I am trying to compute
\begin{align} \int_{\Gamma} \frac{1}{(z-a)^m (z-b)^n} dz \end{align} where $\Gamma$ is a circle whose radius is bigger than $a$ and $b$ with positive $m,n$.
Let $f(z) = \frac{1}{(z-a)^m (z-b)^n}$, and recalling residue theorem, I noticed \begin{align} &Res(f;a) = \frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}} \left(z-b\right)^{-n} |_{z=a} = (-1)^{m-1} {}_{n+m-2} C{}_{n-1} (a-b)^{-(n+m-1)} \\ &Res(f;b) = \frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}} \left(z-a\right)^{-m} _{z=b} = (-1)^{n-1} {}_{n+m-2} C_{m-1} (b-a)^{-(n+m-1)} \\ &\phantom{Res(f;b)} = (-1)^{n-1}{}_{n+m-2} C_{n-1} (-1)^{-(n+m-1)} (a-b)^{-(n+m-1)} = -Res(f;a) \end{align} where $C$ means binomial. ${}_n{}C_m = \frac{n!}{m!(n-m)!} = binomial[n,m]$.
Hence combine all, the integral becomes zero.
Am I right? Is there any other way to compute this kind of integral?
If I am right, and If there are some easier way to compute(?) or show this I want to extend this in more general form
\begin{align} \int_{\Gamma} \frac{1}{(z-a_1)^{m_1} \cdots (z-a_n)^{m_n}} dz \end{align} where $\Gamma$ is a cirlce containing $a_1, \cdots a_n$.
Substitute $z=1/w$. If $|z|=r$ on $\Gamma$, then you get $$\int_\Gamma\frac{dz}{(z-a)^m(z-b)^n}=\int_{|w|=1/r}\frac{w^{m+n-2}~dw}{(1-aw)^m(1-bw)^n}.$$ Now there's no singularities, thus the integral is zero. [The same for the extension.]