Evaluate $$\int_{0}^{2\pi}\frac{1}{\rho^2+r^2-2r\rho \cos(t-\theta)}dt.$$
I found this under some exercises about Poisson's integral formula, to my surprise the problem looks simple but I do not have a single idea of how to go with it. Can somebody help?
Best Answer
Let $z = \mathrm{e}^{i (t - \theta)}$, and assuming $\rho \not= r$, we get $$ \int_0^{2 \pi} \frac{1}{\rho^2 + r^2 - 2 r \rho \cos(t - \theta) } \mathrm{d}t = \oint \frac{1}{ r^2 + \rho^2 - r \rho \left( z + z^{-1} \right)} \cdot \frac{1}{i} \frac{\mathrm{d} z}{z} = \oint \frac{1}{ (r^2 + \rho^2) z - r \rho \left( z^2 + 1 \right)} \cdot \frac{\mathrm{d} z}{i} = \oint \frac{1}{ (r z - \rho)(r - \rho z)} \cdot \frac{\mathrm{d} z}{i} = \frac{2 \pi}{| \rho^2 - r^2 |} $$