If the improper integral $\int_a^{+\infty}f(x)dx$ is finite, and the function $f(x)$ is a monotonic function, can we get that the function $xf(x)$ is monotonic? I assume there may be some counterexamples, but I just can't find one.
Update: I have found a counterexample. Just take$f(x)=\frac1{n^2},n\leq x<n+1,n\in \mathbb{Z}$.
Still, I have another question. Can we prove ${\lim\limits_{x \to +\infty}xf(x)lnx}=0$ with the condition that $\int_a^{+\infty}f(x)dx$ is finite and $f(x)$ is monotonic? I'm just unable to find a counterexample.
Take $$f(x)=e^{-x}$$
$f$ is decreasing at $ [0,+\infty) $ and
$$\int_0^{+\infty}e^{-x}dx=1$$
but
$$(xf)'(x)=(1-x)e^{-x}$$
so $x\mapsto xf(x)$ is not monotonic .