Integral of product of modified bessel functions I need help solving the following integral:
$ \int_{0}^\infty dx \ x^{-1} K_{i a}( x) K_{i b} (x) $
I got it in the middle of a physics problem. I know that it must be proportional to $ \delta(a-b)$, but I couldn't find the proportionality constant. I thought about using the integral 6.576.4 from Gradshteyn (2007), but the conditions there are not met.
 A: This integral usually appears when normalising the wave functions for a quantum mechanical particle in an exponential potential. The result is :
$$\int_{0}^{\infty}dx~ x^{-1}K_{ia}(x)K_{ib}(x) = 2\pi~\Gamma(i a)\Gamma(-ia)~\delta(a-b) = \frac{2\pi^2~\delta(a-b)}{\sinh{\pi a}} $$
One direct (but rather convoluted) way of showing this is to write $\sinh(\pi a)=2\sinh{\frac{\pi a}{2}}\cosh{\frac{\pi a}{2}}$, move the $\sinh$ factor to the left hand side, integrate both sides with respect to $a$ and use the following integral identities:
$$
\int_{0}^{\infty}da~\sinh\left(\frac{\pi a}{2}\right) K_{ia}(x)=\frac{\pi x}{2}
$$
$$
\int_{0}^{\infty}dx~K_{ib}(x) = \frac{\pi}{2\cosh{\frac{\pi b}{2}}}
$$
Which can be found in Gradshteyn. Of course this is not the only way. You could treat this as a scattering problem and normalize the wave function at $x\to-\infty$ where it looks like a plain wave and rely on the fact that Schrodinger's equation preserves the normalization or massage directly some other formulas in Gradshteyn.
