Integration – Proving a Definite Integral Involving Gamma and Zeta Functions

Question

$$\int_0^\infty \frac{x^n\mathrm dx}{e^{nx}-1}= \frac{\Gamma(n)}{n^n}\zeta{(n+1)}$$

I have evaluated the integral by turning denominator into a geometric series and by switching integral and summation. However, my actual answer was in terms of factorials and I had assumed that $n\in\mathbb N$ while solving to make it into this form. Using an online calculator, I had verified it for $n=1,2,3$.

Questions:
Can we solve it by any other method ?
For what (if any) does it holds for some non-natural $n$ ?

Answer 1

We can use a slight extension of Ramanujan's master theorem. I will include a proof below.

Assume $f$ and $h$ have an expansion of the form, $$f(x)=\sum_{m=1}^\infty h(mx),\quad h(x)=\sum_{k=0}^\infty\frac{\varphi(k)}{k!}(-x)^k,$$ then the Mellin transform of $f(x)$ is given by, $$\int_0^{+\infty} x^{n-1}f(x)\ dx=\Gamma(n)\zeta(n)\varphi(-n).$$

In our case, $$f(x)=\frac{1}{e^{nx}-1}=\frac{1}{e^{nx}}\frac{1}{1-e^{-nx}}=\frac{1}{e^{nx}}\sum_{k=0}^\infty e^{-knx}=\sum_{k=1}^\infty e^{-knx}$$ where, $$h(kx)=e^{-knx}=\sum_{m=0}^\infty \frac{n^m}{m!}(-kx)^m$$ hence the Mellin transform, $$\int_0^{+\infty} \frac{x^{s-1}}{e^{nx}-1}\ dx=\Gamma(s)\zeta(s)n^{-s}$$ take $s=n+1$, $$\int_0^{+\infty} \frac{x^n}{e^{nx}-1}\ dx=\frac{\Gamma(n+1)}{n^{n+1}}\zeta(n+1)=\frac{\Gamma(n)}{n^n}\zeta(n+1).$$

Proof. By Ramanujan's master theorem, $$\int_0^{+\infty}x^{n-1} h(x)\ dx=\Gamma(n)\varphi(-n)$$ subbing $x\mapsto mx$, $$m^{n}\int_0^{+\infty} x^{n-1}h(mx)\ dx=\Gamma(n)\varphi(-n)$$ rearranging and summing over $m\in\mathbb{N}$, $$\int_0^{+\infty} x^{n-1}\sum_{m=1}^\infty h(mx)\ dx=\sum_{m=1}^\infty m^{-n}\Gamma(n)\varphi(-n)=\Gamma(n)\zeta(n)\varphi(-n)$$ hence, $$\int_0^{+\infty} x^{n-1}f(x)\ dx=\Gamma(n)\zeta(n)\varphi(-n).$$