I stumbled over the following problem related to the Coulomb kernel that I hoped some of you might understand.
I am interested in the function, that is given by the cartesian integral:
$$f(x',y')=\int_{-R}^R \mathrm{d}x\int_{-\infty}^\infty \mathrm{d}y \frac{\mathrm{sgn}(y)}{\sqrt{x^2+y^2}}\frac{1}{\sqrt{(x'-x)^2+(y'-y)^2}},\tag{1}$$ for large $x'^2+y'^2\gg R^2$. Numerical evaluation shows a smooth function with a regular maximum when crossing the $y'$-axis.
Now I decided to deal with this integral using polar coordinates, where we might read:
$ f(r',\phi')=\int_0^\infty \mathrm{d}r \int_{-\arcsin(R/r)}^{\arcsin(R/r)}\mathrm{d}\phi \left(\frac{1}{\sqrt{r'^2+r^2-2r r' \cos(\pi/2-\phi-\phi')}}-\frac{1}{\sqrt{r'^2+r^2-2r r' \cos(\pi/2+\phi+\phi')}}\right). \tag{2} $ For said limit, this can be appoximated by ignoring the $\phi$ dependence in the integrand and using $\arcsin(R/r)\approx R/r$, which gives the analytic form $f(r',\phi')=\frac{2R}{r'}\log\left(\frac{1+\sin\phi'}{1-\sin\phi'}\right)$, which is supported by numerical evaluation of $(2)$. However, this expression as well as the second integral appear to be divergent along $\phi=\pm\pi/2$ which is completely regular in the first one.
Has anyone spotted the error or seen a similar behavior? I am happy for anyone that can point me towards the problem here.
Mathematically, there shouldn't be a singularity since both poles (at $0$ and $(x',y')$) are of order one and the integration is $2$d. The logarithm arises from $r\sim rr$ and I do not see how one should deal with it. Any ideas anyone?
Thanks a lot in advance.
Regards, Michael