How can this integral be convergent? According to ${\tt Mathematica}$, the following integral converges if
$\beta < 1$.
$$
\int_{0}^{1 - \beta}\mathrm{d}x_{1}
\int_{1 -x_{\large 1}}^{1 - \beta}\mathrm{d}x_{2}\,  \frac{x_{1}^{2} + x_{2}^{2}}{\left(1 - x_{1}\right)\left(1 - x_{2}\right)}
$$
How is this possible ?. For $x_{1} = 0$ the integration over
$x_{2}$ hits $1$ on the boundary, so the denominator vanishes and hence the whole expression should diverge.
How can this integral be convergent ?.
 A: We can prove this integral converges for $0 < \beta < 1$ without evaluation.
Write this as
$$\int_0^{1-\beta}\int_{1-x_1}^{1-\beta} \frac{x_1^2 + x_2^2}{(1-x_1)(1-x_2)}\, dx_2 \, dx_1\\ = \underbrace{\int_0^{\beta}\int_{1-x_1}^{1-\beta} \frac{x_1^2 + x_2^2}{(1-x_1)(1-x_2)}\, dx_2 \, dx_1}_{I_1}+ \underbrace{\int_\beta^{1-\beta}\int_{1-x_1}^{1-\beta} \frac{x_1^2 + x_2^2}{(1-x_1)(1-x_2)}\, dx_2 \, dx_1}_{I_2}$$
The integrand is continuous over the region of integration for $I_2$.  If there is a problem with convergence it will arise with the integral $I_1$.
When $0 \leqslant x_1 \leqslant \beta $, we have $1- \beta  \leqslant 1- x_1 \leqslant 1$ and, making the variable change $u = 1- x_2$, we get
$$I_1 = -\int_0^{\beta}\int_{1-\beta}^{1-x_1} \frac{x_1^2 + x_2^2}{(1-x_1)(1-x_2)}\, dx_2 \, dx_1 = \int_0^{\beta}\int_{x_1}^{\beta} \frac{x_1^2 + (1-u)^2}{(1-x_1)u}\, du \, dx_1$$
Introducing polar coordinates $(r,\theta)$ where $u = r \cos \theta$ and $x_1 = r \sin \theta$, the integral becomes
$$I_1 = \int_0^{\pi/4}\int_0^{\beta/\cos \theta}  \frac{r^2\sin^2 \theta + (1 - r\cos \theta)^2}{(1- r\sin \theta)r \cos \theta}\, r \, dr\, d\theta \\ =  \int_0^{\pi/4}\int_0^{\beta/\cos \theta}\frac{r^2\sin^2 \theta + (1 - r\cos \theta)^2}{(1- r\sin \theta)\cos \theta} \, dr\, d\theta $$
With $0 \leqslant r \leqslant \beta/\cos \theta$ and $0 \leqslant \theta \leqslant \pi/4$ the denominator satisfies (when $\beta < 1$)
$$(1- r\sin\theta)\cos \theta \geqslant \left(1 - \frac{\beta}{\cos \theta} \sin \theta\right) \cos  \theta \geqslant \frac{1 - \beta \tan \theta}{\sqrt{2}} \geqslant \frac{1- \beta}{\sqrt{2}} > 0,$$
and the integral $I_1$ is finite.
A: Changing a little notations (and trying to show the steps), considering
$$I=\int^{1-\beta}_{0}dx \int^{1-\beta}_{1-x}   \frac{x^2+y^2}{(1-x)(1-y)}\,dy$$
$$\int   \frac{x^2+y^2}{(1-x)(1-y)}\,dy=\frac{\left(x^2+1\right) \log (y-1)+\frac{1}{2} (y-1)^2+2 (y-1)}{x-1}$$
$$J(x)=\int^{1-\beta}_{1-x}   \frac{x^2+y^2}{(1-x)(1-y)}\,dy$$ $$J(x)=-\frac{-2 \left(x^2+1\right) \log (-\beta )+2 \left(x^2+1\right) \log (-x)+(x-\beta
   ) (\beta +x-4)}{2 (x-1)}$$ Integrating this last one with respect to $x$ gives
$$2 \int J(x)\,dx=-4 \text{Li}_2(x)+\log (1-x) \left(\beta ^2-4 \beta +4 \log (-\beta )-4 \log
   (-x)+3\right)-x (-(x+2) \log (-\beta )+(x+2) \log (-x)-5)$$
$$2\int^{1-\beta}_{0} J(x)\,dx= -\log (\beta -1) \left(\beta ^2-4 \beta +4 \log (\beta )+3\right)+\log (-\beta )
   \left(\beta ^2-4 \beta +4 \log (\beta )+3\right)+(\beta -1) ((\beta -3) \log
   (\beta )-5)-4 \text{Li}_2(1-\beta )$$ which does not exist (at least as a real if $\beta >1$.
Now, assuming $\beta < 1$, this reduces to
$$2I=\left(\beta ^2-4 \beta +4 \log (\beta )+3\right) \log \left(\frac{\beta }{1-\beta
   }\right)+(\beta -1) ((\beta -3) \log (\beta )-5)-4 \text{Li}_2(1-\beta )$$
