Does there exist a simple expression for integrals of the form,

$I = \int_{-\infty}^0 H_n(u) H_m(u)\, \mathrm{e}^{-u^2}\,du$,

where $m$ and $n$ are nonnegative integers and $H_n$ is the $n$'th (physicists') Hermite polynomial?

When $n+m$ is even, the symmetry of the integrand and the orthogonality of $H_n$ imply,

$I = \sqrt{\pi} \,2^{n-1} n! \,\delta_{n,\,m}$ (for $n+m$ even).

For $n+m$ odd, $I$ is nonzero and increases in magnitude with $n+m$, but I have been unable to find a general formula.

  • $\begingroup$ Check this problem. $\endgroup$ – Mhenni Benghorbal Feb 1 '13 at 9:18
  • $\begingroup$ @MhenniBenghorbal, the question you linked to was concerned with integrals from $-\infty$ to $\infty$, while my question is about integrals from $-\infty$ to $0$. What I stated about the $m+n$ even case follows from considering the $-\infty$ to $\infty$ integrals. But I don't see how to infer something about the $m+n$ odd case. $\endgroup$ – Ted Pudlik Feb 6 '13 at 4:56

It looks to me like we have exponential generating functions

$$\sum_{n=0}^\infty I(n,n+2k+1) t^n/n! = \dfrac{(-1)^{k+1}(2k)!}{k! (1-2t)^{k+3/2} (1+2t)^{k+1/2}}$$

EDIT: Hmm, these can be combined into a bivariate exponential generating function

$$ \sum_{n=0}^\infty \sum_{k=0}^\infty I(n,n+2k+1) \frac{s^k t^n}{k! n!} = \frac{1}{(-1+2t) \sqrt{1+4s-4t^2}}$$

  • $\begingroup$ I can't quite connect the dots here: what does $I(n,n+2k+1)$ stand for? How is $k$ related to $m$ and $n$? $\endgroup$ – Ted Pudlik Feb 6 '13 at 5:11
  • $\begingroup$ $I(n,m)$ is your integral $\int_{-\infty}^0 H_n(u) H_m(u)\ e^{-u^2}\ du$. So I'm taking $m=n+2k+1$ where $k$ is a nonnegative integer. $\endgroup$ – Robert Israel Feb 6 '13 at 19:28
  • $\begingroup$ For example, with $k = 3$ we have $I(0,7) = 120, I(1,8) = 240, I(2,9) = 4320, I(3,10) = 25920, I(4,11) = 570240$, corresponding to $$ \frac{6!}{3! (1-2t)^{9/2} (1+2t)^{7/2}} = \frac{120}{0!} + \frac{240}{1!} t + \frac{4320}{2!} t^2 + \frac{25920}{3!} t^3 + \frac{570240}{4!} t^4 + \ldots $$ $\endgroup$ – Robert Israel Feb 6 '13 at 19:37
  • $\begingroup$ Now I see---a very neat relationship, thank you! $\endgroup$ – Ted Pudlik Feb 8 '13 at 2:56

This question is getting a little old now, but I feel I can add something here, for my own conscience, if nothing else.

My take on this problem is - simply define $u = \sqrt{v}$ in the integral and take advantage of the relation between the Hermite and Laguerre polynomials. i.e.

$$ H_{2n}(\sqrt{v}) = (-1)^n 2^{2n}n! L^{\left(-\frac{1}{2}\right)}_n(v)\\ H_{2n+1}(\sqrt{v}) = (-1)^n 2^{2n+1}n! \sqrt{v} L^{\left(\frac{1}{2}\right)}_n(v) $$

So, assuming that $m$ and $n$ are both even we get: \begin{eqnarray} \int^0_{-\infty}du e^{-u^2} H_n(u) H_m(u)= (-1)^{n+m}\int^{\infty}_0 dx e^{-x^2} H_n(x) H_m(x) \;\; \textrm{changing order of integration and changing variables $u \rightarrow -x$}\\ = \frac{1}{2} \int^{\infty}_0 \left(dv v^{-\frac{1}{2}} \right)e^{-v} H_n(\sqrt{v}) H_m(\sqrt{v}) \;\; \textrm{changing variables $x = \sqrt{v}$}\\ = \frac{1}{2} \int^{\infty}_0 \left(dv v^{-\frac{1}{2}} \right)e^{-v} (-1)^{\frac{n}{2}} 2^{n}\left(\frac{n}{2}\right)! L^{\left(-\frac{1}{2}\right)}_{\frac{n}{2}}(v)(-1)^{\frac{m}{2}} 2^{m}\left(\frac{m}{2}\right)! L^{\left(-\frac{1}{2}\right)}_{\frac{m}{2}}(v) \;\; \textrm{applying the above relation between Hermite and Laguerre polys. - the even case}\\ =\frac{(-1)^{\frac{n}{2}+\frac{m}{2}}2^{m+n}}{2} \left(\frac{n}{2}\right)!\left(\frac{m}{2}\right)!\int^{\infty}_0 dv v^{-\frac{1}{2}} e^{-v}L^{\left(-\frac{1}{2}\right)}_{\frac{n}{2}}(v)L^{\left(-\frac{1}{2}\right)}_{\frac{m}{2}}(v) \;\; \textrm{decluttering algebra}\\ =(-1)^{\frac{n}{2}+\frac{m}{2}}2^{m+n-1} \left(\frac{n}{2}\right)!\left(\frac{m}{2}\right)! \frac{\Gamma\left( \frac{n+1}{2}\right)}{\left(\frac{n}{2}\right)!}\delta_{n,m} \;\; \textrm{applying Laguerre orthogonality}\\ =(-1)^{n}2^{2n-1} \left(\frac{n}{2}\right)! \Gamma\left( \frac{n+1}{2}\right)\delta_{n,m} \end{eqnarray}

Since $n$ is even the expression is positive. Equivalent expressions exist for all other odd/even combinations of $m$ and $n$. I have probably screwed up the algebra in places, but I think the general argument is valid.

  • $\begingroup$ Where does the assumption that $m + n$ is even come into play? I think there must be a typo somewhere, because the result should always be nonnegative in this case, but I'll examine it more closely. $\endgroup$ – Ted Pudlik Oct 22 '15 at 18:46
  • $\begingroup$ Apologies Ted - I knew I'd mess up the algebra, there is a rogue minus sign, I will fix up the workings. $\endgroup$ – user132949 Oct 22 '15 at 22:20

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