How can I calculate the integral? $$ \int_{\left| z \right| = r} \frac{dz}{(z-a)^n(z-b)^n} $$ for $ \left| a \right| < r < \left| b \right|$ and $ m, n > 1$
I tried to use the cauchy integral forula but given the variable m and n, there should be probably another aproach?
For $|a| < r < |b|$, the function $f(z) := 1/(z - b)^n$ is analytic inside and on the circle $|z| = r$, and $a$ lies inside $|z| = r$. Hence, by Cauchy's differentiation formula,
\begin{align}\int_{|z| = r} \frac{1}{(z - a)^n (z - b)^n}\, dz &= \frac{2\pi i}{(n-1)!} f^{(n-1)}(a)\\ &= \frac{2\pi i}{(n-1)!}\frac{(-1)^{n-1}n(n+1)\cdots (2n-2)}{(a - b)^{2n-1}}\\ &= (-1)^{n-1} \binom{2n-2}{n-1}\frac{2\pi i}{(a - b)^{2n-1}}. \end{align}