So we're given this probability function $$p(x) = e^{{-a(x-b)}^2}$$ and we have to find the expectation value of $x$ which would be $$ \left<x\right> = \int_{-\infty}^{\infty}p(x)^*x p(x) = \int_{-\infty}^{\infty}xe^{{-2a(x-b)}^2} $$
The problem for me is when I do integration by parts I need the indefinite integral of $p(x)$ to put into the integration formula. All proofs I've found use trig identities that I'm not sure work with more than just $x^2$ in the exponent. Anyone willing to explain how to find the indefinite integral of $p(x)$?
Substitute $t=x-b$: \begin{align*} &\int_{-\infty}^{\infty} \left(t+b\right)e^{-2at^2} \; \mathrm{dt}\\ &=\int_{-\infty}^{\infty} te^{-2at^2} \; \mathrm{dt}+b\int_{-\infty}^{\infty} e^{-2at^2} \; \mathrm{dt} \\ &=b\int_{-\infty}^{\infty} e^{-2at^2} \; \mathrm{dt} \\ \end{align*} Where the first integral disappears since it is odd (you can also use a substitution), and the second integral is just the familiar Gaussian distribution (substitue $\xi = t \sqrt{2a}$).