I'm interested in evaluating the integral $$ \int_{a}^\infty e^{-x\cosh\alpha}\,K_{\nu}(x\sinh\alpha)\,\frac{dx}{x}, $$ where $a>0$ and $\nu$ is purely imaginary. Here $K$ denotes the MacDonald function (modified Bessel function of the second kind).

This integral is quite close integral 6.628.5 from Gradsheyn & Ryzhik (7th edition) $$ \int_0^\infty e^{-x\cosh\alpha}\,K_{\nu}(x\sinh\alpha)\,x^{\mu}\,dx=\frac{\sin(\mu\pi)}{\sin((\nu+\mu)\pi)}\,\Gamma(\mu-\nu+1)\,Q_{\mu}^{\nu}(\cosh\alpha), $$ which holds provided $\Re(\mu+1)>|\Re(\nu)|$. Here $Q$ denotes the associated Legendre function of the second kind. G&R took the integral from Watson's 1922 "Theory of Bessel Functions". When $\mu=-1$ and $\Re(\nu)=0$, the integral diverges because of the behavior of the integrand around 0. However, the lower limit of integration in the integral I'm after is a positive number, and the integral is convergent. Any help with finding it would be appreciated.

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    $\begingroup$ Old MacDonald had a function, E-I-E-I-O ! :-$)$ $\endgroup$ – Lucian Jul 23 '15 at 15:15
  • $\begingroup$ On a more serious note, for $a=0$, our function has a very short and beautiful closed form $(\pm\infty)$. For $a\neq0$, it may indeed be convergent, but I'm afraid that all magic comes at a price. In this, case, the sacrifice we have to make is abandoning all hope or expectation of it ever possessing a closed form, and, in exchange, we are granted a finite value. $\endgroup$ – Lucian Jul 23 '15 at 15:22

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