# Mathematica 10 cannot solve definite integral [duplicate]

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?

• 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] Nov 7, 2014 at 13:13
• Please report this at [email protected]. Nov 7, 2014 at 13:21
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• Has this been fixed in 10.0.1? Dec 8, 2014 at 16:10
• Seems to be OK in 10.0.2. Dec 12, 2014 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] *)

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

(*36864000 Sqrt[\[Pi]]*)

• "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?
– Domi
Nov 7, 2014 at 17:27