Bug introduced in 10.0 and fixed in 10.0.2

Mathematica 10 fails to solve the following integral, saying that it does not converge.

Integrate[HermiteH[5, x] HermiteH[6, x] HermiteH[5,x] Exp[-x^2], {x, -Infinity, Infinity}]

This is clearly wrong. The same evaluated in Mathematica 9 returns

36864000 Sqrt[Pi]

Is there a solution for this issue, except going back to version 9?

  • 1
    $\begingroup$ int = Integrate[HermiteH[5, x] HermiteH[6, x] HermiteH[5, x] Exp[-x^2], x]; Limit[int, x -> Infinity] - Limit[int, x -> -Infinity] gives 36864000 Sqrt[Pi] $\endgroup$ – Nasser Nov 7 '14 at 13:13
  • 2
    $\begingroup$ Please report this at support@wolfram.com. $\endgroup$ – Sjoerd C. de Vries Nov 7 '14 at 13:21
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  • $\begingroup$ Has this been fixed in 10.0.1? $\endgroup$ – user58955 Dec 8 '14 at 16:10
  • $\begingroup$ Seems to be OK in 10.0.2. $\endgroup$ – murray Dec 12 '14 at 4:24

One can use Expectation on a NormalDistribution, as I've shown before in Mathematica complaints that convergent integral diverges. (Indeed this question is nearly a duplicate of that one.) The expectation of polynomials in a normally distributed variable is a special case in Expectation and are evaluated very quickly.

The argument is integrated against the pdf:

PDF[NormalDistribution[0, 1/Sqrt[2]], x]
(* E^-x^2/Sqrt[π] *)

To get the OP's integral we have to multiply by the constant factor Sqrt[π]:

Sqrt[π] Expectation[HermiteH[5, x] HermiteH[6, x] HermiteH[5, x], 
  x \[Distributed] NormalDistribution[0, 1/Sqrt[2]]]
(* 36864000 Sqrt[π] *)
| improve this answer | |
 HermiteH[5, x] HermiteH[6, x] HermiteH[5, 
   x] Exp[-x^2], {x, -Infinity, Infinity}, PrincipalValue -> True]

(*36864000 Sqrt[\[Pi]]*)
| improve this answer | |
  • 1
    $\begingroup$ "PrincipalValue is an option for Integrate that specifies whether the Cauchy principal value should be found for a definite integral." - Does that make sense here? $\endgroup$ – Domi Nov 7 '14 at 17:27

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