A calculus problem by Guru Prasaadh

Here is the very usual solution to this problem.

$\displaystyle I=\int_0^\infty \dfrac{x^3}{e^x-1} dx \\ \displaystyle \zeta (n) = \dfrac{1}{\Gamma(n)}\int_0^\infty \dfrac{x^{n-1}}{e^x - 1} dx \\ \displaystyle \zeta(4) = \dfrac{1}{\Gamma(4)}\int_0^\infty \dfrac{x^{4-1}}{e^x - 1} dx = \dfrac{1}{\Gamma(4)}\int_0^\infty \dfrac{x^{3}}{e^x - 1} dx\\ \displaystyle \int_0^\infty \dfrac{x^3}{e^x-1} dx = \zeta(4) \times \Gamma(4) = \dfrac{\pi^4}{90} \times 6 = \dfrac{\pi^4}{15}$