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I have just learned about the residue theorem so when I encountered the integral $$ I = \int_{0}^{+\infty} \frac{x^3}{e^x-1}dx $$ I tried to evaluate it with this method, which failed but produced something interesting.

Define the complex logarithm on $\mathbb C \setminus [0,+\infty)$ and consider the function $$ f(z) = \frac{z^3 \log z}{e^z-1} $$ with the path $\gamma$ given by this drawing

gamma

Provided that both $\oint_{C_\epsilon} fdz$ and $ \oint_{C_R}fdz$ goes to zero as $R\to\infty$ and $\epsilon \to 0$ the value of the original integral is given by the residue theorem

$$ I = -\frac{1}{2\pi i}\oint_\gamma fdz = -\sum_{k} \text{Res}(f, z_k) $$

$f$ has simple poles for $e^z=1$, $z=2k\pi i$ where $k \in \mathbb Z$ so the residues are

\begin{align*} \text{Res}(f, z_k) &= 0 && k=0 \\ \text{Res}(f, z_k) &= z^3 \log{z} \big|_{z=2k\pi i} = (2k\pi)^3 \log (2k\pi i) && k\neq 0 \end{align*}

The sum of the residue can be written as $$ \sum_{k=1}^{+\infty} (a_k + a_{(-k)}) $$

where \begin{align*} a_k &= i(2k\pi)^3\left(\log(2k\pi) + i\frac{\pi}{2}\right) \\ a_{(-k)} &= i(2k\pi)^3\left(-\log(2k\pi) - i\frac{3\pi}{2}\right) \end{align*}

Cancelling the terms out $$ I = \sum_{k=1}^{+\infty} i(2\pi k)^3(-i\pi) = 8\pi^4 \sum_{k=1}^{+\infty} k^3 $$

which diverges. Clearly something is wrong: maybe the integral on the circumference is not zero? I'm not sure. However if I replace the infinite sum with $\zeta(-3)=\frac{1}{120}$ I actually get the correct result $I = \frac{\pi^4}{15}$. Why does this work?

P.S. I managed to evaluate it in a more familiar way without involving complex analysis.

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    $\begingroup$ Your integral is actually a representation of the Riemann Zeta function up to a factor involving the Gamma function. It is exactly $\zeta(4)\Gamma(4)$. By the Riemann functional equation, you can relate it to $\zeta(-3)$. $\endgroup$ May 19 '17 at 21:18
  • $\begingroup$ Your integral can be solved by means of gamma functions and infinite series (which is indeed Riemann zeta function, as commented by Raskolnikov). If you want to use residue, this type of integrals can be solve by using Bromwich contour but not like you said. $\endgroup$ May 20 '17 at 0:05
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Let $f_s(z) = \frac {\exp(-s\log(z))\log (z)}{\exp(z)-1}$ and $\gamma$ be a keyhole curve going twice along the positive real axis and making a small loop around $0$, picking the branch of $\log$ where $\log(i) = i\pi/2$ (and so $\log(-i) = 3i\pi/2$ and not $-i\pi/2$ because the branch cut is on $\Bbb R^+$)

Since $f_s(z)$ gets smaller exponentially when $\Re(z)$ gets large, $J(s) = \int_\gamma f_s(z) dz$ is a well-defined function, and is in fact an entire function in $s$.


If $\Re(s) < 0$, then $f_s(z) \to 0$ when $z$ gets closer to $0$, so $J(s)$ can be evaluated by computing the difference between the two horizontal segments, and so $J(s) = \int_0^\infty \frac {\exp(-s\log(z))\log (z) - \exp(-s(\log(z)+2i\pi))(\log (z)+2i\pi)}{\exp(z)-1} dz$.

And so $J(-3) = \int_0^\infty \frac { -2i\pi \exp(3\log(z))}{\exp(z)-1} dz = -2i\pi \int_0^\infty \frac {z^3}{\exp(z)-1} dz$.


On the other hand, when $\Re(s) > 1$, we can approximate $J(s)$ by truncating the segment and closing with a large circle.

If $R_n = (2n+1)\pi$, then $\exp(R_n\exp(i\theta))-1$ stays away from $0$ independantly of $n$, and so $f_s(R_n\exp(i\theta)) = \exp(-s\log R_n-is\theta)(\log R_n +i \theta) / (\exp(z) - 1) = O(n^{- \Re(s)})$.

And so, the integral of $f_s dz$ on the large circle is a $O(n^{1-\Re(s)})$ and converges to $0$, and therefore
$J(s) = \lim _{n \to \infty} \int_{\gamma_n} f_s(z) dz$ where $\gamma_n$ is the closed curve you described with a radius $R_n$.

Now we can use the residue theorem to get $J(s) = 2i\pi \sum_1^\infty \exp(-s\log(2ik\pi))\log(2ik\pi) + \exp(-s(\log(2ik\pi)+i\pi))(\log(2ik\pi)+i\pi) \\ = 2i\pi \exp(-s\log(2i\pi))(\zeta'(s) + \zeta(s)\log(2i\pi)) + \\ 2i\pi \exp(-s(\log(2i\pi)+i\pi))(\zeta'(s) + \zeta(s)(\log(2i\pi)+i\pi))$


Now, we use uniqueness of analytic continuation to deduce that this equality is valid for all $s \neq 1$, and so

$J(-3) =\\ 2i\pi \exp(3\log(2i\pi))(\zeta'(-3) + \zeta(-3)\log(2i\pi)) + \\ 2i\pi \exp(3(\log(2i\pi)+i\pi))(\zeta'(-3) + \zeta(-3)(\log(2i\pi)+i\pi)) \\ = 2i\pi (\exp(3\log(2i\pi))(\zeta'(-3) + \zeta(-3)\log(2i\pi) - \zeta'(-3) - \zeta(-3)(\log(2i\pi)+i\pi))) \\ = - 2i\pi (\exp(3\log(2i\pi))( \zeta(-3)i\pi) \\ = -2i\pi (2i\pi)^3 i\pi \zeta(-3) = (-2i\pi)(8\pi^4 \zeta(-3)) $

Finally, putting both results together we get $\int_0^\infty \frac {z^3}{\exp(z)-1} dz = 8\pi^4 \zeta(-3)$

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  • $\begingroup$ Iirc Riemann gave a proof of the functional equation using the same principle, except the integrand had no $\log(z)$ factor. But then when you try to evaluate it at an integer, since you travel along the real axis twice, it cancels with itself and it doesn't give you anything at first. To get information you have to differentiate the functional equation at that integer, and this computation here is the equivalent to that. so now I can finally shake off this uneasy feeling I was having. $\endgroup$
    – mercio
    May 20 '17 at 14:45
  • $\begingroup$ And $\zeta(-3)$ can be evaluated from $\zeta(s)\Gamma(s)-\sum_{k=0}^K \frac{B_k}{k!}\frac{1}{s+k} = \int_0^\infty x^{s-2} (\frac{x}{e^x-1}-\sum_{k=0}^K \frac{B_k}{k!}x^k 1_{x < 1})dx$ for $\Re(s) > -K$ $\endgroup$
    – reuns
    May 20 '17 at 17:41

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