I have seen an answer (in terms of BesselY and StruveH functions) to this integral:


However, it seems that the Mathematica cannot do this integration. Have you any idea?

  • 1
    $\begingroup$ Did you mean to integrate over $z$ instead of $f$? $\endgroup$ – Roman Jul 19 at 16:39
  • $\begingroup$ @Roman No, it is just a 2Pi. I have written the original form :) $\endgroup$ – Perfect Fluid Jul 19 at 16:41
  • $\begingroup$ Should z be a function of f? $\endgroup$ – mikado Jul 19 at 16:41
  • $\begingroup$ @mikado No, the $f$ is actually $\phi$ in cylindrical coordinates. $\endgroup$ – Perfect Fluid Jul 19 at 16:43
  • 3
    $\begingroup$ So why are you integrating over $f$ then, instead of just multiplying by $2\pi$? Please condense your problem to minimal form. $\endgroup$ – Roman Jul 19 at 16:49

The integral in minimal form would be

Integrate[(E^-x x)/Sqrt[x^2 + y^2], {x, 0, ∞}, Assumptions -> y > 0]
(*    -(1/2) π y (BesselY[1, y] + StruveH[-1, y])    *)

More specifically for your case,

Integrate[(E^(-(R1/Rd)) R1)/Sqrt[R1^2 + z^2], {R1, 0, ∞},
  Assumptions -> Rd > 0 && z > 0]
(*    -(1/2) π z (BesselY[1, z/Rd] + StruveH[-1, z/Rd])    *)

For $z<0$ we should get the same answer if we replace $z$ with $\lvert z \rvert$ in the answer:

(*    -(1/2) π Abs[z] (BesselY[1, Abs[z]/Rd] + StruveH[-1, Abs[z]/Rd])    *)
  • $\begingroup$ The general answer is true, but $z$ is a coordinate and maybe negative too. $\endgroup$ – Perfect Fluid Jul 19 at 17:01
  • $\begingroup$ The integrand only depends on $z^2$ so the result can only depend on $\lvert z\rvert$. I'll update the answer. $\endgroup$ – Roman Jul 19 at 17:08
  • $\begingroup$ indeed, but it's weird that MMA cannot evaluate the integral if you assume that $z$ is negative, i.e., Assumptions -> z < 0. $\endgroup$ – AccidentalFourierTransform Jul 19 at 17:11
  • $\begingroup$ Yes @AccidentalFourierTransform and Assumptions -> Element[z, Reals] doesn't work either. $\endgroup$ – Roman Jul 19 at 17:12

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