If $f$ is a Schwartz function, then $\int_{-\infty}^{\infty}f(-x)dx=\int_{-\infty}^{\infty}f(x)dx$

Question

Let $\mathcal S$ be the Schwartz space on $\mathbb R$, i.e., the set of functions $f\in\cal C^{\infty}$ on $\mathbb R$ such that for every $k,l\ge 0$ $$ \sup |x|^k|f^{(l)}(x)|<\infty. $$ If $f\in\cal S$, then $$ \int_{-\infty}^{\infty}f(-x)dx=\int_{-\infty}^{\infty}f(x)dx. $$

I don't know how to use this rapid decay of $x^kf^{(l)}$ property to show this. If I use the change of variable $u=-x$ I seem to get that the sides are opposite. Why this isn't correct?

Any help you'll be appreciated.

Answer 1

If $f$ is an even function then $f(-x)=f(x)$ so the equation holds. If $f$ is odd then both sides are $0$. In general, $f=g+h$ with $g$ odd and $h$ even.