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Anyone with any ideas on this one? Mathematica and Matlab can't do it.

$\int_0^\infty \frac{\sin(a y) \coth(y)}{(1+9y^2)^2}dy$

Here coth is the hyperbolic cotangent, and a is a positive parameter which will figure in the answer. It's giving me fits.

Greg

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    $\begingroup$ What kind of answer do you want, and what do you want to do with it? @Aaron's answer is correct, but since Ci and SSi are themselves integrals, in some sense this is more complicated than the question... $\endgroup$
    – Igor Rivin
    Commented Jun 22, 2011 at 14:43

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Maple says "undefined". But if you replace $\coth(y)$ with the equivalent $\frac{e^y+e^{-y}}{e^y-e^{-y}}$ then an answer is produced in terms of the cosine integral Ci and the shifted sign integral Ssi

$$\frac{\alpha}{18}\cosh (\frac{\alpha}3) {\it Ci}( \frac{\alpha i}3) - \frac{\alpha \pi i}{36}\cosh (\frac{\alpha}3) -\frac16\,\sinh (\frac{\alpha}3) {\it Ci} (\frac{\alpha i}3) +\frac{\pi i}{12}sinh(\frac{\alpha}3) +$$ $$\frac{\alpha i}{18}\sinh (\frac{\alpha}3) {\it Ssi} \left(\frac{\alpha i}{3}\right) +\frac{\alpha \pi i}{36}\sinh (\frac{\alpha}3) -\frac{i}6 \cosh (\frac{\alpha}3) {\it Ssi} (\frac{\alpha i}3) -\frac{\pi i}{12}\cosh ( \frac{\alpha}3) $$

I can't say how helpful this is, however, for actual values of the parameter $\alpha,$ the imaginary portion is $0.$

later For some reason I can now only get Maple to return the integral unevaluated even though this result is correct. The integrand goes very rapidly to zero so it is possible to get a very accurate numerical integral and this agrees. Although it is indeed uglier, it has the advantage that there is some routine to evaluate the answer.

Here is an alternate form which does not invoke complex numbers.

$$\frac{\alpha}{18}\left(\cosh\frac{\alpha}{3}{\it Chi}\frac{\alpha}{3}-\sinh\frac{\alpha}{3}{\it Shi}\frac{\alpha}{3}\right)+\frac{1}{6}\left(\cosh\frac{\alpha}{3}{\it Shi}\frac{\alpha}{3}-\sinh\frac{\alpha}{3}{\it Chi}\frac{\alpha}{3}\right)$$

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  • $\begingroup$ Ye gods. How strange that the change you made gave the answer, Mathematica still can't do it, even with the change. I had reasons for expecting $e^{\alpha/3}$ to appear in the answer, so no doubt what you give is correct, but what a mess. Probably this reduces somehow, no? Thank you very much, I think this is progress in any event. $\endgroup$
    – Greg Markowsky
    Commented Jun 23, 2011 at 8:21

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