Need help finding solution to $ \int_0^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx$

Question

Recently I came across this integral:

$$ \int_0^{\infty} \frac{1}{1+e^{\sqrt{x}}}\mathrm dx$$

I know that its solution is $\frac{\pi^2}{6}$, through numerical approaches but I'm not entirely sure how you would go about solving the integral analytically.

Initially I tried the trick of adding the exponential term and taking it away from the numerator as this works in the case where there is no square root in the exponent, however I don't believe this to be a fruitful approach. Thanks for any help/advice any of you can offer.

I don't have paper near me, but I would try $x = y^2$ and see where that goes. Perhaps there is a series hiding too. — Sean Roberson, Dec 23 '24 at 15:18
using $\sqrt u \mapsto u$, $I=\int \frac{2u e^{-u}}{1+e^{-u}}du= -2\sum_1\int ue^{-nu} du =-2\sum_1 \frac{(-1)^n}{n^2}=\frac{\pi^2}6$ — Sine of the Time, Dec 23 '24 at 15:24
Make the change of variable $x = t^2$ and the obtained result will involve Dirichlet eta function (see https://en.wikipedia.org/wiki/Dirichlet_eta_function). — Abezhiko, Dec 23 '24 at 15:36

Quanto · Answer 1 · 2024-12-26T00:10:52.333

6

Substitute $t=e^{-\sqrt x}$ $$ \int_0^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx=-2\int_0^1 \frac{\ln t}{1+t}dt =-2(-\frac{\pi^2}{12})=\frac{\pi^2}6 $$ where $\int_0^1 \frac{\ln t}{1+t}dt\overset{ibp}=- \int_0^1 \frac{\ln (1+t)}{t}dt = -\frac{\pi^2}{12} $.

edited Dec 26 '24 at 00:10

answered Dec 24 '24 at 16:35

Quanto

120,125

score 3 · Answer 2 · answered Dec 23 '24 at 15:23

The integral equals $$\sum_{n=1}^\infty(-1)^{n-1}\int_0^\infty e^{-n\sqrt x}dx.$$ Substitute $y=n^2x$ which gives $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\int_0^\infty 2ye^{-y}dy.$$ Now combine Alternating Reciprocal of Squares with $$\int_0^\infty ye^{-y}dy=\Gamma(2)=1.$$

Need help finding solution to $ \int_0^{\infty} \frac{1}{1+e^{\sqrt{x}}}dx$

2 Answers2