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!
Best Answer
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.