Evaluation of the first three terms of $e^x $ in terms of the Hermite polynomials

Question

Q: Evaluate of the first three terms of $e^x $ in terms of the Hermite polynomials.

The coefficients $a_n $ are given by $$a_n = \frac {1}{2^n n! \sqrt {\pi}}\int_{-\infty}^{\infty} e^{-x^2}e^x H_n (x) dx$$

So for $a_0$, we have to solve the integral $\int_{-\infty}^{\infty} e^{-x^2}e^x H_0 (x) dx=\int_{-\infty}^{\infty} e^{x-x^2} dx$

Can someone tell me how I can evaluate this integral $\int_{-\infty}^{\infty} e^{x-x^2} dx$? Or perhaps suggest a different way to calculate the terms?

Answer 1

$\int_{-\infty}^{\infty} e^{x-x^2} dx=e^{\frac{1}{4}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}(x\sqrt{2}-\frac{1}{\sqrt{2}})^2} dx$

$=\frac{1}{\sqrt{2}}e^{\frac{1}{4}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}(u)^2} du$

$=\frac{\sqrt{2\pi}}{\sqrt{2}}e^{\frac{1}{4}}(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}(u)^2} du)$

We know that $P(X\in{\mathbb{R}})=(\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}(u)^2} du)=1$ where X is centered and normally distributed

$\int_{-\infty}^{\infty} e^{x-x^2} dx=\sqrt{\pi}e^{\frac{1}{4}}$

Answer 2

Just note that

$$ x-x^2 = -(x^2 -x + 1/4) + 1/4 = -(x-1/2)^2 + 1/4 $$

And then call $t = x - 1/2$, so that

$$ \int_{-\infty}^{+\infty}{\rm d}x\;e^{x - x^2} = \int_{-\infty}^{+\infty}{\rm d}x\;e^{-(x - 1/2)^2}e^{1/4} = e^{1/4} \int_{-\infty}^{+\infty}{\rm d}t\;e^{-t^2} = e^{1/4}\sqrt{\pi} $$