Find $\int_{0}^{\infty}\frac{x^n}{e^{\pi x}+1}dx$.

Question

$$\int_{0}^{\infty}\frac{x^n}{e^{\pi x}+1}dx$$

I have closed form $${\pi }^{-n-1}\Gamma \left( n+1 \right) \left(1- {2}^{-n} \right) \zeta \left( n+1 \right)$$ but I dont know how to solve the integral step by step

Answer 1

Use $x\mapsto x\pi^{-1}$, to get $$\frac{1}{\pi^{n+1}}\int_{0}^{\infty}\frac{x^n}{e^{ x}+1}dx$$

Then use that $$ \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{{e^x} + 1}}dx} = \left( {1 - {2^{1 - s}}} \right)\Gamma \left( s \right)\zeta \left( s \right) =\eta(s)\Gamma(s) $$

For the proof, expand $$\frac{1}{{{e^x} + 1}}=\frac{e^{-x}}{1+e^{-x}}$$ on $x>0$ using the usual geometric series and integrate term-wise. That is:

$$\frac{{{e^{ - x}}}}{{1 + {e^{ - x}}}} = \sum\limits_{n \geqslant 0} {{{\left( { - 1} \right)}^n}{e^{ - \left( {n + 1} \right)x}}} $$

Whence $$\int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{1 + {e^x}}}dx} = \sum\limits_{n \geqslant 0} {{{\left( { - 1} \right)}^n}\int\limits_0^\infty {{e^{ - \left( {n + 1} \right)x}}{x^{s - 1}}dx} } $$and by $x\mapsto \dfrac{x}{n+1}$, $$\begin{align} \int\limits_0^\infty {\frac{{{x^{s - 1}}}}{{1 + {e^x}}}dx} &= \sum\limits_{n \geqslant 0} {{{\left( { - 1} \right)}^n}\frac{1}{{{{\left( {n + 1} \right)}^s}}}\int\limits_0^\infty {{e^{ - x}}{x^{s - 1}}dx} } \cr &= \Gamma \left( s \right)\sum\limits_{n \geqslant 0} {{{\left( { - 1} \right)}^n}\frac{1}{{{{\left( {n + 1} \right)}^s}}}} \cr &= \Gamma \left( s \right)\sum\limits_{n \geqslant 1} {{{\left( { - 1} \right)}^{n - 1}}\frac{1}{{{n^s}}}} \cr &= \Gamma \left( s \right)\eta \left( s \right) \end{align} $$

Answer 2

Given integral, $$I=\int_{0}^{\infty}\frac{x^n}{e^{\pi x}+1}dx\\ =\int_{0}^{\infty}\frac{x^ne^{-\pi x}}{e^{-\pi x}+1}dx\\ =\int_{0}^{\infty}x^ne^{-\pi x}\sum_{k=0}^{\infty}(-1)^ke^{-k\pi x}dx$$ Now since (by using the definitions of $\Gamma$ function and $\zeta$ function)$$\sum_{k=0}^{\infty}\int_{0}^{\infty}x^ne^{-(k+1)\pi x}dx=\sum_{k=0}^{\infty}\int_{0}^{\infty}x^ne^{-(k+1)\pi x}dx=\pi^{-n-1}\Gamma(n+1)\zeta(n+1)<\infty$$for $n\ge 1$, by Fubini's theorem, we can interchange the summation and integration in the integral to get, $$I=\sum_{k=0}^{\infty}(-1)^k\int_{0}^{\infty}x^ne^{-(k+1)\pi x}dx=\pi^{-n-1}\Gamma(n+1)\sum_{m=1}^{\infty}\frac{(-1)^m}{m^{n+1}}$$ The sum is recognized as $$(1-2^{-n})\zeta(n+1)$$ and hence $$I=\pi^{-n-1}\Gamma(n+1)(1-2^{-n})\zeta(n+1)$$

Answer 3

This also can be related to the generalized poly-logarithm function by

$$\tag{1}\int^{\infty}_0 \frac{t^s}{e^{t-r}+1}\, dt=-\Gamma(1+s)\operatorname{Li}_{1+s}(-e^{r})$$

For $r=0$ and by

$$\operatorname{Li}_{s}(-1)=\sum_{n\geq 1}\frac{(-1)^n}{n^{s}}=-\mu(s)$$

we get

$$\tag{2}\int^{\infty}_0 \frac{t^{s-1}}{e^{t}+1}\, dt=\Gamma(s)\mu(s)$$

Also by $r=\pi i$ we have

$$\tag{3}\int^{\infty}_0 \frac{t^{s-1}}{e^{t}-1}\, dt=\Gamma(s)\zeta(s)$$