Integral involving the modified Bessel function of the second kind K_0 The following integral
$$
\int_0^{+\infty} K_0(\alpha\sqrt{x^2+z^2})\, dx, \alpha>0,
$$
can be computed according to Gradshteyn-Ryzhik 6.596 (3) taking $\nu=0$ and $\mu=-1/2$. Its value is $\frac{\pi}{2\sqrt{\alpha}} e^{-\alpha|z|}$.
My question is: what about the integral
$$
 \int_0^{+\infty} \cosh(\beta x) K_0(\alpha\sqrt{x^2+z^2})dx, \alpha>0, \beta \geq 0?
$$
 A: Assume that $\alpha>\beta$ and $|\arg z|<\frac{\pi}{4}$. Then using http://dlmf.nist.gov/10.32.E10 and interchanging the order of integrations,
\begin{align*}
& \int_0^{ + \infty } {\cosh (\beta x)K_0 (\alpha \sqrt {x^2  + z^2 } )dx} 
\\ &
 = \frac{1}{2}\int_0^{ + \infty } {\exp \left( { - t - \alpha ^2 \frac{{z^2 }}{{4t}}} \right)\int_0^{ + \infty } {\cosh (\beta x)\exp \left( { - \alpha ^2 \frac{{x^2 }}{{4t}}} \right)dx} \frac{{dt}}{t}} 
\\ &
 = \frac{{\sqrt \pi  }}{{2\alpha }}\int_0^{ + \infty } {\exp \left( { - t - \alpha ^2 \frac{{z^2 }}{{4t}} + \left( {\frac{\beta }{\alpha }} \right)^2 t} \right)\frac{{dt}}{{t^{1/2} }}} 
\\ &
 = \frac{1}{2}\sqrt {\frac{\pi }{{\alpha ^2  - \beta ^2 }}} \int_0^{ + \infty } {\exp \left( { - s - \frac{{(\alpha ^2  - \beta ^2 )z^2 }}{{4s}}} \right)\frac{{ds}}{{s^{1/2} }}} 
\\ &
 = \sqrt {\frac{{\pi z}}{{2\sqrt {\alpha ^2  - \beta ^2 } }}} K_{ - 1/2} (\sqrt {\alpha ^2  - \beta ^2 } z)
\\ &
 = \frac{\pi }{{2\sqrt {\alpha ^2  - \beta ^2 } }}e^{ - \sqrt {\alpha ^2  - \beta ^2 }z } .
\end{align*}
In the last step, we employed http://dlmf.nist.gov/10.39.E2. Note that your simplification in terms of an exponential when $\beta=0$ is not correct.
