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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?

physnolimits
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  • How do you define the principal value of an integral $\int_0^{\infty}$? Isn't it defined only for integrals $\int_{- \infty}^{\infty}$? Anyway, the integrand function has a removable discontinuity at $0$. – Crostul Aug 21 '16 at 10:02
  • I don't think it is correct, since numerical integration gives $6.4939$, while $8\pi^4$ is much bigger. However, there is no nice antiderivative (it involves several polylogarithm functions). – wythagoras Aug 21 '16 at 10:06
  • In fact I guess that my result is much bigger than the correct one that is $$pi^4/15$$ but I'm looking for a way to be able to solve Problem like this using "just" complex analysis. P.V. value in general is define when you ave integral from -inf to inf. – physnolimits Aug 21 '16 at 10:29

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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 $$\sum_{k=1}^{\infty} \int_0^{\infty} x^3 e^{-kx}dx = \sum_{k=1}^{\infty} \dfrac6{k^4} = 6 \zeta(4) = \dfrac{\pi^4}{15}$$

Adhvaitha
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  • Thank you, but it's possible to solve with complex analysis? (Just to know) also because I'm not confident with Gamma and Zeta functions – physnolimits Aug 21 '16 at 10:23