Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|improve this question
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. – bill s Aug 20 '13 at 17:07
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. and – Nasser Aug 20 '13 at 18:04
up vote 8 down vote accepted

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]
share|improve this answer
Maybe worth adding: if you apply FunctionExpand to this, you can also get rid of the Pochhammer symbol. – Jens Aug 21 '13 at 3:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.