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 $I=-\pi/w$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.