An integral of a modified Bessel function: $\int_{0}^{\infty} K_{0} \left(\sqrt{a(k^{2}+b)} \right) dk$ 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!
 A: OK, I think I have a way to make an approximation that holds up pretty well by using Laplace's method twice. 
Begin by using the representation
$$K_0(u) = \int_0^{\infty} dt \, e^{-u\cosh{t}}$$
Then the integral you seek may be written, when we reverse the order of integration, as
$$\int_0^{\infty} dk \, K_0\left(\sqrt{a (k^2+b)}\right) = \int_0^{\infty} dt \, \int_0^{\infty} dk \, e^{-\sqrt{a} \cosh{t} \sqrt{k^2+b}}$$
This is justified in that both integrals clearly converge absolutely.  Now, to apply Laplace's method on the integral over $k$, we may assume that $a \cosh{t}$ is sufficiently large, and we use the approximation that 
$$\sqrt{k^2+b} \sim \sqrt{b} + \frac{k^2}{2 \sqrt{b}}$$
We end up with a familiar integral that may be evaluated immediately, and we are left with a single integral:
$$\int_0^{\infty} dk \, K_0\left(\sqrt{a (k^2+b)}\right) \sim \sqrt{\frac{\pi \sqrt{b}}{2 \sqrt{a}}} \int_0^{\infty} \frac{dt}{\sqrt{\cosh{t}}} e^{-\sqrt{a b} \cosh{t}} \quad (a \to \infty)$$
At this point I should say that I am not providing an error estimate, although deriving one should be fairly straightforward.  (This involves providing an additional term of the Taylor expansion of the square root above, and then Taylor expanding the exponential in the integrand.)
Now we are left with the problem of evaluating the above integral, which looks only a little less problematic than the original one.  Nevertheless, we may apply Laplace's method again, as we are claiming that $a$ is sufficiently large, and $b \gt 0$.  In this case, we use the Taylor expansion of $\cosh{t} \sim 1+t^2/2$ and we again end up with a familiar integral.  The final result is
$$\int_0^{\infty} dk \, K_0\left(\sqrt{a (k^2+b)}\right) \sim \frac{\pi}{2 \sqrt{a}} e^{-\sqrt{a b}} \quad (a \to \infty)$$
Some random numerical samples - even with moderate values of $a=1$ and $b=1$ in WA have verified the correctness of this approximation.
A: 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.
A: Using some of the ideas mentioned in Ron Gordon's answer we can evaluate this integral. Change the variable to $x=\sqrt{a} k$, so that the integral becomes
$$ \int_0^\infty K_0(\sqrt{ak^2+ab})\,dk = \frac1{\sqrt{a}} \int_0^\infty K_0(\sqrt{x^2+ab})\,dx, $$
and introduce the function
$$ I(b) = \int_0^\infty K_0(\sqrt{x^2+b^2})\,dx, $$
so that the integral is $I(\sqrt{ab})/\sqrt{a}$.
First, making the substitution $x=\sqrt{s^2-b^2}$, we get that
$$ I(b) = \int_b^\infty K_0(s)\frac{s\,ds}{\sqrt{s^2-b^2}}, $$
and we can use $K_0'(s) = -K_1(s)$ to write $K_0(s)=\int_s^\infty K_1(u)\,du$, so that
$$ I(b) = \int_b^\infty \frac{s\,ds}{\sqrt{s^2-b^2}}\int_s^\infty K_1(u)\,du. $$
The integral has the range $b<s<u<\infty$, and we can do the integral over $s$ explicitly, giving
$$ I(b) = \int_b^\infty\sqrt{u^2-b^2}K_1(u)\,du. $$
Second, using the formula
$$ K_0(u) = \int_0^\infty e^{-u\cosh t}\,dt, $$
we can write
$$ I(b) = \int_b^\infty \frac{s\,ds}{\sqrt{s^2-b^2}}\int_0^\infty e^{-s\cosh t}\,dt = \int_0^\infty b K_1(b\cosh t)\,dt, $$
where the integral over $s$ can be done in closed form in Mathematica. Substituting $t = \text{arccosh}(u/b)$ we get
$$ I(b) = \int_b^\infty \frac{b\,du}{\sqrt{u^2-b^2}}\,K_1(u). $$
From the two expressions above we can easily see that
$$ \frac{dI}{db} = -I(b), $$
and Mathematica will tell us that
$$ I(0) = \int_0^\infty K_0(x)\,dx = \frac\pi2. $$
Therefore,
$$I(b) = \frac\pi2 e^{-b}, $$
and
$$ \int_0^\infty K_0(\sqrt{a x^2+a b})\,dx = \frac{\pi}{2\sqrt{a}}e^{-\sqrt{a b}}. $$
This matches Ron Gordon's answer, even though his was only an asymptotic calculation.
A: $\int_0^\infty K_0\left(\sqrt{a(k^2+b)}\right)dk$
$=\int_0^\infty\int_0^\infty e^{-\sqrt{a(k^2+b)}\cosh t}~dt~dk$
$=\int_0^\infty\int_0^\infty e^{-(\sqrt{a}\cosh t)\sqrt{k^2+b}}~dk~dt$
$=\int_0^\infty\int_0^\infty e^{-(\sqrt{a}\cosh t)\sqrt{(\sqrt{b}\sinh u)^2+b}}~d(\sqrt{b}\sinh u)~dt$
$=\sqrt{b}\int_0^\infty\int_0^\infty e^{-\sqrt{a}\sqrt{b}\cosh t\cosh u}\cosh u~du~dt$
$=\sqrt{b}\int_0^\infty K_1(\sqrt{a}\sqrt{b}\cosh t)~dt$
