Integral involving $\coth (x)$: Maple and Mathematica disagree 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. 
 A: 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:

evalf(%);

-.2435876565
Now can Maple guess what that number is?

identify(%);

$-2\,{\frac {\zeta  \left( 3 \right) }{{\pi }^{2}}}$
A: 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 $$\zeta^'(-2)=-4\frac{\zeta(3)}{\pi^2}.$$
