Integral from $-\infty$ to $\infty$ of a function: $\int_{-\infty}^{+\infty} x e^{-x^2}\ \text{d}x$?

Question

Integral from $-\infty$ to $\infty$ of $xe^{-x^2}$.

Now I know if the integral from $a$ to $\infty$ of $f(x) dx$ and $-\infty$ to $a$ are convergent, then I could find the integral by summing the integrals from $-\infty$ to $a$ and the one from $a$ to $\infty$. But I tried to start the problem and I got $\infty$ when trying to evaluate from $t$ to $0$, using limits to replace $-\infty$ with $t$. How do I solve this??

$$\int_{-\infty}^{+\infty} x e^{-x^2}\ \text{d}x$$

Answer 1

You shouldn't get infinity at either limit. Maybe you used $e^{(-x)^2}$. You should be able to use the substitution $u=x^2$ successfully.

Answer 2

The function

$$f(x) = xe^{-x^2}$$

is an odd function, that is

$$f(-x) = -f(x)$$

or still

$$\int_{-a}^{a} f(x)\ \text{d}x = 0$$

Thence

$$\int_{-\infty}^{+\infty} x e^{-x^2}\ \text{d}x = 0$$

Remark

This is not a really general result. Not every odd functions are integrable.