Sorry in advance if this formatting comes out strange, this is my first question! I can't find a way to integrate, e.g., a function of the Hermite polynomials for general (still integer) order. For example,

Integrate[E^-x^2 HermiteH[m,x] x HermiteH[n,x ], {x, -\[Infinity], \[Infinity]}, 
    Assumptions->{m \[Element] Integers, n \[Element] Integers, m >= 0, n >= 0]

will not calculate. But if I give specific m and n,

Integrate[E^-x^2 HermiteH[m,x] x HermiteH[n,x ]/.{m->2,n->1}, {x, -\[Infinity], \[Infinity]}]

the answer comes right out. I know there is a general rule for certain classes of these integrals, and have often come across situations where I end up having to do them by hand.

Does anyone know how to make Mathematica recognize/do these integrals? Thanks in advance.

  • $\begingroup$ When I ask for HermiteH[m, x] it does not return a general form. But if I ask for HermiteH[2, x] it does. This might be the place to look. $\endgroup$
    – bill s
    Commented Aug 20, 2013 at 17:07
  • 1
    $\begingroup$ Because when you give numbers, you are now integrating polynomials with known length and terms. Mathematica can integrate combinations of polynomials much easier. When you do not specify $m,n$ values, Mathematica has to use the formal definition of $H(n)$ and it can't expand it to an actual polynomial. en.wikipedia.org/wiki/Hermite_polynomials and mathworld.wolfram.com/HermitePolynomial.html $\endgroup$
    – Nasser
    Commented Aug 20, 2013 at 18:04

1 Answer 1


In cases like this, a little help to Mathematica can often go a long way. You can notice that for these functions, the integral is zero unless Abs[n-m]==1. So you only need to generate a 1D table:

 tab=Table[Integrate[E^-x^2 HermiteH[n-1,x] x HermiteH[n,x],{x,-\[Infinity],\[Infinity]}],{n,1,5}];

and this result can be fed to FindSequenceFunction

 FindSequenceFunction[tab, n]

which returns:

 2^(-1 + n) Sqrt[\[Pi]] Pochhammer[1, n]
  • 1
    $\begingroup$ Maybe worth adding: if you apply FunctionExpand to this, you can also get rid of the Pochhammer symbol. $\endgroup$
    – Jens
    Commented Aug 21, 2013 at 3:37

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