As the title says, I'm interested in the proof that $$ \int_1^{\infty} \frac{e^{-2at}}{\sqrt{{t-\frac{1}{t}}}} dt = \sqrt{\frac{a}{\pi}}\,\mathrm{K}_{\frac34}(a) \,\mathrm{K}_{\frac14}(a) $$ for $a>0$, where $\mathrm K$ is the modified Bessel function of the second kind.
The motivation of this integral is related to solving this other integral, but I suppose the above is interesting on its own.
By $(10.32.17)$, $(10.39.2)$ and the change of variables $t=-1+\cosh s$, we find \begin{align*} K_{3/4} (a)K_{1/4} (a) & = 2\int_0^{ + \infty } {K_{1/2} (2a\cosh s)\cosh s\,{\rm d}s} = \sqrt {\frac{\pi }{a}} \int_0^{ + \infty } {{\rm e}^{ - 2a\cosh s} \sqrt {\cosh s} \,{\rm d}s} \\ & = \sqrt {\frac{\pi }{a}} \int_0^{ + \infty } {{\rm e}^{ - 2a(1 + t)} \sqrt {1 + t} \frac{{{\rm d}t}}{{\sqrt {t(t + 2)} }}} = \sqrt {\frac{\pi }{a}} \int_1^{ + \infty } {{\rm e}^{ - 2at} \sqrt t \frac{{{\rm d}t}}{{\sqrt {t^2 - 1} }}} \\ & = \sqrt {\frac{\pi }{a}} \int_1^{ + \infty } {\frac{{{\rm e}^{ - 2at} }}{{\sqrt {t - \frac{1}{t}} }}{\rm d}t} \end{align*} for $\operatorname{Re}(a)>0$. For $(10.32.17)$, see page $440$ of G. N. Watson's book A Treatise on the Theory of Bessel Functions.