Evaluate $ \int_0^{\infty} \int_0^{\infty} \frac{1}{(1+x)(1+y)(x+y)}\,dy dx$

CW because some ideas were taken from other posts, which I have linked to.

I'll integrate first in $y$ and then reduce the problem to a known integral. Using partial fractions, we have$$ \frac{1}{(1+y)(x+y)} = \frac{1}{(x-1)}\left(\frac{1}{1+y}-\frac{1}{x+y}\right) $$Thus $$ \int_0^{\infty}\int_0^{\infty}\frac{1}{(1+x)(1+y)(x+y)} \,dydx $$ $$ =\int_0^{\infty}\frac{1}{x^2-1}\int_0^{\infty}\frac{1}{1+y}-\frac{1}{x+y} \,dydx $$ $$ =\int_0^{\infty}\frac{1}{x^2-1}\left(\left.\log\left|\frac{1+y}{x+y}\right|\right|_0^{\infty}\right)\,dx $$ $$ =\int_0^{\infty}\frac{1}{x^2-1}\left(-\log\left|\frac{1}{x}\right|\right)\,dx $$ $$ =\int_0^{\infty}\frac{\log|x|}{x^2-1}\,dx $$Break into $[0,1)$ and $(1,\infty)$ (the function has a removable singularity at $x=1$, is integrable near zero by the MVT and at infinity by direct comparison, so we're good) and enforce the substitution $z=1/x$. You'll see that they are equal; hence it suffices to integrate over $(0,1)$ and double it. This allows us to use the Maclaurin series of log, convergent on $(0,1)$: $$ \int_0^{\infty}\frac{\log|x|}{x^2-1}\,dx =2 \int_0^1 \frac{\log(x)}{(x+1)(x-1)}\,dx $$It's a nice exercise to work out that this last integral gives $\pi^2/8$ by reducing it to $\sum_{n\ge 1}(2n-1)^{-2}$. More generally, we have $\int _0^{\infty} \frac{\log(x)}{x^2+\alpha^2}\,dx = \frac{\pi\log \alpha}{2\alpha}$, and interpreting $\log(i) = \pi i/2$ gives us $\pi^2/4$, as above.

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}