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

  • 1
    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 5:44
up vote 3 down vote accepted

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.

  • i got this: $$\frac{6}{(n1)^{4}}-\frac{6}{(n2)^{4}}$$ n1=neven number n2=Odd number – Ganesh Oct 29 '16 at 8:06
  • 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 5:54

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}

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.