In a recent paper I've read, the author expressed the integral below as $$ \int_b^{\infty} d x \, \sqrt{x^2 - b^2} \, K_{i 2a} (2x) = \frac{1}{2} \int_b^{\infty} dx \, x K_{i a}^2 (x) \, , $$ where $K_{\nu}(z)$ is the modified Bessel function of the second kind, and $a$, $b$ are positive real numbers.
Using Mathematica, I verified that both sides of the equation indeed give the same result, but I just could not figure out how to go from the left-hand side to the right-hand side.
Hope someone here can lend a helping hand. Thanks a lot in advance!
We can use the integral representation $$K^{2}_{\mu}(x) = 2 \int_{0}^{\infty} K_{2 \mu} (2 x \cosh t) \, dt \, , \quad x >0,$$ make the substitution $u = x \cosh (t)$ , and then change the order of integration.
$$ \begin{align} \frac{1}{2} \int_{b}^{\infty} x K^{2}_{ia}(x) \, \mathrm dx &= \int_{b}^{\infty} x \int_{0}^{\infty} K_{2ia}(2x \cosh t) \, \mathrm dt \, \mathrm dx \\ & = \int_{b}^{\infty} x \int_{x}^{\infty} \frac{K_{2ia} (2u)}{\sqrt{u^{2}-x^{2}}} \, \mathrm du \, \mathrm dx \\ &= \int_{b}^{\infty} K_{2ia}(2u) \int_{b}^{u} \frac{x}{\sqrt{u^{2}-x^{2}}} \, \mathrm dx \, \mathrm du \\ &= \int_{b}^{\infty} \sqrt{u^{2}-b^{2}} K_{2ia}(2u) \, \mathrm du \end{align}$$