How to Integrate a Physicists’ Hermite Polynomial with e^(-x²)

If the integrand is the physicists’ Hermite polynomial multiplied by its standard Gaussian weight, the integral has a simple antiderivative. The key is to differentiate the product of the weight and the Hermite polynomial of one lower degree.

For n ≥ 1,

$$\int H_n(x)e^{-x^2}\,dx=-H_{n-1}(x)e^{-x^2}+C.$$

The case n = 0 must be treated separately: its integral involves the error function. The result also gives definite integrals over finite intervals and, by taking limits, over the whole real line.

The physicists’ convention

The physicists’ Hermite polynomials are defined by Rodrigues’ formula

$$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}.$$

The first few are $H_0(x)=1$, $H_1(x)=2x$, and $H_2(x)=4x^2-2$. They satisfy the derivative relation

$$H_n'(x)=2nH_{n-1}(x)$$

and the recurrence $H_n(x)=2xH_{n-1}(x)-2(n-1)H_{n-2}(x)$ for $n\ge 2$.

Different conventions exist for Hermite polynomials. The formula here uses the physicists’ version, whose Gaussian weight is $e^{-x^2}$. The probabilists’ version uses a different scaling and the weight $e^{-x^2/2}$.

Derive the antiderivative

For $n\ge 2$, differentiate the product $e^{-x^2}H_{n-1}(x)$:

$$\begin{aligned} \frac{d}{dx}\left(e^{-x^2}H_{n-1}(x)\right) &=e^{-x^2}\left(H_{n-1}'(x)-2xH_{n-1}(x)\right)\\ &=e^{-x^2}\left(2(n-1)H_{n-2}(x)-2xH_{n-1}(x)\right)\\ &=-e^{-x^2}H_n(x). \end{aligned}$$

The last equality follows from the Hermite recurrence. For $n=1$, the identity follows directly from $H_1(x)=2x$ and $H_0(x)=1$, so no $H_{-1}$ is needed. Integrating the derivative gives the stated antiderivative.

For $n=0$, the integrand is just $e^{-x^2}$. Its antiderivative is not an elementary function; using the error function,

$$\int e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)+C,$$

where $\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\,dt$.

Worked example: integrate $H_3(x)e^{-x^2}$

The physicists’ polynomial of degree three is $H_3(x)=8x^3-12x$. Applying the formula with $n=3$ uses $H_2(x)=4x^2-2$:

$$\int (8x^3-12x)e^{-x^2}\,dx=-(4x^2-2)e^{-x^2}+C.$$

You can check it by differentiating the right-hand side. The result is $(8x^3-12x)e^{-x^2}$, as required.

For a finite interval $[a,b]$ and $n\ge 1$, the same antiderivative gives

$$\int_a^b H_n(x)e^{-x^2}\,dx =H_{n-1}(a)e^{-a^2}-H_{n-1}(b)e^{-b^2}.$$

For example, the integral for $n=1$ from $0$ to $1$ is $1-e^{-1}$, since $H_0=1$.

Integrating over the whole real line

For each fixed polynomial $H_{n-1}(x)$, the product $H_{n-1}(x)e^{-x^2}$ tends to zero as $x\to\pm\infty$. Therefore, for every $n\ge 1$,

$$\int_{-\infty}^{\infty}H_n(x)e^{-x^2}\,dx=0.$$

For $n=0$, the Gaussian integral instead gives

$$\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}.$$

These statements concern the integral against the constant function. They are consistent with Hermite orthogonality: $H_n$ is orthogonal to lower-degree polynomials under the weight $e^{-x^2}$.

Common mistakes

• Using the wrong Hermite convention. The physicists’ convention has $H_1(x)=2x$ and weight $e^{-x^2}$. The probabilists’ convention has $He_1(x)=x$ and weight $e^{-x^2/2}$.
• Applying the formula at $n=0$. There is no polynomial $H_{-1}$ in this formula. Integrate the Gaussian with the error function instead.
• Dropping the minus sign. The derivative of $e^{-x^2}H_{n-1}(x)$ is $-e^{-x^2}H_n(x)$, not its positive.
• Confusing an indefinite and a definite integral. For finite limits, evaluate the endpoint expression; the integral is not generally zero. The zero result applies to the whole real line when $n\ge 1$.

If the exponential in your integrand is not $e^{-x^2}$—for example, if it is $e^{-x^2/2}$—this product identity does not directly give the same formula. Check both the polynomial convention and the precise exponential before applying it.