Solving Gaussian Integral

Question

I learned that a Gaussian Integral is \begin{equation} \int_{-\infty}^{\infty}xe^{-2ax^2}dx=0 \end{equation} Because of the odd function symmetry. But if I shift the $x$ to \begin{equation} \int_{-\infty}^{\infty}xe^{-2a(x-x_0)^2}dx \end{equation} then how to solve this integral?

Answer 1

$$\int_{-\infty}^{\infty} xe^{-2a(x-x_0)^2}dx$$
$$=\int_{-\infty}^{\infty} (x-x_0)e^{-2a(x-x_0)^2}dx + x_0 \int_{-\infty}^{\infty} e^{-2a(x-x_0)^2}dx$$
$$=\int_{-\infty}^{\infty} xe^{-2ax^2}dx + x_0 \int_{-\infty}^{\infty} e^{-2ax^2}dx$$
$$=x_0\sqrt{\frac{\pi}{2a}}$$