Is there any approximate solution for this integral, $$\int_0^\infty v^{m-1} K_0\bigg(2 \sqrt{\frac{v}{yz}}\bigg) \exp\bigg[-\frac{a v}{b}\bigg]\,\Bbb dv.$$
Please provide some reference with the answers if possible. Below are some reference books which have similar integrals but I couldn’t find anything:
I will assume that $a/b > 0$, $yz > 0$ and $m > 1/2$. By a simple change of integration variables and 6.631/3. from Gradshteyn and Ryzhik's book Table of Integrals, Series, and Products (seventh edition), we have $$ \int_0^\infty {v^{m - 1} K_0 \!\left( {2\sqrt {\frac{v}{{yz}}} } \right)\exp \left( { - \frac{{av}}{b}} \right)\mathrm{d}v} = \frac{{(yz)^{m - 1/2} }}{{4^{m - 1} }}\int_0^\infty {t^{2m - 2} K_0 (t)\exp ( - \alpha t^2 )\mathrm{d}t} \\ = \left( {\frac{{yz}}{4}} \right)^{m - 1/2} \alpha ^{1 - m} \Gamma ^2\! \left( {m - \frac{1}{2}} \right)\exp \left( {\frac{1}{{8\alpha }}} \right)W_{1 - m,0} \!\left( {\frac{1}{{4\alpha }}} \right), $$ where $\alpha = ayz/(4b)$ and $W$ is the Whittaker function.