I have seen an answer (in terms of BesselY and StruveH functions) to this integral:
Integrate[E^(-R1/Rd)R1/Sqrt[R1^2+z^2],{R1,0,Infinity},{f,0,2Pi}]
However, it seems that the Mathematica cannot do this integration. Have you any idea?
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1 Answer
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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]) *)
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$\begingroup$ The general answer is true, but $z$ is a coordinate and maybe negative too. $\endgroup$ Commented Jul 19, 2019 at 17:01
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$\begingroup$ The integrand only depends on $z^2$ so the result can only depend on $\lvert z\rvert$. I'll update the answer. $\endgroup$– RomanCommented Jul 19, 2019 at 17:08
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$\begingroup$ indeed, but it's weird that MMA cannot evaluate the integral if you assume that $z$ is negative, i.e.,
Assumptions -> z < 0. $\endgroup$ Commented Jul 19, 2019 at 17:11 -
$\begingroup$ Yes @AccidentalFourierTransform and
Assumptions -> Element[z, Reals]doesn't work either. $\endgroup$– RomanCommented Jul 19, 2019 at 17:12
2Pi. I have written the original form :) $\endgroup$