Hermite Polynomials Homework: Integral w/ p>r & p=r

[ultimateguy]

Homework Statement

Show that

$$\int_{-\infty}^{\infty} x^r e^{-x^2} H_n(x) H_{n+p} dx = 0$$ if p>r and $$= 2^n \sqrt{\pi} (n+r)!$$ if p=r.

with n, p, and r nonnegative integers.
Hint: Use this recurrence relation, p times:

$$H_{n+1}(x) = 2xH_n(x) - 2nH_{n-1}(x)$$

Homework Equations

Normalization:
$$\int_{-\infty}^{\infty} e^{-x^2} [H_n(x)]^2 dx = 2^n \sqrt{\pi} n!$$

The Attempt at a Solution

I've written :

$$H_{n+p}(x) = 2xH_{n+p-1}(x) - 2nH_{n+p-2}(x)$$

I'm confused as to what the $$x^r$$ in the equation is. I know that I have to use the normalization and that will give me that extra r in the factorial, but can't quite see how.

 

[Jhero]

To solve this problem, you can use the given hint to rewrite the integral in terms of H_n(x) and H_{n+p-1}(x). Then, you can use the normalization equation to simplify the integral and eliminate the x^r term.

First, rewrite the integral using the given recurrence relation:

\int_{-\infty}^{\infty} x^r e^{-x^2} H_n(x) H_{n+p} dx = \int_{-\infty}^{\infty} x^r e^{-x^2} H_n(x) [2xH_{n+p-1}(x) - 2(n+p-1)H_{n+p-2}(x)] dx

Next, use the normalization equation to simplify the integral and eliminate the x^r term:

\int_{-\infty}^{\infty} x^r e^{-x^2} H_n(x) H_{n+p} dx = 2^{n+p-1} \sqrt{\pi} (n+p-1)! \int_{-\infty}^{\infty} e^{-x^2} H_n(x) H_{n+p-1}(x) dx - 2(n+p-1)\int_{-\infty}^{\infty} x^r e^{-x^2} H_n(x) H_{n+p-2}(x) dx

Since p>r, the second integral on the right-hand side will be equal to 0, leaving:

\int_{-\infty}^{\infty} x^r e^{-x^2} H_n(x) H_{n+p} dx = 2^{n+p-1} \sqrt{\pi} (n+p-1)! \int_{-\infty}^{\infty} e^{-x^2} H_n(x) H_{n+p-1}(x) dx

Now, using the normalization equation again, we can simplify the remaining integral:

\int_{-\infty}^{\infty} x^r e^{-x^2} H_n(x) H_{n+p} dx = 2^{n+p-1} \sqrt{\pi} (n+p-1)! 2^n \sqrt{\pi} n! = 2^n \sqrt{\pi} (n+r)!

Therefore, the integral is equal to 2