Proving an Integral with Cauchy Residue Theorem I need help proving this. The clue given is that Cauchy residue theorem can be used:
$${1 \over {2\pi j}}\int_{c\ -\ j\infty}^{c\ +\ j\infty} x^{-s}\sigma^{s-1} {{\left(1-\beta^{s+1}\right)}\over{\left(1-\beta^2\right)}}\left[{\Gamma\left(s\right)\over{\Gamma\left(2+s\right)}}\right]ds=\left\{\begin{matrix}
\frac{1}{\sigma\left(1+\beta\right)} & \text{if }|x|\leq \beta\\ 
\frac{\sigma - |x|}{\sigma^2\left(1-\beta^2 \right )} & \text{if }\beta<|x|\leq \sigma\\
 0& \text{otherwise}
\end{matrix}\right.$$
where $x$, $\beta$ and $\sigma$ are real. Both $\beta$ and $\sigma$ are positive whereby $\beta<\sigma$
I have tried solving this and the furthest i have reached is at this point:
$${1 \over {2\pi j}}\int_{c\ -\ j\infty}^{c\ +\ j\infty} x^{-s}\sigma^{s-1} {{\left(1-\beta^{s+1}\right)}\over{\left(1-\beta^2\right)}}\left[{\Gamma\left(s\right)\over{\Gamma\left(2+s\right)}}\right]ds$$
$$={1 \over {2\pi j}}{1 \over \sigma \left(1-\beta^2\right)}\int_{c\ -\ j\infty}^{c\ +\ j\infty} \left(x^{-1}\sigma\right)^s{{\left(1-\beta^{s+1}\right)}\over{s\left(s+1\right)}}ds$$
This means that the singularity occurs at point $s=0$ and $s=-1$. I am not sure how to carry on beyond this point. Could someone help please? Thanks
 A: The integral looks messy but it all comes down to proving the following.($c>0$) (Which is used in the analytic proof of PNT)
$$I_k(u)=\frac1{2\pi i}\int_{c-\infty i}^{c+\infty i}\frac{u^{-z}}{z(z+1)\cdots(z+k)}dz=\cases{\frac{(1-u)^k}{k!}&if $0<u\le1$\\0&if $u>1$}$$
To prove this, apply the residue theorem on the integral
$$\frac1{2\pi i}\int_{C(R)}\frac{u^{-z}\Gamma(z)}{\Gamma(z+k+1)}dz$$
Divide the circle into two arcs by the chord $c+is$. Choose $C(R)$ as the left arc if $0<u\le1$, the right arc if $u>1$. Since $c>0$, left arc contains all the poles and the right one has none. Setting $R\to\infty$, The circular part of the integral tends to zero.
If $u>1$ then $C(R)$ has no poles inside so our integral equals $0$.
If $0<u\le1$, then by the residue theorem we have $$\frac1{2\pi i}\int_{C(R)}\frac{u^{-z}\Gamma(z)}{\Gamma(z+k+1)}dz=\sum_{n=0}^k\operatorname{Res}_{z=-n}\frac{u^{-z}\Gamma(z)}{\Gamma(z+k+1)}=\sum_{n=0}^k\frac{u^n(-1)^n}{(k-n)!n!}=\frac{(1-u)^k}{k!}$$
Now back to the original problem. If $x$ is positive,
\begin{align}&{1 \over {2\pi i}}{1 \over \sigma \left(1-\beta^2\right)}\int_{c\ -\ i\infty}^{c\ +\ i\infty} \left(x^{-1}\sigma\right)^s{{\left(1-\beta^{s+1}\right)}\over{s\left(s+1\right)}}ds\\=&\frac1{\sigma(1-\beta^2)}I_2(x/\sigma)-\frac{\beta}{\sigma(1-\beta^2)}I_2(x/(\sigma\beta))\\=&\cases{\frac1{\sigma(1+\beta)}&if $x\le\beta$\\\frac{\sigma-x}{\sigma^2(1-\beta^2)}&if $\beta<x\le\sigma$\\0&if $\sigma<x$}\end{align}
