I previously posted a similar problem here and I have solved many of the problems from the answers given with explanations. This time however I am at this point of integration where:
$$\int_{c\ -\ i\infty}^{c\ +\ i\infty} \left(x^{-1}\sigma\right)^{s}\Gamma\left(\,{\beta +s-1}\,\right)\,{\rm d}s$$
where $\beta$, $\sigma$ and $x$ are real numbers.
I know that Cauchy's residue theorem is applicable for the evaluation but I cant figure out how can the simplification be made.
You may just use the previous answer given to the question in reference. If you perform the change of variable $s \to t$, with $$ \beta+s-1=\frac{t}{2}, \quad ds=\frac12 dt, $$ your initial integral is now equal to $$ \begin{align} &\frac{\left(x^{-1}\sigma\right)^{1-\beta}}{2} \int_{2(c+\beta-1)\ -\ i\infty}^{2(c+\beta-1)\ +\ i\infty} \left(\left(x^{-1}\sigma\right)^{1/2}\right)^{t}\Gamma\left(\frac t2\right)\,{\rm d}t \\\\&=\frac{\left(x^{-1}\sigma\right)^{1-\beta}}{2} \int_{c'\ -\ i\infty}^{c'\ +\ i\infty} u^{t}\:\Gamma\left(\frac t2\right)\,{\rm d}t\\\\ &=\frac{\left(x^{-1}\sigma\right)^{1-\beta}}{2} 4\pi i\cdot e^{-\dfrac{1}{u^2}}\\\\ &=2\pi i\:\left(x^{-1}\sigma\right)^{1-\beta} \cdot e^{-\dfrac{x}{\sigma}}. \end{align} $$