Solve the integral $\int_{(0,\infty)\times(0,\infty)}\frac{1}{\alpha}\left ( \frac{x}{x+y} \right )^{\alpha-1}e^{-\frac{x+y}{\alpha}}dm_2(x,y)$ Can I have a hint to solve this integral? 
$$
\int_{(0,\infty)\times(0,\infty)}\frac{1}{\alpha}\left ( \frac{x}{x+y} \right )^{\alpha-1}e^{-\frac{x+y}{\alpha}}dm_2(x,y)
$$
 A: By direct calculation, we see that
\begin{align}
\int^\infty_0 \int^\infty_0 \frac{1}{\alpha}\left(\frac{x}{x+y}\right)^{\alpha-1}e^{-(x+y)/\alpha} dydx =&\ \int^\infty_0 \int^\infty_0 \frac{1}{\alpha}\left(\frac{1}{1+y/x}\right)^{\alpha-1}e^{-x(1+y/x)/\alpha} dydx\\
=&\ \int^\infty_0 \frac{1}{\alpha}\left(\frac{1}{1+u}\right)^{\alpha-1}\int^\infty_0 e^{-x(1+u)/\alpha} x\ dx du\\
=&\ \int^\infty_0 \frac{1}{\alpha}\left(\frac{1}{1+u}\right)^{\alpha-1} \frac{\alpha^2}{(1+u)^2}\ du\\
=&\ \int^\infty_0  \frac{\alpha}{(1+u)^{\alpha+1}}\ du = 1.
\end{align} 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}\int_{0}^{\infty}{1 \over \alpha}
\pars{x \over x + y}^{\alpha - 1}\expo{-\pars{x + y}/\alpha}\dd y\,\dd x}} =
{1 \over \alpha}\int_{0}^{\infty}\int_{x}^{\infty}
\pars{x \over y}^{\alpha - 1}\expo{-y/\alpha}\dd y\,\dd x
\\[5mm] = &\
{1 \over \alpha}\int_{0}^{\infty}y^{1 - \alpha}
\expo{-y/\alpha}\int_{0}^{y}x^{\alpha - 1}\,\dd x\,\dd y =
{1 \over \alpha}\int_{0}^{\infty}y^{1 - \alpha}
\expo{-y/\alpha}\pars{y^{\alpha} \over \alpha}\dd y =
{1 \over \alpha^{2}}\int_{0}^{\infty}y\expo{-y/\alpha}\dd y
\\[5mm] = &\
\int_{0}^{\infty}y\expo{-y}\dd y = \bbx{1}
\end{align}
