I previously posted a similar problem here and here. This time however I am dealing with multiple gamma functions. This is the problem I am dealing with right now:
$$ \int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(\,x^{-1}\sigma\,\right)^{s}\, {\Gamma\left(s\right) \over \Gamma\left(s + 2\right)}\,{\rm d}s $$
where β, σ and x are real number
I know that Cauchy's residue theorem is applicable for the evaluation but I cant figure out how can the simplification be made
The first integral is relatively simple because it is simply
$$\int_{c-i\infty}^{c+i \infty} ds \frac{e^{s t}}{s (s+1)}$$
where $t=\log{\left ( x^{-1} \sigma \right )}$. This is a well-known inverse Laplace transform and may be evaluated using a Bromwich contour closed to the left or right according to the sign of $t$. When $t \gt 0$, the integral is simply $i 2 \pi$ times the sum of the residues at the poles (assuming $c \gt 0$, which I am); when $t \lt 0$, the integral is simply zero. Thus, the integral is
$$i 2 \pi \left (1-e^{-t} \right ) \theta(t) = i 2 \pi \left (1-\frac{x}{\sigma} \right ) \theta(\sigma-x)$$
where $\theta$ is the Heaviside step function.