Intuition for $\int_{-\infty}^\infty f(x) f'(x) \,dx=0$ where $f(x) \to 0$ when $x \to \pm \infty$ Assume $$\lim_{x\rightarrow \pm\infty}f(x) = 0,\:\:\: f(x)\in C^{1}(\mathbb{R})$$
For this case, I saw a theorem that says that: $$\int_{-\infty}^{{\infty}}f(x)\cdot \frac{d}{dx}f(x)\:dx = 0$$ (not hard to prove with integration by parts)
Is there any intuition for this result? I can see the intuition only in the case where f(x) is symmetric. We have: $$df \approx \frac{d}{dx}f(x)\:dx,$$ so the integrand is the product of f(x) and the infinitesimal addition to f at x's neighborhood, df(x). Thanks to the symmetry we have: $$df(-x)f(-x)=df(-x)f(x)=-df(x)f(x)$$ so the values of the integrand at x,-x cancel out and we have a total sum of 0.
Thanks!
The conditions imply that $f$ is smooth enough to use integration by parts, for which we find that
$$
\int_{-\infty}^\infty f(x)f'(x)dx = [f^2(x)]_{x=-\infty}^{x=\infty} - \int_{-\infty}^\infty f'(x)f(x)dx = -\int_{-\infty}^\infty f(x)f'(x)dx
$$
so the integral is equal to the negative of itself, and hence it is zero. Note here that you really need to replace all the $\pm \infty$ with finite limits and then take the limit at the end.
The conditions imply that $f$ is smooth enough to use integration by parts, for which we find that
$$
\int_{-\infty}^\infty f(x)f'(x)dx = [f^2(x)]_{x=-\infty}^{x=\infty} - \int_{-\infty}^\infty f'(x)f(x)dx = -\int_{-\infty}^\infty f(x)f'(x)dx
$$
so the integral is equal to the negative of itself, and hence it is zero. Note here that you really need to replace all the $\pm \infty$ with finite limits and then take the limit at the end.