Complex integral on unit disk I am trying to calculate the following integral :
$\int_{r=0}^{1}\int_{t=0}^{2\pi} \frac{r}{re^{it}-a} drdt$ where $a$ is a complex number of modulus 1.
According to computation websites, this integral converges ; but I don't manage to make it by hand. I tried to use integration by parts, but the singularity at $a$ makes me feel it will not work... Any ideas?
 A: Hint: Calculate $\int_{|z|=1} \frac  1 {z (z-a)}dz$ using Residue Theorem. [There is a simple pole at $0$].  This integral can be written as $\int_0^{2\pi} \frac 1 {re^{it}-a} idt$. Rest is clear.
A: Make the substitution $$z = e^{it},$$ $$dz = ie^{it}\,dt = iz\,dt,$$
$$dt = \frac{1}{iz}\,dz,$$
to get
$$\int_{0}^{2\pi}\frac{r}{re^{it} - a}dt = \int_{S^1}\frac{r}{(rz - a)iz}\,dz =: f(r).$$
$f(r)$ can be computed using the residue theorem. Then you want $\int_{0}^1 f(r)\,dr$.
A: This integral can also be computed without using the residue theorem.
Suppose that $r \in (0, 1)$. For every $t \in [0, 2 \pi]$, we have $$ \frac{r}{r e^{i t} -a} = \frac{-r}{a} \frac{1}{1 -\frac{r}{a} e^{i t}} = \frac{-r}{a} \sum_{n = 0}^{+\infty} \left( \frac{r}{a} \right)^{n} e^{i n t}$$ since $\left\lvert \frac{r}{a} \right\rvert = r < 1$. Moreover, the series $\sum_{n \geq 0} \left( \frac{r}{a} \right) e^{i n t}$ converges normally (and hence uniformly) on $[0, 2 \pi]$. Therefore, we have $$\int_{0}^{2 \pi} \frac{r}{r e^{it} -a} = \frac{-r}{a} \sum_{n = 0}^{+\infty} \int_{0}^{2 \pi} \left( \frac{r}{a} \right)^{n} e^{i n t} \, dt = \frac{-2 \pi r}{a}$$ since, for every integer $n \geq 0$, we have $$\int_{0}^{2 \pi} \left( \frac{r}{a} \right)^{n} e^{i n t} \, dt = \begin{cases} 2 \pi & \text{if } n = 0\\ 0 & \text{otherwise} \end{cases} \, \text{.}$$
Therefore, we have $$\int_{0}^{1} \left( \int_{0}^{2 \pi} \frac{r}{r e^{i t} -a} \, dt \right) \, dr = \int_{0}^{1} \frac{-2 \pi r}{a} \, dr = \frac{-\pi}{a} \, \text{.}$$
That being said, I like very much using the residue theorem here.
