I asked a similar but different question earlier here (this is not a duplicate)
I'm interested in solving this problem in closed form, if such a solution exists, in $2d$.
$$\int_{\Gamma_R} G\; d\Gamma_R$$
Where $\Gamma_R$ is a circle of radius $R$ centered at the origin.
Let $G$ be the fundamental solution of the laplace equation ($\Delta u = 0$), in $2d$.
$G(x, \bar{x}) = -\frac{1}{2 \pi} ln\left(\lVert x - \bar{x}\rVert\right)$
Putting it together:
$$\int_{\Gamma_R} G(x, \bar{x}) \; d\Gamma_R = -\frac{1}{2 \pi} \int_{\Gamma_R} ln\lVert x - \bar{x}\rVert\; d\Gamma_R $$
I think the right approach could be to parameterize a circle and solve this in polar coordinates, but I'm stuck here. I used $R \; d\theta = d\Gamma_R$
$$\int_{\Gamma_R} G(x, \bar{x}) \; d\Gamma_R = -\frac{R}{2 \pi }\int_{0}^{2\pi} ln\sqrt{((R\;cos(\theta) - \bar{x}_0)^2 - (R\;sin(\theta) - \bar{x}_1)^2}d\theta$$
Also the point $\bar{x}$ should be inside the circle, if that makes any difference.
This integral can be computed either using complex analysis or calculus and the result is
$$I=-R\ln\max(\|\bar{x}\|,R)$$
I denote $\bar{x}=(x_0,y_0)$ and $z_0=x_0+iy_0$. Then, for the complex analysis solution, first notice that $\ln|x-\bar{x}|=\frac{1}{2}\ln|z-z_0|^2$ and the integral can be rewritten as a contour integral on the complex plane.
$$I=-\frac{R}{4\pi}\oint_{|z|=R}\frac{\ln(z-z_0)+\ln(\bar{z}-\bar{z_0})}{iz}dz$$
However since $\bar{z}=R^2/z$ on the circle $\Gamma_R$, changing coordinates $z\to R^2/z$ in the second term only yields an integral that can be done using Cauchy's lemma
$$-4\pi I=R\oint_{|z|=R}\frac{\ln(z-z_0)+\ln(z-\bar{z_0})}{iz}dz$$
When $|z_0|>R$, the branch cut of the logarithm is outside the circle, and the only singularity is the pole at $z=0$, so by the residue theorem
$$-\frac{4\pi}{R} I=2\pi \ln (z_0\bar{z_0})\Rightarrow I=-R\ln|\bar{x}|$$
For $|z_0|>R$ the proof was easy, and it turns out that the proof is equally simple for $|z_0|<R$, with a change of variables $z\to 1/z$ which shows the following string of equalities
$$\oint_{\Gamma_R(0)}\frac{\ln(z-z_0)}{iz}dz=\oint_{\Gamma_{1/R}(0)}\frac{\ln(1/z-z_0)}{iz}dz=-\oint_{\Gamma_{1/R}(0)}\frac{\ln(z)}{iz}dz+\oint_{\Gamma_{1/R}(0)}\frac{\ln(1-z_0z)}{iz}dz$$
The second integral is zero by residue theorem, since the branch cut is outside the circle of radius $1/R$ and the first integral can be shown by direct computation to be equal to
$$-\oint_{\Gamma_{1/R}(0)}\frac{\ln(z)}{iz}dz=-2\pi \ln R$$
Adding the $\bar{z_0}$ contribution, which is identical, yields: $I=-R\ln R$, which completes the proof.
Alternative complex analysis proof:
When $|z_0|<R$ one can use a keyhole contour with the keyhole shifted to $z_0$, and inner circle radius $\epsilon$ to calculate the integral on the outer circle using Cauchy's theorem, while circumventing the branch cut of the logarithm. One obtains the inequality
$$\oint_{\Gamma_R(0)}\frac{\ln(z-z_0)}{iz}dz+\left[\int_{z-z_0=xe^{i\pi}}-\int_{z-z_0=xe^{-i\pi}}-\oint_{\Gamma_\epsilon(z_0)}\right]dz\frac{\ln(z-z_0)}{iz}=0$$ In the first two line integrals connecting outer and inner circle we have $0<x<x_0+\sqrt{R^2-y_0^2}$. Substituting everything in we finally obtain $$\oint_{\Gamma_R(0)}\frac{\ln(z-z_0)}{iz}dz=2\pi \log(-z_0)-2\pi \int_0^{x_0+\sqrt{R^2-y_0^2}}\frac{dx}{z_0-x}$$
Adding the $\bar{z_0}$ contribution as well and rearranging we obtain the real valued formula
$$-\frac{4\pi}{R} I=4\pi\ln |\bar{x}|+2\pi\int_0^{x_0+\sqrt{R^2-y_0^2}}\frac{2(x-x_0)}{(x-x_0)^2+y_0^2}$$
The second integral is elementary: $$2\pi\int_0^{x_0+\sqrt{R^2-y_0^2}}\frac{2(x-x_0)}{(x-x_0)^2+y_0^2}=4\pi \ln R-4\pi \ln \bar{x}$$
which in turn shows that $I=-R\ln R$.
Calculus proof:
The integral may be written in the form
$$I=-\frac{R}{4\pi}\int_0^{2\pi}\ln(a+b\cos(\theta-\theta_0))d\theta$$
with $a=R^2+r^2, b=-2Rr, e^{i\theta_0}=\frac{x_0+iy_0}{\sqrt{x_0^2+y_0^2}}$.
The $\theta_0$ parameter can be killed by a shift of variables $\theta\to \theta+\theta_0$. Define the integral
$$J(a,b)=\int_0^{2\pi} \ln(a+b\cos\theta)d\theta$$
Use Feynman's trick to simplify the integral and instead solve the ODE:
$$\frac{\partial J}{\partial b}=\int_{0}^{2\pi}\frac{d\theta}{b}\left(1-\frac{a}{a+b\cos\theta}\right)~~, ~ J(a,0)=2\pi \ln a$$
Using the result $\int_{0}^{2\pi}(a+b\cos\theta)^{-1}d\theta=2\pi(a^2-b^2)^{-1/2}~, |a|>|b|$, it is left as an exercise to the reader to show that
$$J(a,b)=2\pi\ln\left(\frac{a+\sqrt{a^2-b^2}}{2}\right)$$
from which the desired result follows.