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In a certain list of exercises, contained the question

$$\int_{0}^{\infty}{\left(x^3\cdot(e^x-1)^{-1} \right)}dx$$

How do you solve?

benjamin_ee
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    Generally speaking, $$\int_0^\infty\frac{x^n}{e^x-1}dx=\Gamma(n+1)\cdot\zeta(n+1)$$ where $\Gamma(n+1)=n!$ , and $\displaystyle\zeta(n)=\sum_{k=1}^\infty\frac1{k^n}$ , which, for even values of n, is a rational multiple of $\pi^n$ . This is due to the fact that $\displaystyle\Gamma(n+1)=n!=\int_0^\infty\frac{x^n}{e^x}dx$. – Lucian Jan 06 '14 at 23:08

1 Answers1

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Let $I$ be your integral.

$$I=\int_0^{\infty}\frac{x^3}{e^x-1}dx=\int_0^{\infty}x^3\sum_{k=1}^{\infty}e^{-kx}dx.$$ Here, you can use $$\begin{align}\int_0^{\infty}x^3e^{-nx}dx&=\int_{0}^{\infty}x^3\left(-\frac{1}{n}e^{-nx}\right)'dx\\&=\left[-\frac{x^3}{ne^{nx}}\right]_{0}^{\infty}-\left(-\frac 3n\right)\int_{0}^{\infty}x^2\left(-\frac 1ne^{-nx}\right)'dx\\&=0+\frac 3n\left[-\frac{x^2}{ne^{nx}}\right]_{0}^{\infty}-\frac 3n\cdot\left(-\frac 2n\right)\int_{0}^{\infty}x\left(-\frac 1ne^{-nx}\right)'dx\\&=0+\frac{3!}{n^2}\left[-\frac{x}{ne^{nx}}\right]_{0}^{\infty}-\frac{3!}{n^2}\cdot\left(-\frac 1n\right)\left[-\frac{1}{ne^{nx}}\right]_{0}^{\infty}\\&=\frac{3!}{n^4}.\end{align}$$ Hence, $$I=3!\sum_{k=1}^{\infty}\frac{1}{k^4}=3!\times \frac{π^4}{90}=\frac{π^4}{15}.$$

P.S. You can see the proof for $\sum_{k=1}^{\infty}1/k^4=\zeta (4)=\pi^4/90$ here. Also, here.

Community
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mathlove
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  • Incrible, because $$\frac{1}{e-1}=\left(e^{-x}+e^{-2x}+\dots \right)?$$ – benjamin_ee Jan 06 '14 at 13:09
  • No. $\frac{1}{e^x-1}=e^{-x}+e^{-2x}+\cdot.$ If you know geometric progression, then you can sum each term at the right hand side. – mathlove Jan 06 '14 at 14:42
  • The right hand side of my comment above is $$\lim_{n\to \infty}\sum_{k=1}^{n}e^{-kx}=\lim_{n\to\infty}\frac{e^{-x}-e^{-(n+1)x}}{1-e^{-x}}=\frac{e^{-x}}{1-e^{-x}}=\frac{1}{e^x-1}.$$ This is equal to the left hand side of it. – mathlove Jan 06 '14 at 16:05