It is known that


for integer values of $n$ (this is also generalises, but that's not important for this question). I have also had a look at how to arrive at this expression, starting from the series representation of the Riemann Zeta function.

However, just out of interest, I've tried to see if I could manage to derive the series representation for the case that sparked my interest in the first place $(n=3)$ on my own - that is, working backwards from this formula to the series that initially defines the Riemann Zeta function. I did manage to arrive at a series representation, but it's not the correct series.

Here's what I did:

\begin{align} \int_0^\infty\frac{x^3}{e^x-1}dx&=\int_0^\infty\frac{x^3(e^x+1)}{e^{2x}-1}dx\\ &=\int_0^\infty\frac{x^3}{e^{2x}-1}dx+\int_0^\infty\frac{x^3e^x}{e^{2x}-1}dx\\ &=\frac{1}{16}\int_0^\infty\frac{x^3}{e^x-1}dx+\int_0^\infty\frac{x^3e^x}{e^{2x}-1}dx \end{align} Subtracting the first term on the RHS, we have \begin{align} \frac{15}{16}\int_0^\infty\frac{x^3}{e^x-1}dx=\int_0^\infty\frac{x^3e^x}{e^{2x}-1}dx \end{align} Multiplying across the fraction, we have \begin{align} \int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{16}{15}\int_0^\infty\frac{x^3e^x}{e^{2x}-1}dx\\ &=\frac{1}{15}\int_0^\infty\frac{x^3e^{\frac{x}{2}}}{e^x-1}dx \end{align}

Letting $z:=e^x$, then $x=\log z\implies dx=\frac{dz}{z}$.

\begin{align} \int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\int_1^\infty\frac{\log^3z\sqrt{z}}{z(z-1)}dz \end{align}

Letting $u:=\frac{1}{z}$, we have $z=\frac{1}{u}\implies dz=-\frac{du}{u^2}$, $u(1)=1$ and $u(z\rightarrow\infty)=0$. Thus

\begin{align} \int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\int_0^1\frac{\log^3\left(\frac{1}{u}\right)\sqrt{u}}{u^2}\frac{1}{1-\frac{1}{u}}du\\ &=-\frac{1}{15}\int_0^1\frac{\log^3u\sqrt{u}}{u}\frac{1}{u-1}du\\ &=\frac{1}{15}\int_0^1\frac{\log^3u}{\sqrt{u}}\frac{1}{1-u}du\\ &=\frac{1}{15}\int_0^1\frac{\log^3u}{\sqrt{u}}\sum_{k=0}^\infty u^k du\\ &=\frac{1}{15}\sum_{k=0}^\infty\int_0^1u^{k-\frac{1}{2}}\log^3u du \end{align}

Letting $t:=\log u$, then $u=e^t\implies du=e^tdt$, $t(u\rightarrow 0)=-\infty$ and $t(1)=0$.

\begin{align} \int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\sum_{k=0}^\infty\int_{-\infty}^0 t^3e^{\left(k+\frac{1}{2}\right)t}dt \end{align}

Letting $p:=-t$, then $dt=-dp$, $p(t\rightarrow -\infty)=\infty$ and $p(0)=0$.

\begin{align} \int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\sum_{k=0}^\infty\int_0^\infty p^3e^{-\left(k+\frac{1}{2}\right)p}dp \end{align}

Letting $s:=\left(k+\frac{1}{2}\right)p$, then $dp=\frac{ds}{k+\frac{1}{2}}$. The bounds of integration are unchanged, and thus

\begin{align} \int_0^\infty\frac{x^3}{e^x-1}dx&=\frac{1}{15}\sum_{k=0}^\infty\frac{1}{\left(k+\frac{1}{2}\right)^4}\int_0^\infty s^3e^{-s}ds\\ &=\frac{\Gamma(4)}{15}\sum_{k=0}^\infty\frac{1}{\left(k+\frac{1}{2}\right)^4}\\ &=\frac{2}{5}\sum_{k=0}^\infty\frac{1}{\left(k+\frac{1}{2}\right)^4} \end{align}

EDIT: Apparently, this is simply a convoluted way of deriving the correct answer, as the series evaluates to $\frac{\pi^4}{6}$, making the final expression equal to the known value of

$$\int_0^\infty\frac{x^3}{e^x-1}dx = 6\sum_{k=1}^\infty\frac{1}{k^4} = \frac{\pi^4}{15}$$

Many thanks to Simple Art for verifying this result!

  • $\begingroup$ The mistake is to not write $\zeta(s) \Gamma(s) = \sum_{n=1}^\infty n^{-s} \int_0^\infty y^{s-1} e^{-y}dy \underset{y = nx}= \sum_{n=1}^\infty\int_0^\infty x^{s-1} e^{-nx}dx$ $=\int_0^\infty x^{s-1} \sum_{n=1}^\infty e^{-nx}dx = \int_0^\infty \frac{x^{s-1}}{e^x-1}dx$. Also $\Gamma(s+1) = s \Gamma(s),\Gamma(1)= 1 \implies \Gamma(n+1) = n!$. $\endgroup$ – reuns Jan 4 '17 at 21:41
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    $\begingroup$ In line "Letting $z:=e^x$, then $x=\log z$" why you did not change lower limit. $\endgroup$ – Nosrati Jan 4 '17 at 21:52
  • $\begingroup$ You should probably write $\log^3 z\sqrt{z}$ as $(\log z)^3 \sqrt{z}$ to avoid confusion. $\endgroup$ – eyeballfrog Jan 4 '17 at 21:55
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    $\begingroup$ When you do the substitution $t=\log z$ you are missing the $e^t$ factor before the $dt$ $\endgroup$ – b00n heT Jan 4 '17 at 21:55
  • $\begingroup$ I've corrected both of the mistakes that you point out (thanks!). Unfortunately, the end result is still wrong. I understand that there are easier ways to derive this, but I'd still like to know where the error in my calculations is. $\endgroup$ – Tom Jan 4 '17 at 22:10

Your so called proof is very much unorthodox. A much shorter proof can be found in this post and it basically goes as follows:

$$\begin{align}\int_0^\infty\frac{x^k}{e^x-1}\ dx&=\int_0^\infty\frac{x^ke^{-x}}{1-e^{-x}}\ dx\\&=\int_0^\infty x^ke^{-x}\sum_{n=0}^\infty e^{-nx}\ dx\\&=\int_0^\infty\sum_{n=1}^\infty x^ke^{-nx}\ dx\\&=\int_0^\infty\sum_{n=1}^\infty\frac1{n^{s+1}}y^{k+1}e^{-y}\ dy\tag{$y=nx$}\\&=\sum_{n=1}^\infty\frac1{n^{s+1}}\int_0^\infty y^{k+1}e^{-y}\ dy\\&=\zeta(s+1)\Gamma(s+1)\end{align}$$

The last line comes by definition. And when $s\in\mathbb N$, $\Gamma(s+1)=s!$

Around the middle part, where you do substitutions, there appears to be missing factors (see comments).

In essence, I think the above is a much simpler derivation that uses the exact same geometric expansion, just in a far simpler way.

As you have edited your question, I give you the relief that your solution is correct. Particularly, scroll down quite a ways and you will find a nice surprise.

  • $\begingroup$ You're right, it is far simpler. But I'd still be interested in where my mistake lies. I'm currently going over the missing factors in the substitutions, maybe that will fix that. $\endgroup$ – Tom Jan 4 '17 at 22:05
  • $\begingroup$ @Tom Yes, I'm looking through and will notify you if anything I find is of error. $\endgroup$ – Simply Beautiful Art Jan 4 '17 at 22:21
  • $\begingroup$ @Tom Done, you have correctly done everything, see my last lines. $\endgroup$ – Simply Beautiful Art Jan 4 '17 at 23:45
  • $\begingroup$ Interesting! Thanks for checking everything - I hadn't thought it possible that two so different series evaluated to the same value, so I hadn't even checked the final form that I had found. Do you think that it is possible, in general, to find a connection between $\sum_{n=0}^\infty\frac{1}{n^k}$ and $\sum_{n=0}^\infty\frac{1}{(n+\frac{1}{2})^k}$ for fixed values of $k$? $\endgroup$ – Tom Jan 4 '17 at 23:54
  • $\begingroup$ @Tom Yes. This is why we have the Hurwitz zeta function. $\endgroup$ – Simply Beautiful Art Jan 4 '17 at 23:55

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