Integral representation of the modified Bessel function involving $\sinh(t) \sinh(\alpha t)$

Question

I've come across this peculiar integral representation for $K_\alpha(x)$:

$\frac{\alpha}{x}K_\alpha(x) = \int_0^\infty dt \sinh(t) \sinh(\alpha t) e^{-x \cosh(t)}$

How does it come about? Are there peculiar conditions of validity? Where do I find this in the literature?

The closest in the literature I've found is:

$K_\alpha(x) = \int_0^\infty dt \cosh(\alpha t) e^{-x \cosh(t)}$

but they don't seem compatible at first glance. Thanks.

Answer 1

$K_\alpha(x)=\int_0^\infty e^{-x\cosh t}\cosh\alpha t~dt$

$\alpha K_\alpha(x)=\alpha\int_0^\infty e^{-x\cosh t}\cosh\alpha t~dt$

$=\int_0^\infty e^{-x\cosh t}~d(\sinh\alpha t)$

$=[e^{-x\cosh t}\sinh\alpha t]_0^\infty-\int_0^\infty\sinh\alpha t~d(e^{-x\cosh t})$

$=\int_0^\infty e^{-x\cosh t}x\sinh t\sinh\alpha t~dt$

$\therefore\dfrac{\alpha}{x}K_\alpha(x)=\int_0^\infty e^{-x\cosh t}\sinh t\sinh\alpha t~dt$