# Definite Gaussian/exponential integral

I've been revising some quantum mechanics, and I was wondering how I would calculate a specific standard deviation.

The wave function I am working with is $\psi(x)=C(a,b)\exp(\frac{-(x-x_0)^2}{4a^2})e^{ibx}$ in $L^2(\mathbb{R})$ for $a,b\in\mathbb{R}$ and the integral I need is:

$\int_{-\infty}^{\infty}x^2\psi(x)\overline{\psi(x)}$ $dx = \int_{-\infty}^{\infty}x^2\exp(\frac{-(x-x_0)^2}{2a^2})$ $dx$.

I know that this has something to do with Gaussian integrals, but can't find the actual solution to any integral of this form anywhere! (Most have either no $x$ terms in front of the exponential, or at least certainly not an $x^2$.) Does this definite integral exist, and what is it?

Any help would be greatly appreciated! Thanks!

PS: In case it is relevant, I was working out ${\langle Q^2 \rangle}_\omega$ where $Q$ is the position operator given by $Q\psi(x) = x\psi(x)$, in order to then calculate the standard deviation which is given by ${\sigma}^2_\omega (Q) = {\langle Q^2 \rangle}_\omega - {{\langle Q \rangle}_\omega}^2$.

• your last integral is $a^2+x_0^2$. – Math-fun May 24 '16 at 14:54
• Thank you! Is it just a form I should just learn, or is there some method to reaching that solution? – ScuffleIt May 24 '16 at 14:56
• ups I have missed some $\pi$'s here and there ... :-/ – Math-fun May 24 '16 at 15:13

\begin{align} \int_{-\infty}^{\infty} x^2\exp\Big(-\frac{(x-x_0)^2}{2a^2}\Big)dx&=\sqrt{2\pi}a\int_{-\infty}^{\infty} x^2 \times \frac{1}{\sqrt{2\pi}a}\exp\Big(-\frac{(x-x_0)^2}{2a^2}\Big)dx\\ &=\sqrt{2\pi}aEX^2 \end{align} where $X$ is normally distributed with mean $x_0$ and variance $a^2$. Since $Var(X)=EX^2-E^2X$ we have that $EX^2=a^2+x_0^2$ Therefore your integral becomes $$\sqrt{2\pi}a(a^2+x_0^2)$$
Note that $f(x)=\frac{1}{\sqrt{2\pi}a}\exp\Big(-\frac{(x-x_0)^2}{2a^2}\Big)$ is the pdf of a normally distributed random variable with mean $x_0$ and variance $a^2$. See also this link.