$\int dx \; f(\sqrt{x^2+a^2})$ by substitution $u=\sqrt{x^2+a^2}$

Question

I need to calculate the following integral:

$I = \int x\arctan(\frac{a}{x}) \; K_0(\sqrt{x^2+a^2})dx$,

where $K_0$ is a Modified Bessel Function of the Second Kind. I think there is a good chance to be able to calculate it using integration by parts because quite a lot is known about integrals of both $\arctan(x)$ and $K_0(x)$ (see http://personalpages.to.infn.it/~zaninett/pdf/abramovitz2.pdf, page 248).

I started by doing the substitution $u = \sqrt{x^2 + a^2}$, but then I realized that $x(u)= \pm \sqrt{u^2-a^2}$, so there is no function $x(u)$, which I need in $arctan(\frac{a}{x})$.

So how does one deal with integrals of the kind $\int dx \; f(\sqrt{x^2+a^2})$?

Of course, it would AMAZING if someone knows how to actually calculate this integral?

Answer 1

With $u=\sqrt {x^2+a^2}\;$ and $0\leq x=\sqrt {u^2-a^2}$ and $a>0$ we have $K_0(\sqrt {x^2+a^2}\;)=K_0(u)$. And $x\cdot dx=x\frac {u}{\sqrt {u^2-a^2}}du=u\cdot du$. And $\arctan (a/x)=\arcsin (a/u)$. So the integral in the Q is $$\int uK_0(u)\arcsin (a/u)\;du$$.

This is all I can say. I dk if this will help.