Convergence of an integral and the asymptotic of the function

Question

Let $f(x):[0,\infty) \to [0,\infty)$. Is it true that $\int_0^\infty f(x)dx<\infty$ implies $xf(x) \to 0,$ as $ x\to \infty?$

Answer 1

No. It isn't even true, that $f(x) \to 0$ for $x \to \infty$. To see that, define $f \colon [0,\infty) \to [0, \infty)$ by $$ f(x) = \begin{cases} 1 & x \in \left[n, n+\frac 1{n^2}\right), n \in \mathbf N - \{0\}\\ 0 & \text{otherwise} \end{cases} $$ Then $$ \int_0^\infty f(x) \, dx = \sum_{n=1}^\infty n^{-2} = \frac{\pi^2}6 < \infty $$ but $$ 1 = f(n) \not\to 0, \qquad n \to \infty $$