If $\lim_{x\to\infty}\int_0^x a(t)\,dt=L$, is it true that $\lim_{x\to\infty}a(x)=0?$ Let $s(x)=\int_0^xa(t)\,dt$. I wonder:
If $\displaystyle \lim_{x\to\infty}s(x)=L$ is it true that $\displaystyle\lim_{x\to\infty}a(x)=0?$

The statement is false even for continuous non-negative functions.
As an example, consider the function $a: [0, \infty) \to \mathbb R$ whose
graph consists of triangles with baseline $[n, n + 1/n^2]$ and height $2$
for each $n \in \mathbb N$ and is zero otherwise.
Then $\int_0^\infty a(x)dx = \sum_{n=1}^\infty 1/n^2 = \pi^2/6$,
but $\lim_{x \to \infty} a(x)$ does not exist.
