# Integral of Hermite polynomial multiplied by $\exp(-x^2/2)$

What is the value of $\int_{-\infty}^{\infty}e^{-\frac{x^2}{2}}H_n(x)dx$ where $H_n(x)$ is the $n^{\small\mbox{th}}$ Hermite Polynomial (physicist's convention)?

For probabilists' Hermite polynomials: The Hermite polynomials are the orthogonal polynomials corresponding to the weight function $w(x) = e^{-x^2/2}$. This means that $\int_{-\infty}^{\infty} H_n(x)H_m(x)e^{-x^2/2} \, dx = 0$ whenever $n \not= m$ (or equivalently, $\int_{-\infty}^{\infty} H_n(x) P(x) e^{-x^2/2} \, dx = 0$ for any polynomial $P$ of degree less than $n$). Since $H_0(x) = 1$, it follows that $$\int_{-\infty}^{\infty} H_n(x)e^{-x^2/2} \, dx = \int_{-\infty}^{\infty} H_n(x)H_0(x)e^{-x^2/2} \, dx = 0$$ for all $n > 0$. The only time this integral is non-zero is when $n = 0$, in which case $$\int_{-\infty}^{\infty} H_0(x)e^{-x^2/2} \, dx = \int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{2\pi}.$$

• I guess it depends on which "Hermite polynomials" Tarek is asking about. Jan 17, 2013 at 22:04
• I am asking about the physicists' Hermite polynomials. Jan 17, 2013 at 23:04
• Ok, then disregard this and go Mhenni's route. I mistakenly assumed you were using the probabilists' polynomials because you were using the corresponding weight. Jan 17, 2013 at 23:12
• @Tarek, then you should have edited your question to include the normalization you're working with... Mar 23, 2013 at 13:05

Notice that, if $n$ is odd , then the integrand is an odd function which implies that the integral equals to $0$. If $n$ is even, then the integral equals to

$${2}^{2\,n+\frac{5}{2}}\Gamma \left( n+ \frac{3}{2} \right),\quad n=0,1,2,\dots.$$

Note this, in the above formula, $n=0$ corresponds to the case $H_{2}(x)$, $n=1$ correspons to the case $H_{4}(x)$ and so on.

One can have instead, the formula which include the case $n=0$

$${4}^{n}\sqrt {2}\,\Gamma \left( n+\frac{1}{2} \right), \quad n=0,1,2,\dots.$$

Again, in the above formula, $n=0$ corresponds to the case $H_{0}(x)$, $n=1$ corresponds to the case $H_{2}(x)$ and so on.

• How did you calculate this? Jan 17, 2013 at 20:24

Using the generating funtion $$e^{2xt-t^2}=\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}$$ we can obtain two recurrence relations.

Differentiating with respect to $$t$$ we obtain $$H_{n+1}(x)=2xH_n(x)-2nH_{n-1}(x)$$ for $$n\geq 1$$ with $$H_0(x)=1$$ and $$H_1(x)=2x$$. Similarly, differentiating with respect to $$x$$ we obtain $$\frac{d}{dx}H_n(x)=2nH_{n-1}(x)$$ for $$n\geq 1$$.

Now, define $$I_n=\int_{-\infty}^{\infty}e^{-x^2/2}H_n(x)dx$$ then \begin{align}I_{n+1}&=\color{blue}{\int_{-\infty}^{\infty}2xe^{-x^2/2}H_{n}(x)dx}-2nI_{n-1}\\ &=\color{blue}{-2e^{-x^2/2}H_n(x)\Big|_{-\infty}^{\infty}+4nI_{n-1}}-2nI_{n-1}=2nI_{n-1}. \end{align} where the integral in blue was solved by parts and since that $$H_n$$ is a polynomial of degree $$n$$ we have that $$\lim_{x\to\pm\infty}e^{-x^2/2}H_n(x)=0$$, therefore $$-2e^{-x^2/2}H_n(x)\Big|_{-\infty}^{\infty}=0$$.

To finish, since $$I_0=\int_{-\infty}^{\infty}e^{-x^2/2}H_0(x)dx=\int_{-\infty}^{\infty}e^{-x^2/2}dx=\sqrt{2\pi}$$ and $$I_1=\int_{-\infty}^{\infty}e^{-x^2/2}H_1(x)dx=\int_{-\infty}^{\infty}2xe^{-x^2/2}dx=0$$ we conclude that $$I_{2n+1}=0$$ for all $$n\in\mathbb{N}$$ and \begin{align}I_{2n}&=2(2n-1)I_{2n-2}\\ &=2^2(2n-1)(2n-3)I_{2n-4}\\ &~~\vdots\\ &=2^n(2n-1)(2n-3)\cdots1\cdot I_0\\ &=2^n\frac{2n(2n-1)(2n-2)(2n-3)\cdots1}{2n(2n-2)\cdots2}I_0\\ &=2^n\frac{(2n)!}{2^nn!}I_0\\ &=\frac{(2n)!}{n!}I_0\\ &=\frac{(2n)!}{n!}\sqrt{2\pi}.\end{align}

Since the weighting function corresponds to the probabilists Hermite polynomials, the substitution $$\operatorname{H}_n(x)=2^{\frac{n}{2}}\operatorname{He}_n(\sqrt{2}x)$$ and the integral becomes $$I_n = 2^{\frac{n}{2}}\int_{-\infty}^\infty\operatorname{He}_n(\sqrt{2}x)e^{-\frac{x^2}{2}}dx.$$

We can use the Hermite multiplication theorem $$\operatorname{He}_n(\gamma x)=n!\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor }\frac{1}{2^kk!(n-2k)!}\gamma^{n-2k}\left(\gamma^2-1\right)^k \operatorname{He}_{n-2k}(x)$$ with $$\gamma=\sqrt{2}$$ to obtain $$I_n = 2^{\frac{n}{2}}n!\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor }\frac{1}{2^kk!(n-2k)!}2^{\frac{n}{2}-k}\int_{-\infty}^\infty\operatorname{He}_{n-2k}(\sqrt{2}x)e^{-\frac{x^2}{2}}dx.$$ The integral is zero whenever $$n$$ is odd, so we enforce this by setting $$n=2s$$. Using orthogonality of the probabilists Hermite polynomials $$\int_{-\infty}^\infty \operatorname{He}_n(x)\operatorname{He}_m(x)e^{-\frac{x^2}{2}}dx = n!\sqrt{2\pi}\delta_{n,m}$$ the desired integral becomes $$I_{2s} = \frac{(2s)!}{s!}\sqrt{2\pi}.$$