Feynman's trick VII

Integral involving the eta Dirichlet function

We repost here the solution of this integral previously posted on Twitter:
\[ \int_{0}^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx = \frac{\pi^2}{6}\]

Proof
\begin{align*} \int_{0}^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx =& 2\int_{1}^{\infty} \frac{\ln(w)}{w(1+w)}dw \quad (w \mapsto e^{\sqrt{x}})\\ =& -2 \int_{0}^{1} \frac{\ln(r)}{(1+r)}dr \quad (r \mapsto \frac{1}{w})\\ =& -2 \int_{0}^{1} \ln(r) \sum_{n=0}^{\infty} (-1)^n r^n dr \\ =& 2 \sum_{n=0}^{\infty}(-1)^{n+1} \int_{0}^{1} \ln(r) r^n dr \\ =& 2 \sum_{n=0}^{\infty}(-1)^{n+1} \int_{0}^{1} \frac{d}{dt}\Big|_{t=0+} r^t r^n dr \\ =& 2 \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty}(-1)^{n+1} \int_{0}^{1} r^{t+n} dr \\ =& 2 \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty}(-1)^{n+1} \frac{1}{(n+t+1)} \\ =& 2 \lim_{t \to 0+ }\sum_{n=0}^{\infty}(-1)^n \frac{1}{(n+t+1)^2} \\ =& 2 \sum_{n=0}^{\infty}(-1)^n \frac{1}{(n+1)^2} \\ =& 2 \eta(2) \quad (\textrm{ Dirichlet eta function})\\ =& 2 \left(\frac{\pi^2}{12}\right) = \frac{\pi^2}{6} \end{align*} If we do not know the value of $\eta(2)$ we can obtain it through the zeta function with this relation: \[\eta(v) = (1-\frac{1}{2^{v-1}})\zeta(v)\] Then, if $v=2$, \[\eta(2) = \frac{1}{2} \zeta(2) = \frac{\pi^2}{12} \] Hence
\[ \boxed{\int_{0}^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx = \frac{\pi^2}{6}}\]