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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?

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  • 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
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    $\begingroup$ Please report this at support@wolfram.com. $\endgroup$ – Sjoerd C. de Vries Nov 7 '14 at 13:21
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – Michael E2 Nov 7 '14 at 13:43
  • $\begingroup$ 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
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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] *)
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Integrate[
 HermiteH[5, x] HermiteH[6, x] HermiteH[5, 
   x] Exp[-x^2], {x, -Infinity, Infinity}, PrincipalValue -> True]

(*36864000 Sqrt[\[Pi]]*)
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  • 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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