Asymptotic behavior of the integral of Hermite functions/polynomials on half-lines

Question

I would like to understand the asymptotic behaviour of the following integrals with fixed $x_0>0$: $$J_m=\int^{+\infty}_{x_0}|H_m(x)|^2 e^{-x^2}dx,$$ where $H_m(x)$ is the $m-$th Hermite polynomial. For instance, I would like to understand if $$\frac{J_m}{\sqrt{2^m m!}}=O(m^p)$$ with $p\in\mathbb{R}^−$. Thank you in advance!

Answer 1

Your conjecture is not true. Moreover, $$J_m(y):=\int_y^\infty dx\, e^{-x^2}H_m(x)^2\sim c_m:=\frac12\,\pi^{1/2}\,2^m m! \tag{1}\label{1}$$ whenever $$0\le y=o(m^{1/2}).$$

Indeed, recalling that $H_m^2$ is an even function, we have $$J_m(0)=c_m. \tag{2}\label{2}$$ Also, by Theorem 8.22.9(a) in Szegő's book Orthogonal Polynomials, for any real $h\in(0,1)$, $$e^{-x^2/2}H_m(x)=O(c_m^{1/2}m^{-1/4})$$ (as $m\to\infty$) uniformly in $x\in[0,(2m)^{1/2}h]$. So, $$J_m(0)-J_m(y)=\int_0^y dx\, (e^{-x^2/2}H_m(x))^2=O((c_m^{1/2}m^{-1/4})^2y)=o(c_m).$$ So, in view of \eqref{2}, we do have \eqref{1}. $\quad\Box$

Answer 2

Let me first consider the case $x_0=0$, when the integral has a closed form expression: $$\int^{\infty}_{0}|H_m(x)|^2 e^{-x^2}\,dx=\sqrt{\pi } \,2^{m-1} m!\;\;(m\in\mathbb{N}).$$ See, for example, this MSE post for a derivation.

Considering next a nonzero $x_0$, to evaluate the integral from $0$ to $x_0$ (and subtract it from the answer given above) I substitute the large-$m$ asymptotics of the Legendre polynomials, $$e^{-\frac{x^2}{2}}H_m(x) \rightarrow \frac{2^m}{\sqrt \pi}\Gamma\left(\tfrac{1}{2}+m/2\right) \cos \left(x \sqrt{2 m}- \tfrac{1}{2}m\pi \right),$$ to obtain the asymptotics $$\int^{x_0}_{0}|H_m(x)|^2 e^{-x^2}\,dx\rightarrow \left[\frac{2^m}{\sqrt \pi}\Gamma(\tfrac{1}{2}+m/2)\right]^2\int^{x_0}_0 \cos^2 \left(x \sqrt{2 m}- \tfrac{1}{2}m\pi \right)\,dx\rightarrow\frac{x_0}{2}\left[\frac{2^m}{\sqrt \pi}\Gamma(\tfrac{1}{2}+m/2)\right]^2,$$ which is smaller than the integral from $0$ to $\infty$ by a factor $\sqrt m$.

Hence I conclude that $$\lim_{m\rightarrow\infty}\frac{1}{2^m m!}\int_{x_0}^{\infty}|H_m(x)|^2 e^{-x^2}\,dx=\tfrac{1}{2}\sqrt\pi.$$

_{Note that the OP uses the convention that $H_m(x)$ is the socalled physicists Hermite polynomial.}