I must find the following definite integral $$\phi(x,y) = \int_{0}^{\alpha} f(\theta)\frac{-1}{2\pi} \ln \sqrt{(x-\rho\cos\theta)^2+(y-\rho\sin\theta)^2} \rho d\theta$$ for a given $f(\theta)$ which is assumed to be a polynomial (for simplicity, we can start with $f=constant$).
It is a fact that the function $\phi(x,y)$ is potential and harmonic in all infinite space. Indeed, it is the solution of potential problem using the method of fundamental solution. Therefore, conformal mapping method can be used theoretically (?)
The integral is over a circular arc with the center of origin, radius $\rho$ and angle $0<\theta<\alpha$. By imposing a logarithmic conformal map $$u+iv = \ln(x+iy)$$ $$u = \ln\sqrt{x^2+y^2}$$ $$v = \arctan(y,x)$$ where $u+iv$ is the computational domain, the arc is mapped to a straight vertical line at $u=\ln\rho$ and $0<y<\alpha$ in the $u+iv$ domain. The potential in $u-v$ plane is obtained via $$\psi(u,v) = \int_{0}^{\alpha} f(y)\frac{-1}{2\pi} \ln \sqrt{(u-\ln\rho)^2+(v-v')^2} dv'$$
Now using the fact that conformal map preserves the harmonic property, we should have $$\phi(x,y) = \psi(u(x,y),v(x,y))$$
I see no error in these reasonings, however, I observe errors in numerical examples. I suspect the error stems from the map I use and also what happens to $f$ under the map.