I was posed an interesting integral.

$$ \int_{-\infty}^{\infty}\left(\frac{\coth(x)}{x^{3}}-\frac{1}{3x^{2}}-\frac{1}{x^{4}}\right)dx .$$

The integral evaluates to $\displaystyle\frac{-2}{{\pi}^{2}}\zeta(3)$ (Mathematica confirms this), but for some reason Maple says it is divergent (gives infinity).

Is there a way to evaluate this using real analysis? I have seen a nice complex analysis solution resulting in the aforementioned zeta solution.

I looked at the Taylor series for $\dfrac{\coth(x)}{x^{3}}$ and it is :

$$ \frac{1}{x^{4}}+\frac{1}{3x^{2}}-\frac{1}{45}+\frac{2x^{2}}{945}-\frac{x^{4}}{4725}+\cdots $$

It would appear the first two terms have been subtracted from both sides and made part of the integrand. Yet, upon integrating the Taylor series, it appears divergent.

$$\int\left(\frac{\coth(x)}{x^{3}}-\frac{1}{3x^{2}}-\frac{1}{x^{4}}\right)=-\frac{x}{45} + \frac{2x^{3}}{2035}-\frac{x^{5}}{23625}+\frac{2x^{7}}{654885}+\cdots$$

Any ideas on a good method to evaluate this? Series? Double integrals? etc.

I just pointed out the Taylor series because I found it interesting. Often times, Taylor series will converge to a result, but here it does not appear to. I suppose I am overlooking something.

  • $\begingroup$ I still want to see your solution with zeta functions! $\endgroup$ – Eric Naslund Oct 30 '11 at 23:06
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    $\begingroup$ I wouldn't call what you've written a Taylor series for that function. I'd say it doesn't have a Taylor series around zero. $\endgroup$ – Gerry Myerson Oct 31 '11 at 0:22
  • $\begingroup$ Maybe Mapple calculates it by splitting into 2 or 3 integrals... $\endgroup$ – N. S. Oct 31 '11 at 2:13
  • $\begingroup$ I had also a similar problem to yours and this $\endgroup$ – user13838 Oct 31 '11 at 2:22
  • $\begingroup$ Maple 15 does not say the integral diverges, it merely fails to evaluate it. $\endgroup$ – GEdgar Oct 31 '11 at 13:22

First recall that $\coth(x)=i\cot(ix).$ Since $$\pi\cot(\pi x)=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^{2}-n^{2}},$$ we have that $$\pi\coth(\pi x)=\frac{1}{x}+\sum_{n=1}^{\infty}\frac{2x}{x^{2}+n^{2}}.$$ Rearrange and let $u=\pi x.$ Then $$\frac{\coth(u)}{u^{3}}=\frac{1}{u^{4}}+\sum_{n=1}^{\infty}\frac{2}{u^{2}\left(u^{2}+n^{2}\pi^{2}\right)}=\frac{1}{u^{4}}+2\sum_{n=1}^{\infty}\frac{1}{\pi^{2}n^{2}}\left(\frac{1}{u^{2}}+\frac{-1}{u^{2}+n^{2}\pi^{2}}\right).$$ Splitting up the sum this is $$\frac{1}{u^{4}}+\frac{2}{u^{2}\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\left(\frac{1}{u^{2}+n^{2}\pi^{2}}\right)=\frac{1}{u^{4}}+\frac{1}{3u^{2}}-\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\left(\frac{1}{u^{2}+n^{2}\pi^{2}}\right).$$ Here we used the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}.$ This last series converges absolutely, which explains why substracting off the $\frac{1}{u^{4}}$ and $\frac{1}{3u^{2}}$ is so important. (the Taylor series tells us this as well) This also means we can rearrange things freely. Thus our integral is $$\int_{-\infty}^{\infty}-\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\left(\frac{1}{u^{2}+n^{2}\pi^{2}}\right)du=\frac{-2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\int_{-\infty}^{\infty}\frac{1}{u^{2}+n^{2}\pi^{2}}du.$$ Substituting $u=n\pi x$ we see that $$\int_{-\infty}^{\infty}\frac{1}{u^{2}+n^{2}\pi^{2}}du=\frac{1}{n\pi}\int_{-\infty}^{\infty}\frac{1}{x^{2}+1}du=\frac{1}{n}$$ where we note that $\int_{-\infty}^\infty \frac{1}{1+x^2}dx=\pi$. This gives us the final answer of $$\int_{-\infty}^{\infty}\left(\frac{\coth(u)}{u^{3}}-\frac{1}{3u^{2}}-\frac{1}{u^{4}}\right)du=\frac{-2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{3}}=\frac{-2\zeta(3)}{\pi^{2}}.$$

I hope that helps,

Remark: Consider $\frac{\coth\left(u\right)}{u^{2m-1}}$ for integers $m$. Let $$\left[\frac{\coth\left(u\right)}{u^{2m-1}}\right]$$ refer to the same series, except that we cut off the terms in the Laurent expansion around zero where the power is negative. Then it is possible to prove a generalization that $$\int_{-\infty}^{\infty}\left[\frac{\coth\left(u\right)}{u^{2m-1}}\right]=(-1)^{m+1}\frac{2\zeta(2m-1)}{\pi^{2m-2}}.$$ This can be proven using the ideas above.

Zeta function: There is some connection between your integral and $\zeta^'(-2).$ In a similar way the integral of $\left[\frac{\cosh(x)}{x^{2m-1}}\right]$ is connected to $\zeta^'(-2m+2)$. I have some ideas, but cannot pin it down exactly the way I want. Just note in particular that $$\zeta^'(-2)=-4\frac{\zeta(3)}{\pi^2}.$$

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    $\begingroup$ Aw man. My first thought when reading the question was "I wish I could use that cotangent identity." Then I read the first line of your answer and slap myself. $\endgroup$ – anon Oct 30 '11 at 23:09
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    $\begingroup$ @anon: Haha, that comment made me laugh, I know that feeling! I am still trying to think of how to use something like $$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1}dx$$ and partial integration. I mean we have $$\zeta(3)=\frac{1}{2}\int_0^\infty \frac{x^{2}}{e^x-1}dx,$$ Things look like they are almost there, and we can write $$\coth(x)=1+\frac{2}{e^{2x}-1},$$ which looks good too. Just not quite there though... $\endgroup$ – Eric Naslund Oct 30 '11 at 23:53
  • $\begingroup$ Your solution is awesome. Thank you very much. $\endgroup$ – Cody Oct 31 '11 at 18:47

In Maple 15 I get:

int(coth(x)/x^3 - 1/3/x^2 - 1/x^4, x = -infinity .. infinity);

$\int _{-\infty }^{\infty }\!{\frac {\coth \left( x \right) }{{x}^{3}}} -1/3\,{x}^{-2}-{x}^{-4}{dx}$

So Maple doesn't know how to do this symbolically. Try it numerically:



Now can Maple guess what that number is?


$-2\,{\frac {\zeta \left( 3 \right) }{{\pi }^{2}}}$

  • $\begingroup$ I have Maple 10. An older version may be the reason. I entered the integral in just as you have and it gave infinity. $\endgroup$ – Cody Oct 31 '11 at 18:47
  • $\begingroup$ I was looking at this and got to thinking there should be a way to do this using the Psi function. Eric's method above kind of resembles the series for Psi. If we take the series for Psi $\psi(x)=-\gamma + \displaystyle \sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{x+k-1}\right)$ and differentiate twice, we arrive at $\psi^{2}(x)=\displaystyle -\sum_{k=1}^{\infty}\frac{2}{(k+x-1)^{3}}$. Now, letting x=1, gives $\displaystyle-\sum_{k=1}^{\infty}\frac{2}{k^{3}}=-2\zeta(3)$. There it is except for dividing by ${\pi}^{2}$. How can we relate this to the integral at hand?. $\endgroup$ – Cody Oct 31 '11 at 23:23

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