How to evaluate $\int_0^\infty \frac{x^5(e^{3x}-e^x)}{(e^x-1)^4}\,\mathrm{d}x$

Question

Today the integral that got me triggered is $$\int_0^\infty \frac{x^5(e^{3x}-e^x)}{(e^x-1)^4}\,\mathrm{d}x$$ and it seems to be linked to Euler $\Gamma(n)$ function since we can get close to something similar to its integral definition but then the integrand seems to be quite difficult to solve… Have you any idea? (The answer is $5!\zeta(4)$)

Answer 1

Observe that $e^{3x}-e^x=e^x(e^{2x}-1)=e^x(e^x-1)(e^x+1)$ an then

$$\int_0^\infty \frac{x^5(e^{3x}-e^x)}{(e^x-1)^4}\,\mathrm{d}x=\int_0^\infty \frac{x^5 e^x(e^{x}+1)}{(e^x-1)^3}\,\mathrm{d}x$$

Now consider the following integral representation of Zeta function Zeta function, DLMF

$$\zeta\left(s\right)=\frac{1}{\Gamma\left(s+1\right)}\int_{0}^{\infty}\frac{e^{% x}x^{s}}{(e^{x}-1)^{2}}\mathrm{d}x,$$

and integrating by parts you obtain

$$\zeta\left(s\right)=\frac{1}{\Gamma\left(s+1\right)}\int_{0}^{\infty}\frac{e^{% x}x^{s}}{(e^{x}-1)^{2}}\mathrm{d}x=\frac{1}{\Gamma\left(s+2\right)}\int_0^\infty \frac{x^{s+1} e^x(e^{x}+1)}{(e^x-1)^3}\,\mathrm{d}x$$

Finally, later integral is your desired integral ($s=4$) and then

$$\int_0^\infty \frac{x^5(e^{3x}-e^x)}{(e^x-1)^4}\,\mathrm{d}x=5!\,\zeta\left(4\right)=\frac{4}{3}\pi^4$$