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.
Thank you for your help.