Closed form for volume under $ f(x,y)=\exp\bigg(\frac{1}{\ln(x^2+y^2)}\bigg) $

Question

I would like to find a closed form for the volume underneath $f(x,y)$ with $0<f(x,y)<1$ and $-1<x,y<1$: $$ f(x,y)=\exp\bigg(\frac{1}{\ln(x^2+y^2)}\bigg)$$

I know how to derive the closed form for a similar looking integral from this answer Closed form for $\int_0^1 e^{\frac{1}{\ln(x)}}dx$?: $$ I:=\int_0^1 \exp\bigg(\frac{1}{\ln(x)}\bigg) ~dx =2K_1(2)$$ where $K_1$ is a modified bessel function of the second kind.

I can't seem to get the techniques found in the linked post to work for this function of two variables. Any help is appreciated.

Answer 1

This is a response to the comments to the original question. First, consider the switch to polar coordinates

$$x=r\cos(t)$$

$$y=r\sin(t)$$

Then if the integral outside the unit circle converges if and only if

$$\int_0^{2\pi}\int_1^2\exp\left(\frac{1}{\log(r^2)}\right)rdrdt=2\pi \int_1^2\exp\left(\frac{1}{2\log(r)}\right)rdr$$

However, this integral does not converge according to Mathematica. Therefore, the original integral does not converge and we are done