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?

  • $\begingroup$ Do you really need the indefinite integral? Or just the integral over $\mathbb{R}^+$? $\endgroup$ – Jack D'Aurizio Mar 15 '17 at 22:22
  • $\begingroup$ There is a neat difference in complexity in dealing with $\int e^{-x^2}\,dx$ or just $\int_{0}^{+\infty}e^{-x^2}\,dx$, for instance. $\endgroup$ – Jack D'Aurizio Mar 15 '17 at 22:23
  • $\begingroup$ Unfortunately, I do need the indefinite integral @JackD'Aurizio $\endgroup$ – Mencia Mar 15 '17 at 22:32
  • $\begingroup$ For which purpose, if I may ask? $\endgroup$ – Jack D'Aurizio Mar 15 '17 at 22:35
  • $\begingroup$ I am a theoretical physicist. Maybe we can start a chat if you want more details? $\endgroup$ – Mencia Mar 15 '17 at 22:36

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.

  • $\begingroup$ Thank very much, but I don't it helps for the indefinite integral. For definite it probably would because the integrand is an even function of $x$, so I could use symmetries to rewrite the integral in terms an integral for $x>0$. $\endgroup$ – Mencia Mar 15 '17 at 23:00

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