# 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 $$\int_0^{+\infty} \cosh(\beta x) K_0(\alpha\sqrt{x^2+z^2})dx, \alpha>0, \beta \geq 0?$$

• You have to assume that $\alpha>\beta$, otherwise the integral does not converge.
– Gary
Commented Apr 21, 2021 at 13:56

Assume that $$\alpha>\beta$$ and $$|\arg z|<\frac{\pi}{4}$$. Then using $$(10.32.10)$$ and interchanging the order of integrations, \begin{align*} & \int_0^{ + \infty } {\cosh (\beta x)K_0 (\alpha \sqrt {x^2 + z^2 } )\,\text{d}x} \\ & = \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)\text{d}x} \frac{{\text{d}t}}{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{{\text{d}t}}{{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{{\text{d}s}}{{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 } }}\text{e}^{ - \sqrt {\alpha ^2 - \beta ^2 }z } . \end{align*} In the last step, we employed $$(10.39.2)$$. Note that your simplification in terms of an exponential when $$\beta=0$$ is not correct.