Does anyone know how to evaluate $\int_0^\infty dx \, x^2 e^{-\frac{a^2}{2} x^2} \sqrt{x^2 + b^2}$ by hand?
I've been relying on Mathematica for these kinds of integrals, which in this case evaluates to $$\frac{b^2}{2 a^2} e^{\frac{a^2 b^2}{4}} K_1\left( \frac{a^2 b^2}{4} \right) \, ,$$ where $K_\nu(z)$ is the modified Bessel function of the second kind. I haven't found integral tables (e.g. Gradshteyn and Ryzhik) to be particularly helpful here either, but I may just need to look harder.
A plausible approach I thought might work would be to start by Fourier transforming the root, since $$\sqrt{x^2 + b^2} = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} -b \sqrt{\frac{2}{\pi}} \frac{K_1(b |q|)}{|q|} e^{-i x q} \, dq \, ,$$ or perhaps more neatly with a Fourier cosine transformation $$\sqrt{x^2 + b^2} = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} -b \sqrt{\frac{2}{\pi}} \frac{K_1(b q)}{q} \cos(x q) \, dq \, ,$$
and then re-ordering integrals, but this doesn't seem to quite work out how I want (or at least I'm not sure how to progress from here). Any help would be much appreciated.
Best Answer
$$ \int_0^\infty dx \, x^2 e^{-\frac{a^2}{2} x^2} \sqrt{x^2 + b^2} $$ is equivalent to $$ b^3\int_0^\infty d\left(\frac{x}{b}\right) \, \left(\frac{x}{b}\right)^2 e^{-\frac{a^2b^2}{2} \left(\frac{x}{b}\right)^2} \sqrt{\left(\frac{x}{b}\right)^2 + 1} = b^3\int_0^\infty dz \, z^2 e^{-k z^2} \sqrt{z^2 + 1} $$ where $k = \frac{a^2b^2}{2}$ we can move $z^2$ under the sqrt $$ \sqrt{z^4 +z^2} $$ this leads to $$ b^3\int_0^\infty dz \, z e^{-k z^2} \sqrt{z^4 + z^2} $$ if we set $z^2 = v$ we have $$ \frac{b^3}{2}\int_0^\infty dv e^{-k v} \sqrt{v^2 + v} $$ we can use completing the square $$ v^2 + v = \left(v + \frac{1}{2}\right)^2 - \frac{1}{4} = \left(\frac{v + \frac{1}{2}}{1/2}\right)^2 - 1 $$ setting $t = \frac{v + \frac{1}{2}}{1/2}$ $$ \frac{b^3}{2}\int_1^\infty 2dt e^{-k \left(\frac{t}{2} - \frac{1}{2}\right)} \sqrt{t^2 - 1} $$ which becomes $$ b^3\mathrm{e}^{\frac{k}{2}}\int_1^\infty dt e^{-\frac{k}{2}t} \sqrt{t^2 - 1} $$ now we can use the identity here
$$ K_n(z) = \dfrac{\sqrt{\pi}}{\left(n-\frac{1}{2}\right)!}\left(\frac{1}{2}z\right)^{n}\int_1^\infty dx\mathrm{e}^{-zx}\left(x^2-1\right)^{n-1/2} $$ setting $n=1, z =\frac{k}{2}$ we find $$ K_1(\frac{k}{2}) = \dfrac{\sqrt{\pi}}{\left(-\frac{1}{2}\right)!}\frac{k}{4}\int_1^\infty dx\mathrm{e}^{-\frac{k}{2}x}\sqrt{x^2-1} $$ with $(-1/2)! = \Gamma(1/2) = \sqrt{\pi}$ thus $$ \frac{K_1(\frac{k}{2})}{k/4} = \int_1^\infty dx\mathrm{e}^{-2kx}\sqrt{x^2-1} $$