Double exponential integral

Evaluate the integral:

$$\int_0^{\infty} \int_0^{\infty} \frac{{\rm d}x \; {\rm d}y}{\left(e^x+e^y\right)^2}$$

(Ovidiu Furdui)

Solution

We have successively:

\begin{align*}
\int_{0}^{\infty}\int_{0}^{\infty} \frac{{\rm d}(x, y)}{\left ( e^x+e^y \right )^2} &= \int_{0}^{\infty} \int_{0}^{\infty} \frac{{\rm d}(x, y)}{e^{2y}\left ( 1+e^{x-y} \right )^2}\\
 &= \int_{0}^{\infty} e^{-2y} \left ( \int_{0}^{\infty} \frac{{\rm d}x}{\left ( 1+e^{x-y} \right )^2} \right ) \, {\rm d}y\\
 &\!\!\!\!\!\!\overset{u=e^{x-y}}{=\! =\! =\! =\! =\!} \int_{0}^{\infty} e^{-2y} \left ( \int_{e^{-y}}^{\infty} \frac{{\rm d}u}{u \left ( u+1 \right )^2} \right )\, {\rm d}y  \\
 &= \int_{0}^{\infty} e^{-2y} \left [ \int_{e^{-y}}^{\infty} \left ( \frac{1}{u} - \frac{1}{u+1}- \frac{1}{\left ( u+1 \right )^2} \right ) \, {\rm d}u \right ] \, {\rm d}y \\
 &=\int_{0}^{\infty} e^{-2y} \left ( \frac{1}{e^y+1} + \log \left ( e^y+1 \right ) -1  \right ) \, {\rm d}y \\
 &=\int_{0}^{\infty} \frac{e^{-2y}}{e^y+1} \, {\rm d}y + \int_{0}^{\infty}e^{-2y} \log \left ( e^y+1 \right ) \, {\rm d}y -\int_{0}^{\infty}e^{-2y} \, {\rm d}y\\
 &=\log 2 -\frac{1}{2} \cancel{ +\frac{1}{2} - \frac{1}{2}}  \\
 &=\log 2 - \frac{1}{2}
\end{align*}

since  

\begin{align*}
\int_{0}^{\infty} \frac{e^{-2y}}{e^y+1} \, {\rm d}y &\overset{u=e^y}{=\! =\! =\! =\! =\!} \int_{1}^{\infty} \frac{{\rm d}u}{u^3 \left ( u+1 \right )} \\
 &= \int_{1}^{\infty} \left [ \frac{1}{u}+ \frac{1}{u^3} - \frac{1}{u^2} - \frac{1}{u+1} \right ] \, {\rm d}u\\
 &=\left [ \frac{1}{u} + \log u - \log (u+1) - \frac{1}{2u^2} \right ]_1^{\infty} \\
 &= \log 2 - \frac{1}{2}
\end{align*}

and 

\begin{align*}
\int_{0}^{\infty}e^{-2y} \log \left ( e^y+1 \right  )\, {\rm d}u &\overset{u=e^x}{=\! =\! =\! =\! =\!} \int_{1}^{\infty} \frac{\log \left ( u+1 \right )}{u^3} \, {\rm d}u \\
 &=\left [ - \frac{\log \left ( u+1 \right )}{2u^2} \right ]_1^{\infty} + \frac{1}{2}\int_{1}^{\infty} \frac{{\rm d}u}{u^2 \left ( u+1 \right )}  \\
 &=  \frac{\log 2}{2}+ \frac{1}{2} \int_{1}^{\infty} \left [ \frac{1}{u^2} + \frac{1}{u+1} - \frac{1}{u} \right ] \, {\rm d}u\\
 &= \frac{\log 2}{2}+ \frac{1}{2} \left ( 1 - \log 2 \right ) \\
 &=\frac{1}{2}
\end{align*}

as well as the (trivial) integral $\displaystyle{\int_{0}^{\infty} e^{-2y} \, {\rm d}y = \left [ -\frac{1}{2} e^{-2y} \right ]_0^{\infty} = \frac{1}{2}}$.