In studying a physics problem, I've encountered the following series of Hermite polynomials
$$ \sum_{n=0}^{\infty} \frac{H_n(x) H_n(y)}{n!2^n(n+1)^2}.$$
As of now, I am completely in a loss on whether anything can be said about this series. When looking for properties of Hermite polynomials, I've found (for example in this question) a similar expression using a Dirac delta distribution (a simple derivation of half of the formula was found here). The full version, found on Wolfram functions site, is
$$ \sum_{n=0}^{\infty} \frac{H_n(x) H_n(y)}{n!2^n}=\sqrt{\pi} \exp\left[(x^2+y^2)/2\right]\delta(x-y),$$
which seems very similar to what I'm trying to evaluate, and even shows some intuition to me that since the Hermie polynomials are orthonormal to eachother I am supposed to get a Dirac delta (with the weight function aptly treated on the right side here).
However, the series I need to evaluate is not this, but instead each term is multiplied by a factor of $(n+1)^{-2}$, and I don't see a way to handle this nicely. I've tried applying the Mehler kernel, but I can't find a choice of $\rho$ that would result in my particular series.
At this point I'm tempted to believe – based on intuition and not mathematics – that my series should evaluate to exactly the same result as without having the extra term, seeing as the Hermite polynomials are still orthonormal no matter how we weigh each of them in the sum. This seems logical to me, however this argument is ad hoc enough that it's probably false.
Does anyone have a clue on how to say anything (or even fully evaluate) about my series?
Best Answer
Citing Wikipedia which cites Mehler, $$\sum_{n\geq 0}\frac{H_n(x)H_n(y)}{n!}\left(\frac{u}{2}\right)^n = \frac{1}{\sqrt{1-u^2}}\,\exp\left(\frac{2u}{1+u}xy-\frac{u^2}{1-u^2}(x-y)^2\right) \tag{1}$$ follows from $$ H_n(x) = (-1)^n e^{x^2}\frac{1}{2\sqrt{\pi}}\int (is)^n e^{isx-\frac{s^2}{4}}\,ds.\tag{2} $$ As a consequence, in order to compute your series it is sufficient to (numerically) evaluate $$ \sum_{n\geq 0}\frac{H_n(x)H_n(y)}{2^n n!(n+1)^2}=\int_{0}^{1}\frac{-\log u}{\sqrt{1-u^2}}\,\exp\left(\frac{2u}{1+u}xy-\frac{u^2}{1-u^2}(x-y)^2\right)\,du. \tag{3}$$