I am trying to evaluate the following integral \begin{align} \int_{\mathbb{R}^2_{+}} \frac{ds_1 ds_2 s_2^{\epsilon} e^{- \frac{s_1 s_2 q^2}{s_1+s_2}}}{(s_1+s_2)^{1-\epsilon}} = \frac{\Gamma(1+2\epsilon)}{(q^2)^{1+2\epsilon}} \int_0^{1} \frac{dx}{x^{1+2\epsilon} (1-x)^{1+\epsilon}} \end{align} Here $\mathbb{R}^{2}_{+} = \{(t_1, t_2) | 0\leq t_1, t_2 \leq \infty \}$.
It seems the following two formulas might be useful \begin{align} &\frac{1}{A^{a_1} A^{a_2}} = \frac{\Gamma(a_1 +a_2)}{\Gamma(a_1) \Gamma(a_2)} \int_{0}^{\infty} \frac{x^{a_2-1}dx}{(A+xB)^{a_1+a_2}} \\ &\frac{1}{A(p)^n} = \frac{1}{\Gamma(n)} \int_0^{\infty} du \phantom{1} u^{n-1} e^{-u A(p)} \end{align} My first trial is given as follows : \begin{align} \int_{\mathbb{R}^2_{+}} \frac{ds_1 ds_2 s_2^{\epsilon} e^{- \frac{s_1 s_2 q^2}{s_1+s_2}}}{(s_1+s_2)^{1-\epsilon}} = \int_{\mathbb{R}^2_{+}} \frac{1}{(1+ \frac{s_1}{s_2})^{1-\epsilon}} s_2^{2\epsilon-1} e^{-\frac{1}{1+\frac{s_2}{s_1}} s_2 q^2} \end{align} Here If I used the second formula then still my $q^2$ factor does not match.
Let $t= \frac{s_1 s_2}{s_1+s_2}$ and $x = \frac{s_1}{s_1+s_2}$, and $ds_1 ds_2 \left| \frac{\partial (t,x)}{\partial (s_1, s_2)}\right| = dtdx = ds_1 ds_2 \frac{s_1 s_2}{(s_1+s_2)^3}$ \begin{align} \int_{\mathbb{R}^2_{+}} \frac{ds_1 ds_2 s_2^{\epsilon} e^{- \frac{s_1 s_2 q^2}{s_1+s_2}}}{(s_1+s_2)^{1-\epsilon}} &= \int ds_1 ds_2 \frac{s_1 s_2}{(s_1+s_2)^3} \frac{(s_1+s_2)^3}{s_1 s_2} \frac{s_2^{\epsilon}}{(s_1+s_2)^{1-\epsilon}} e^{-tq^2} t^{2\epsilon} t^{-2\epsilon} \\ & = \int dt dx \frac{s_1^{-1-2\epsilon} s_2^{-1-\epsilon}}{(s_1+s_2)^{-2-3\epsilon}} t^{2\epsilon} e^{-tq^2} = \frac{\Gamma(1+2\epsilon)}{(q^2)^{1+2\epsilon}} \int_0^{1} \frac{dx}{x^{1+2\epsilon} (1-x)^{1+\epsilon}} \end{align} In the process we used \begin{align} \frac{1}{(q^2)^{1+2\epsilon}} = \frac{1}{\Gamma(1+2\epsilon)} \int_0^{\infty} dt \phantom{1} t^{2\epsilon} e^{-t q^2} \end{align}