$\int_{-\infty}^{+\infty} xf(x)f'(x)dx=-\frac{1}{2}.$

Question

Let $f:\mathbb{R}\to\mathbb{R}, \ f\in C^1(\mathbb{R})$ such that $\lim\limits_{x\to\pm\infty}\sqrt{\lvert x\rvert}f(x)=0$.

Suppose: $$\int_{-\infty}^{+\infty} f^2(x)dx=1,$$

Prove that: $$\int_{-\infty}^{+\infty} xf(x)f'(x)dx=-\frac{1}{2}.$$

Answer 1

Hint:

Use integration by part and the assumption $\lim_{x\to \infty}xf^2(x)=\lim_{x\to -\infty}xf^2(x)=0$

Answer 2

We write, using the chain rule $(f^{2})'=2ff'$:

$\int_{-\infty}^{+\infty} xff'dx=\frac{1}{2}\int_{-\infty}^{+\infty} x(f^{2})'dx$

We now do an integration by parts

$\int_{-\infty}^{+\infty} xff'dx=xf^{2}|_{-\infty}^{\infty}-\frac{1}{2}\int_{-\infty}^{+\infty}f^{2}dx$

Notice that because:

$\lim\limits_{x\to\pm\infty}\sqrt{\lvert x\rvert}f(x)=0$

We have that (this follows easily by an $\epsilon-\delta$ argument):

$\lim\limits_{x\to\pm\infty}{\lvert x\rvert}f^{2}(x)=0$

and thus $\lim\limits_{x\to\pm\infty} xf^{2}(x)=0$

Finally: $\int_{-\infty}^{+\infty} xff'dx=-\frac{1}{2}\int_{-\infty}^{+\infty}f^{2}dx=-\frac{1}{2}$

as $\int_{-\infty}^{+\infty}f^{2}dx=1$ by hypothesis.