This question is in reference to the earlier question asked here.
$$\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} e^{\frac{-1}{1-(x^2+y^2)}}\log\Big(\log\big(\frac{2}{\sqrt{(1-x)^2+(1-y)^2}}\big)\Big)\,dy\,dx.$$ or
$$\int_{0}^{2\pi} \int_{0}^{1} e^{\frac{-1}{1-r^2}}\log\Big(\log\big(\frac{2}{\sqrt{r^2-2r\cos(\theta)-2r\sin(\theta)+2}}\big)\Big)\,r\,dr\,d\theta.$$
The suggestions in the comments are:
To find the numerical approximation.
To convert it into a path integral.
How should one go about these?
Thanks!
What you've asked here are two different questions, which makes this difficult. I will attempt to give some information about one of these, finding a numerical approximation.
The question of numerically approximating $\int_\Omega f(x) dx$ for $x \in \Omega \subset \mathbb{R}^n$ is a fairly well studied thing. In fact, the very definition of such an integral comes from an approximation $-$ the Riemann sum. We simply divide $\Omega$ into $k$ pieces $\Omega_m$ for $m\in \{1,\dots ,k\}$. Then, for each piece, we choose a point $x_m \in \Omega_m$ as a "representative" from that section, and evaluate the function at that point to get $f(x_m)$. Instead of integrating the whole function at every point, we approximate it by a bunch of constant functions on each piece, so we have $\int_\Omega f(x) dx \approx \sum\limits_{m=1}^k \int_{\Omega_m} f(x_m) dx$. Then, the integral of a constant function is "defined" to be that constant times the measure (length, area, volume, etc.) of the piece on which they are defined, $\int_{\Omega_m} f(x_m)dx = f(x_m)\mu(\Omega_m)$. Thus, we have an approximation for our integral: $$ \int_\Omega f(x) dx \approx \sum\limits_{m=1}^k f(x_m)\mu(\Omega_m) $$
If we take your second form of the integral, then we have $\Omega = \{0,2\pi\}\times\{0,1\} \in \mathbb{R}^2$ is just a rectangle in the plane. Then, we can choose our partition to be rectangles as well (ignore what $r$ and $\theta$ mean to you in terms of polar coordinates, they're just like $x$ and $y$ for this). One might choose to use $p$ even slices along the $r$ axis and $q$ even slices along the $\theta$ axis, so that each of the $\Omega_m$ are $\frac{1}{p}$-by-$\frac{2\pi}{q}$ rectangles, each with an area of $\mu(\Omega_m) = \frac{2\pi}{pq}$, for $m \in \{1,\dots,k\}$ and $k=pq$. You can choose any point in these rectangles, but for the sake of example we'll take the "top-right" point of each one, so $x_m = (r_i,\theta_j) = \left(\frac{i}{p},\frac{2\pi j}{q}\right)$ for $i \in \{1,\dots,p\}$ and $j \in \{ 1,\dots, q \}$. Then, the approximation of the integral is
$$\int_0^{2\pi}\int_0^1 f(r,\theta) drd\theta = \int_\Omega f(x) dx \approx \sum\limits_{m=1}^k f(x_m)\mu(\Omega_m) = \frac{2\pi}{pq}\sum\limits_{i=1}^p \sum\limits_{j=1}^q f\left(\frac{i}{p},\frac{2\pi j}{q}\right) $$
where $f(r,\theta)$ is your desired integrand, and $p,q$ are chosen to get your desired accuracy with respect to computational time.
This is far from the best approximation, and if you want to learn some better ones, then you should look into numerical integration or "quadrature" as it's called. However, this will "do the job" if you aren't too concerned about being both highly accurate and having a quick computation.