} B\left(s, \frac{\beta+1}{2}-s\right)\, \mathrm{d}s\\
&= \Gamma\left(\frac{\beta+1}{2}\right) \mathcal{M}^{-1}\left\{B\left(s,\frac{\beta+1}{2}-s\right)\right\}\left[\frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\right],
\end{align}
where $B(u,v)$ represents the Beta function and $\mathcal{M}^{-1}(f(s))[r]$ represents the inverse Mellin transfrom of $f(s)$ evaluated at $r$. Now, for $0 < \text{Re}(s) < \frac{\beta + 1}{2}$, $$B\left(s,\frac{\beta+1}{2}-s\right) = \int_0^\infty \frac{r^{s-1}}{(1 + r)^{s + \left(\frac{\beta + 1}{2} - s\right)}}\, \mathrm{d}r = \int_0^\infty r^{s-1} (1 + r)^{-\frac{\beta+1}{2}}\, \mathrm{d}r,$$
that is, $B\left(s, \frac{\beta+1}{2}-s\right)$ is the Mellin transform of $(1 + r)^{-\frac{\beta + 1}{2}}$ evaluated at $s$. Hence,
$$F(x;\sigma,\beta) = \Gamma\left(\frac{\beta+1}{2}\right)(1 + r)^{-\frac{\beta+1}{2}}\bigg|_{r = \frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2} = \Gamma\left(\frac{\beta+1}{2}\right)\left[1+\frac{1}{\beta}\left(\frac{x}{\sigma}\right)^2\right]^{-\frac{\beta+1}{2}}.$$
