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Aaron Meyerowitz
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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 $0.$

Aaron Meyerowitz
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