TITLE: Integral involving $\coth (x)$: Maple and Mathematica disagree
QUESTION [16 upvotes]: 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.

REPLY [5 votes]: 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(%);

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