orthogonality of hermite polynomials when $x \rightarrow f(x)$

Question

We know that if $m \ne n$:

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

And

$$ \int_{-\infty}^{\infty} e^{-x^2} H_n(x)^2dx = \delta_{nm} 2^n n! \sqrt{\pi} $$

Now I am trying to find this when $x \rightarrow f(x)$ therefore I found here that when $m \ne n$:

$$ \int_{-\infty}^{\infty} f'(x) e^{-f(x)^2}H_n(f(x))H_m(f(x))dx = 0 $$

Now i am trying to find what this is equal to:

$$ \int_{-\infty}^{\infty}f'(x)e^{-f(x)^2}H_n(f(x))^2dx = ? $$

Also i am having a bit of trouble when finding the following:

$$ \int_{-\infty}^{\infty} e^{-x^2}(4x^2-2)^2 dx = 0 $$

But shouldn't it be equal to some numerical value due to the orthogonal property it holds?

Answer 1

Note that the Hermite polynomials form a complete orthogonal set for $\mathcal{L}^2[-\infty,\infty]$. This means we can expand a function $f(x)$ in this basis.

Your first question is about finding the projection of $f'(x)$ on this space. There is no short cut. You have to do the integral. This is similar to finding Fourier coefficients.

With that, let me address your second question. Note that the integral is zero if $m\neq n$. In this case, $m=n=2$. Using the identity , we obtain: $$ 2^{2}\sqrt{\pi}2!=8\sqrt{\pi} $$