TITLE: An integral of a modified Bessel function: $\int_{0}^{\infty} K_{0} \left(\sqrt{a(k^{2}+b)} \right) dk$
QUESTION [6 upvotes]: Let $K_{\alpha}(z)$ be the modified Bessel function of the second kind of order $\alpha$.
I need to compute the following integral:
$$\int_0^\infty\;\;K_0\left(\sqrt{a(k^2+b)}\right)dk$$ 
where $a>0$ and $b>0$. 
I have tried several substitutions and played around a lot in Mathematica, and can't seem to solve this. Perhaps an integral representation of $K_{0}(z)$ would be helpful here.
Even if this can't be done exactly, a sensible approximation strategy would also be useful. 
Any advice would be greatly appreciated. Thanks in advance for your time!

REPLY [3 votes]: The evaluation is a bit easier if we use the integral representation $$K_{0}(x) = \frac{1}{2} \int_{0}^{\infty} \frac{1}{t} \, \exp \left(-t - \frac{x^{2}}{4t} \right) \, dt , \quad x>0,  $$ which can be derived from the integral representation $$K_{0}(x) = \int_{0}^{\infty} \exp(-x \cosh t) \, dt, \quad x>0, $$ by extending the interval of integration to the entire real line and then making the substitution $ e^{t} = \frac{2}{z} u$.
Using this representation of the modified Bessel function of the second kind of order zero, we get
$$\begin{align}\int_{0}^{\infty} K_{o}\sqrt{a^{2}\left(x^{2}+b^{2} \right)} &= \int_{0}^{\infty} K_{0} \left(a\sqrt{x^{2}+b^{2}} \right) \, dx  \\ &= \frac{1}{2} \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{t} \, \exp \left(-t - \frac{a^{2}(x^{2}+b^{2})}{4t} \right) \, dt \, dx \\ &= \frac{1}{2} \int_{0}^{\infty} \frac{1}{t} \, \exp \left(-t-\frac{a^{2}b^{2}}{4t} \right) \int_{0}^{\infty} \exp \left(-\frac{a^{2}x^{2}}{4t} \right) \, dx \, dt \tag{1}\\ &= \frac{\sqrt{\pi}}{2a} \int_{0}^{\infty} \frac{1}{\sqrt{t}} \,  \exp \left(-t-\frac{a^{2}b^{2}}{4t}\right) \, dt \\ &= \frac{\sqrt{\pi}}{a} \int_{0}^{\infty} \exp \left(-u^{2}-\frac{a^{2}b^{2}}{4u^{2}} \,  \right) \, du \\ &= \frac{\sqrt{\pi}}{a} \, \sqrt{\frac{\pi}{4}} \exp \left(-2 \, \sqrt{\frac{a^{2}b^{2}}{4}} \right) \tag{2}\\ &= \frac{\pi}{2a} \,  \exp(-ab). \end{align}$$

$(1)$ Since the integrand is nonnegative, we're allowed to switch the order of the integration.
$(2)$ How to evaluate $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ for $a,b>0$

This particular integral is a limiting case of the more general integral
$$\small  \frac{1}{\beta^{\mu}} \int_{0}^{\infty} J_{\mu}(\beta t) \, \frac{K_{\nu}(a\sqrt{t^{2}+b^{2}}{})}{(t^{2}+b^{2})^{\nu/2}} \, t^{\mu+1} \, dt = \frac{1}{a^{\nu}} \left(\frac{\sqrt{a^{2}+\beta^{2}}}{b} \right)^{\nu-\mu-1}K_{\nu-\mu-1} \left(b \sqrt{a^{2}+\beta^{2}} \right),\tag{3}$$ where $a, \beta >0$, $\Re(b) >0$, and $\Re(\mu) >-1$.
(As $\beta\downarrow 0$, $J_{\mu}(\beta t)\sim \left(\frac{\beta t}{2} \right)^{\mu} \frac{1}{\Gamma(\mu+1)}$.)
Integral $(3)$ is entry (2) in section 47 of chapter 13 of the textbook A Treatise on the Theory of Bessel Functions.