TITLE: Evaluate $ \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x)(1+y)(x+y)}\,dy dx$
QUESTION [5 upvotes]: Evaluate $ \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x)(1+y)(x+y)}\,dy dx$

We can calculate $\int \frac{1}{(1+x)(x+y)} \,dx= \frac{\ln|x+1| - \ln|x+y|}{y-1}$ but that doesn't make it quite easier. Is there a trick to compute this easier?

REPLY [4 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
{\cal J} &\equiv
\bbox[5px,#ffd]{\int_{0}^{\infty}\int_{0}^{\infty}
{\dd x\,\dd y \over
\pars{1 + x}\pars{1 + y}\pars{x + y}}}:\ {\Large ?}.
\end{align}
With the variable changes $\ds{x \equiv 1/a - 1}$ and
$\ds{y \equiv 1/b - 1}$ the above integral becomes
\begin{align}
{\cal J} & = \int_{0}^{1}\int_{0}^{1}
{\dd a\,\dd b \over a + b - 2ab} =
\int_{0}^{1}\ln\pars{1 - b \over b}
{\dd b \over 1 - 2b}
\end{align}
With $\ds{\pars{~1 - 2b = t \implies
b = {1 - t \over 2}~}}$:
\begin{align}
{\cal J} & = \underbrace{\int_{0}^{-1}{\ln\pars{1 - t} \over t}\,\dd t}_{\ds{\pi^{2} \over 12}}\ -\
\underbrace{\int_{0}^{1}{\ln\pars{1 - t} \over t}\,\dd t}_{\ds{-\,{\pi^{2} \over 6}}}
\\[5mm] & = \bbx{\large{\pi^{2} \over 4}} \\&&
\end{align}