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Evaluation of $$\int^{\infty}_{0}\frac{x^3}{e^x-1}dx$$

$\bf{My\; Try::}$ Let $\displaystyle I = \int^{\infty}_{0}\frac{x^3}{e^x-1}dx$

Now put $\displaystyle e^x=\frac{1}{t}\;,$ Then $\displaystyle e^xdx = -\frac{1}{t^2}dt$ and $x=-\ln t$ and changing limits, We get

$$I = \int^{0}_{1}\frac{\ln^3 t}{1-t}dt = -\int^{1}_{0}\frac{\ln^3 t}{1-t}dt=-\int^{1}_{0}\ln^3(t)\sum^{\infty}_{n=0}t^ndt=\sum^{\infty}_{n=0}\int^{1}_{0}\ln^3 (t)t^ndt$$

Now Using By parts, We get $$I = 0+3\sum^{\infty}_{n=0}\frac{1}{n+1}\int^{1}_{0}\ln^2 (t)\cdot t^ndt$$

Again using By parts, We get $$I = 0-6\sum^{\infty}_{n=0}\frac{1}{(n+1)^2}\int^{1}_{0}\ln(t)\cdot t^ndt$$

Againg using by parts, We get $$I = 6\sum^{\infty}_{n=0}\frac{1}{(n+1)^3}\int^{1}_{0}t^ndt = \sum^{\infty}_{n=0}\frac{6}{(n+1)^4}=6\zeta(4)$$

Now how can i solve it, Help required, Thanks

juantheron
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    For a proof of $\zeta(4) = \frac{\pi^4}{90}$, I recommend using $f(z) = \frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}$ so that $f''(1/2) = 6 \sum_{n=-\infty}^\infty \frac{1}{(1/2-n)^4} = 12\ 2^4 \sum_{n=1}^\infty \frac{1}{(2n-1)^4} = 12\ 2^4(1-2^{-4}) \zeta(4)$. On the other hand, $f'(z) = \frac{2\pi \cos(\pi z)}{\sin^3(\pi z)}$ and $f''(z) = \frac{-2\pi^2}{\sin^2(\pi z)} + \frac{6\pi^2 \cos(\pi z)}{\sin^4(\pi z)}$ and $f''(1/2) = \ldots$ – reuns Oct 30 '16 at 05:44
  • This is a Debye function, see https://en.wikipedia.org/wiki/Debye_function . Equation 27.1.3 in the "Handbook of mathematical functions" edited by Abramowitz and Stegung says $\int_0^\infty t^n dt/(e^t-1) = n!\zeta(n+1)$. – R. J. Mathar Nov 06 '23 at 11:42

2 Answers2

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Making the problem more general, consider $$I=\int^{\infty}_{0}\frac{x^n}{e^{ax}-1}dx\qquad (a>0)$$ Changing variable $ax=t$ leads to $$I=\frac 1 {a^{n+1}}\int^{\infty}_{0}\frac{t^n}{e^{t}-1}dx$$ Using what mathlove answered here, we then end with $$I=\frac{\zeta (n+1)\, \Gamma (n+1)}{a^{n+1}}$$ and nice expressions when $n$ is odd.

Community
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Claude Leibovici
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  • i got this: $$\frac{6}{(n1)^{4}}-\frac{6}{(n2)^{4}}$$ n1=neven number n2=Odd number – Ganesh Oct 29 '16 at 08:06
  • I would mention $f(z) = \frac{\pi^2}{\sin^2(\pi z)} = \sum_{n=-\infty}^\infty \frac{1}{(z-n)^2}\implies \zeta(2k) = \frac{f^{(2k-2)}(1/2)}{(2k)!(2^{2k}-1)} $ – reuns Oct 30 '16 at 05:54
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Expanding Claude Leibovici's solution, \begin{align} \int\limits_{0}^{\infty} \frac{x^{n}}{\mathrm{e}^{ax}-1} dx &= \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-ax} x^{n}}{1-\mathrm{e}^{-ax}} dx \\ &= \int\limits_{0}^{\infty} \mathrm{e}^{-ax} x^{n} \sum\limits_{k = 0}^{\infty} \mathrm{e}^{-kax} dx \\ &= \sum\limits_{k = 0}^{\infty} \int\limits_{0}^{\infty} x^{n} \mathrm{e}^{-(a+ak)x} dx \end{align}

To evaluate the integral, let $(a+ak)x = z$ \begin{equation} \int\limits_{0}^{\infty} x^{n} \mathrm{e}^{-(a+ak)x} dx = \frac{1}{a^{n+1}} \frac{1}{(1+k)^{n+1}} \int\limits_{0}^{\infty} z^{n} \mathrm{e}^{-z} dz = \frac{1}{a^{n+1}} \frac{1}{(1+k)^{n+1}} \Gamma(n+1) \end{equation}

Now we have \begin{equation} \int\limits_{0}^{\infty} \frac{x^{n}}{\mathrm{e}^{ax}-1} dx = \frac{1}{a^{n+1}} \Gamma(n+1) \sum\limits_{k = 1}^{\infty} \frac{1}{k^{n+1}} = \frac{1}{a^{n+1}} \Gamma(n+1) \zeta(n+1) \end{equation}

The original problem is \begin{equation} \int\limits_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} dx = \Gamma(4) \zeta(4) = 6\frac{\pi^{4}}{90} = \frac{\pi^{4}}{15} \end{equation}

poweierstrass
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