INTEGRAL \int_0^\infty ln(1+e^{-x})/1+e^{-2x}dx

I came across this beautiful integral today in this twitter. Lets evaluate it


\int_0^\infty \frac{\ln(1+e^{-x})}{1+e^{-2x}}dx=\,\frac{7 \pi^2}{96}-\frac{\ln^2(2)}{8}


\begin{aligned} I&=\int_0^\infty \frac{\ln(1+e^{-x})}{1+e^{-2x}}dx\\ &=\int_0^1\frac{\ln(1+x)}{x(1+x^2)}dx \qquad (e^{-x} \mapsto x)\\ &=\underbrace{\int_0^1\frac{\ln(1+x)}{x}dx}_{J}-\underbrace{\int_0^1\frac{x\ln(1+x)}{(1+x^2)}dx}_{K} \end{aligned}

\begin{aligned} J&=\int_0^1\frac{\ln(1+x)}{x}dx\\ &=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\int_0^1x^{k-1}dx\\ &=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^2}\\ &=\frac{\pi^2}{12} \end{aligned}

\begin{aligned} K&=\int_0^1\frac{x\ln(1+x)}{(1+x^2)}dx\\ &=\int_0^1 \left(\int_0^1 \,\frac{x^2}{(1+x^2)(1+xy)}\,dy\right)dx\\ &=\int_0^1 \left(\int_0^1 \,\frac{x^2}{(1+x^2)(1+xy)}\,dx\right)dy\\ \end{aligned}

Lets focus in the inner integral

\begin{aligned} \int_0^1 \,\frac{x^2}{(1+x^2)(1+xy)}\,dx\\ &=\frac{1}{1+y^2}\int_0^1\left(\frac{yx-1}{1+x^2}+\frac{1}{1+yx} \right)dx\\ &=\frac{y}{1+y^2}\int_0^1 \frac{x}{1+x^2}dx-\frac{1}{1+y^2}\int_0^1 \frac{1}{1+x^2}dx+\frac{1}{1+y^2}\int_0^1\frac{1}{1+yx} dx\\ &=\frac{y}{2(1+y^2)}\ln(1+x^2)\Big|_0^1 -\frac{1}{1+y^2}\arctan(x)\Big|_0^1+\frac{1}{y(1+y^2)}\ln(1+yx)\Big|_0^1\\ &=\frac{\ln 2}{2}\frac{y}{1+y^2} -\frac{\pi}{4}\frac{1}{1+y^2}+\frac{\ln(1+y)}{y(1+y^2)}\\ \end{aligned}

Then

\begin{aligned} K&=\frac{\ln 2}{2}\int_0^1\frac{y}{1+y^2}dy -\frac{\pi}{4}\int_0^1\frac{1}{1+y^2}dy+\int_0^1\frac{\ln(1+y)}{y(1+y^2)} dy\\ &=\frac{\ln 2}{4}\ln(1+y^2)\Big|_0^1-\frac{\pi}{4}\arctan(y)\Big|_0^1+\int_0^1\frac{\ln(1+y)}{y}dy-\int_0^1 \frac{y \ln(1+y)}{1+y^2}dy\\ &=\frac{\ln^2 2}{4}-\frac{\pi^2}{16}+\frac{\pi^2}{12}-K\\ &=\frac{\ln^2 2}{8}+\frac{\pi^2}{96} \end{aligned}

Putting back the values of J and K we obtain

\begin{aligned} \int_0^\infty \frac{\ln(1+e^{-x})}{1+e^{-2x}}dx&=\frac{\pi^2}{12}-\frac{\pi^2}{96}-\frac{\ln^2 2}{8}\\ &=\frac{7\pi^2}{96}-\frac{\ln^2 2}{8} \qquad \blacksquare\\ \end{aligned}

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