# Integral of complex logarithm on a disk in the plane

Let $a$ be a complex number and $D$ the disk centered around $0$ and of radius $R$. I would like to compute the integral

I=$\int_D \log(|z-a|)d^2z$.

I am interested in particular in the case $R\gg |a|$. I expect the result to be $2\pi|a|^2/4+C$ (physicist use this result, but I did not find a derivation), where $C$ depends on $R$ but not on $a$ (the specific form of $C$ is not important to me).

Here is what I have tried so far : I can first write the integral on the disk parametrized with $z=re^{it}$, so that it becomes

$I=\int_{r=0}^{R}\int_{t=0}^{2\pi}\log(|re^{it}-a|) r dr dt$

If $r<|a|$, then I can compute the integral on $t$ because $\log(z-a)$ is analytic on the disk, so I get

$\int_{t=0}^{2\pi}\log(|re^{it}-a|) dt=Re(\int_\gamma \frac{\log(z-a)}{iz} dz)=Re(2\pi i \frac{\log(a)}{i})=2\pi \log{|a|}$, where $\gamma$ is the circle of radius $r$.

The integral up to $r=|a|$ is then $2\pi \log{|a|}\int_{r=0}^{a}r dr =\pi \log{(|a|)}|a|^2$.

But for $r>|a|$, the logarithm has a branch cut inside the disk of integration so I cannot simply compute the integral in the same way. Moreover since I am only interested in the real part, it seems that computing the imaginary part is more complicated than what should be needed.

Note that in the case $a=0$, then the integral is

$\int_D \log(|z|)d^2z=\int_{r=0}^{R}\int_{t=0}^{2\pi}\log(r) r dr dt=2\pi(\frac{R^2}{4}+\frac{R^2\log(R)}{2})$, by integration by part.

Let

$I(\vec{a}) = \int_D d^2 r \; \log|\vec{r}-\vec{a}|$

Using the important identity

$\nabla^2 \log |\vec{r}| = 2 \pi \delta^2(\vec{r})$

for the planar Laplacian, we see that

$\nabla^2 I(\vec{a}) = 2 \pi$

if $\vec{a} \in D$. In this case, we have

$I(\vec{r}) = 2 \pi \frac{1}{4} \pi r^2 + h(\vec{r})$,

where $h(\vec{r})$ is harmonic, $\nabla^2 h = 0$. This is true for any region $D$, but we now specialize to the disk of radius $R$.

Then, evidently the problem is radially symmetric, so that $h$ is a function of $|\vec{r}|$ only. The most general radially symmetric harmonic function in the plane is

$h(r) = c \log r + h_0$

For $\vec{a}=0$, we have $I(0) = h(0) = -\frac{\pi}{2} R^2 + \pi R^2 \log R$ (correcting a typo in OP), which is finite, so that $c=0$. Then $h_0=h(0)$ and we have

$I(\vec{r}) = \frac{\pi}{2} (r^2 - R^2) + \pi R^2 \log R$.

For $|\vec{a}| \geq R$, OP's technique works.

• You say "integrate over the Disk" but don't you really mean integrate about the boundary of the disk, a circle? – user247327 Apr 10 '18 at 23:58
• @user247327 Sorry, I didn't pay attention to the $\color{red}{2}$ in $\mathrm{d}^{\color{red}{2}}z$. Thanks. – Felix Marin Apr 11 '18 at 0:19