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 ?.

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
$$2I=\left(\beta ^2-4 \beta +4 \log (\beta )+3\right) \log \left(\frac{\beta }{1-\beta
   }\right)+(\beta -1) ((\beta -3) \log (\beta )-5)-4 \text{Li}_2(1-\beta )$$
