How to solve this indefinite real integral?

Question

I have to solve the following real integral $$ \int_{0}^{\infty} \frac{x^3} {{e^x -1}{}} dx$$ Using complex analysis I split it in 4 different integral, according to big Circle lemma the integral along the big circle is zero that heads to: $$ P.V. \int_{0}^{\infty} \frac{x^3} {{e^x -1}{}} dx = \int_{\gamma_\epsilon}^{} \frac{x^3} {{e^x -1}{}} $$ Using little circle lemma the second integral gives $$ 8\pi^4 $$ It's this result correct?

Answer 1

The result is incorrect. We have $$\dfrac{x^3}{e^x-1} = \dfrac{x^3e^{-x}}{1-e^{-x}} = \sum_{k=1}^{\infty} x^3e^{-kx}$$ We have $$\int_0^{\infty}x^3e^{-kx}dx = \dfrac1{k^4}\int_0^{\infty}t^3e^{-t}dt = \dfrac{\Gamma(4)}{k^4} = \dfrac{3!}{k^4} = \dfrac6{k^4}$$ Hence, the integral becomes $$\sum_{k=1}^{\infty} \int_0^{\infty} x^3 e^{-kx}dx = \sum_{k=1}^{\infty} \dfrac6{k^4} = 6 \zeta(4) = \dfrac{\pi^4}{15}$$