# Evaluate $\iint_{x^2+y^2 < 1} \frac{dxdy}{x+iy-w}$ using the residue theorem

The goal is to solve the following integral by passing it to polar coordinates and then using the Residue Theorem: $$\iint_{x^2+y^2 < 1} \frac{dxdy}{x+iy-w},$$ where $w\in\mathbb{C}$ and $|w| <1$.

I am ending up with $0$ as my answer which seems slightly questionable. So if I'd love it if someone could point out if they see a mistake in my calculations. Here's what I have so far.

First, passing to polar coordinates we get: \begin{align} \int_0^{2\pi}\int_0^1\frac{rdrd\theta}{r\cos\theta +ir\sin\theta -w} = \int_0^{2\pi}\int_0^1\frac{r}{re^{i\theta}-w}drd\theta = \int_0^1\int_0^{2\pi}\frac{r}{re^{i\theta}-w}d\theta dr. \end{align}

Now I will evaluate $$\int_0^{2\pi}\frac{r}{re^{i\theta}-w}d\theta$$ using the Residue Theorem. First we let $\gamma$ be the unit circle oriented counter clockwise. Then let $z=re^{i\theta}$, so $dz = rie^{i\theta}d\theta = rizd\theta$. Then our integral becomes \begin{align} \int_0^{2\pi}\frac{r}{re^{i\theta}-w}d\theta = \int_{\gamma} \frac{r}{(z-w)}\frac{dz}{irz} = -i\int_{\gamma} \frac{dz}{z(z-w)}. \end{align} Then $f(z) = \frac{1}{z(z-w)}$ has simple poles at $0$ and $w$, which are both inside $\gamma$. So we can calculate the resides of $0$ and $w$: $$Res(f,0) = \frac{1}{f'(0)} = \frac{1}{2\cdot 0 -w} = -\frac{1}{w},$$ and $$Res(f,w) = \frac{1}{f'(w)} = \frac{1}{2\cdot w -w} = \frac{1}{w}.$$ Then by the Residue Theorem, $$\int_{\gamma} \frac{dz}{z(z-w)} = 2\pi i(Res(f,0) + Res(f,w)) = 2\pi i (-\frac{1}{w} + \frac{1}{w}) = 0.$$

Then this means that $$\int\int_{x^2+y^2 < 1} \frac{dxdy}{x+iy-w} = \int_0^1\int_0^{2\pi}\frac{r}{re^{i\theta}-w}d\theta dr = \int_0^1 0 dr = 0.$$

Does this make sense or does anyone spot a mistake?

For $0<|w|<1$, let us consider the following integrals $$I_1:=\iint_{x^2+y^2 < R_1} \frac{dxdy}{x+iy-w}\quad,\quad I_2:=\iint_{R_2<x^2+y^2 < 1} \frac{dxdy}{x+iy-w}$$ where $0<R_1<|w|<R_2<1$. Then using your approach we get $$I_1=\frac{1}{i}\int_0^{R_1}rdr\int_{|z|=r} \frac{dz}{z(z-w)}=2\pi\int_0^{R_1}r\mbox{Res}(f,0) dr=-\frac{\pi R_1^2}{w}$$ and $$I_2=\frac{1}{i}\int_{R_2}^1rdr\int_{|z|=r} \frac{dz}{z(z-w)}=2\pi\int_{R_2}^1r(\mbox{Res}(f,0)+\mbox{Res}(f,w)) dr=0.$$ Hence, by defining the given improper integral ($w$ is inside the domain of integration) as $$I:=\iint_{x^2+y^2 < 1} \frac{dxdy}{x+iy-w}=\lim_{R_1 \to |w|^-}I_1+\lim_{R_2 \to |w|^+}I_2$$ we should conclude that $I=-\pi|w|^2/w=-\pi\overline{w}$ for $0<|w|<1$ (it is $0$ for $w=0$). In a similar way, for $|w|\geq 1$, we get that $I=-\pi/w$.