I am trying to evaluate this integral numerically: $$ \int_0^{\infty } m \exp (-m) J_1(m){}^2 \, dm $$ Everything is OK when only the integration method is specified:

NIntegrate[-m Exp[-m] BesselJ[1, m]^2, {m, 0, Infinity}, Method -> "ClenshawCurtisRule"]

but when I specify the WorkingPrecision, the integral remains unevaluated:

NIntegrate[-m Exp[-m] BesselJ[1, m]^2, {m, 0, Infinity}, Method -> "ClenshawCurtisRule",
  WorkingPrecision -> 10]

What is wrong with this code?

I am using Mathematica v9.0.1


This bug is still present in version

  • $\begingroup$ You can make it easier for others to check your code when you copy it straight from the Mathematica cell (copy as plain text) and paste it in your question with an indentation of 4 spaces. $\endgroup$ Apr 29 '13 at 8:57
  • $\begingroup$ @ThiesHeidecke Codes are replaced with plain text. $\endgroup$
    – M6299
    Apr 29 '13 at 9:20

This is a bug. As a workaround for this specific integral you could use a symbolic solution:

Integrate[-m*Exp[-m]*BesselJ[1, m]^2, {m, 0, Infinity}]

(* (-3*EllipticE[-4] + 5*EllipticK[-4])/(5*Pi) *)

"LevinRule" should work splendidly here, I think:

NIntegrate[-m Exp[-m] BesselJ[1, m]^2, {m, 0, Infinity}, 
           Method -> "LevinRule", WorkingPrecision -> 20]

ruebenko's answer has given a closed form for this particular definite integral. Personally, I prefer it when the parameters of the elliptic integrals are within $[0,1)$, so I apply the imaginary modulus transformations here to yield

N[(EllipticK[4/5] - 3 EllipticE[4/5])/(Sqrt[5] π), 20]
  • $\begingroup$ yes of course, good point. Using a different method is certainly an option. $\endgroup$
    – user21
    Apr 29 '13 at 12:03
  • 2
    $\begingroup$ This bug is apparently not present in version 7 so I added a version-9 tag; can you determine if this is in version 8? $\endgroup$
    – Mr.Wizard
    Apr 30 '13 at 11:29
  • 2
    $\begingroup$ @Mr. Wizard, Yes, it's busted in version 8. $\endgroup$
    – J. M.'s torpor
    Apr 30 '13 at 11:37

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