Improper integral converges then there exists a sequence $x_n\to \infty$ such that $f(x_n)\to 0$.

Question

Let $f:[0,\infty)\to [0,\infty)$ be a continuous function such that the integral $$ \int_0^\infty f(x)dx $$ converges. Then prove that

1. There exists a sequence $x_n\to \infty$ such that $f(x_n)\to 0$
2. If $f$ is decreasing function then $f(x)\to 0$ as $x\to \infty$.

For the first part, I proceed through contradition: If possible, let for every sequence $x_n\to \infty,\ f(x_n)\to \infty$. We have $$ \left|\lim_{t\to \infty} \int_0^tf(x)dx \right|<\infty . $$Intituively it is clear that we will get some contradiction. But I am unable to formalize it.

Answer 1

Since $\int_0^\infty f(x)dx$ converges, $\int_n^{n+1}f(x)dx$ converges to $0$. By the mean value theorem for integrals, there exists $x_n\in (n,n+1)$ such that $f(x_n) = \int_n^{n+1}f(x)dx$.

Then $x_n\to \infty$ and $\lim_n f(x_n) = 0$.

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If $f$ is decreasing, the existence of such a sequence implies that $f$ goes to $0$ at $\infty$.

Answer 2

Since $f$ is non-negative, $\liminf_{x\to\infty}f(x)\ge0$. If there is no sequence $x_n\to\infty$ such that $f(x_n)\to0$, then $\liminf_{x\to\infty}f(x)>0$. This implies that $f(x)\ge\epsilon$ for all $x$ large enough and some $\epsilon>0$. This implies that the integral is not convergent.