Why $\int_{- \infty} ^\infty x e^{-ax^2}dx=0$

Question

I really don't understand why $\int_{- \infty} ^\infty x e^{-ax^2}dx=0$.

I can see $x e^{-ax^2}$ as $\frac{d}{dx}\frac{e^{-ax^2}}{-2a}$ that, integrated, is clearly different from 0...

Answer 1

Without calculating nothing directly, you can seet that the function is symmteric with respect to the origin.

If you want a formal answer, you can split the integral into two integrals (you can do it because you know both of them converge):

$\int_{-\infty}^{\infty}xe^{-ax^2}dx=\int_{-\infty}^{0}xe^{-ax^2}dx+\int_{0}^{\infty}xe^{-ax^2}dx$

and then change variables in the left integral $x\to -x$ and you will get zero.

Answer 2

Just for fun, we can answer the question in the following cumbersome way: $$\begin{aligned} \int_{-\infty}^{\infty} x e^{-ax^2} dx &=\frac{d}{dt}\int_{-\infty}^\infty e^{-ax^2+tx}dx \Big|_{t=0}\\ &=\frac{d}{dt}\int_{-\infty}^\infty e^{-a(x-t/(2a))^2}dx\, e^{t^2/(4a)}\Big|_{t=0}\\ &=\sqrt{\frac{\pi}{a}}\frac{d}{dt}e^{t^2/(4a)}\Big|_{t=0}\\ &=0\,. \end{aligned}$$

Answer 3

The function $f(x)=xe^{-ax^2}$ is an odd function.

Thus, $\int\limits_{-\infty}^{+\infty}f(x)dx=0$