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