# Confusion about convergence of series and improper integral

As it is known that $$a_n\to 0$$ is a necessary condition for a series $$\sum_1^\infty a_n$$ to converge, so does there exist a somewhat similar requirement of the integrand for convergent improper integral?

It clear doesn't hold that $$\int f(x)dx$$ need $$f(x)\to 0$$ to converge. For example, $$\int_r^1\frac{1}{\sqrt x}\to 2$$ and $$\frac{1}{\sqrt x}\to \infty$$ when $$r\to 0$$. But it is kind of like that $$F(x)=\int^1_r \frac{1}{\sqrt x}$$, $$F(x)$$ is constantly increasing and should goes to $$\infty$$ when $$r\to 0$$, since $$F'(x)=\frac{1}{\sqrt{x}}$$ monotonically goes to $$\infty$$, but why $$F(x)$$ is actually convergent beside the calculation showing it does?

And as for $$\int^r_0 f(x), r\to\infty$$, a integral is defined on $$[1, \infty)$$. Does such integrals require that $$f(x)\to 0$$ when $$r\to\infty$$ to converge?

• On thing that holds is that if $\lim_{\alpha\to\infty}\int_0^\alpha f(x)\,dx$ converges, then $\limsup\limits_{x\to\infty} f(x)\ge0$ and $\liminf\limits_{x\to\infty}f(x)\le 0$.
– user239203
Aug 9, 2020 at 20:33
• Confront with the Fresnel integral $\int_0^\alpha\sin x^2\,dx\stackrel{\alpha\to\infty}\longrightarrow \sqrt{\frac2\pi}$.
– user239203
Aug 9, 2020 at 20:37
• Same with $\int_{1}^{+\infty}\cos x^2 \,dx$ Aug 9, 2020 at 20:45
• Also $\int_0^\infty x \sin(x^3)\; dx = \sqrt{3} \Gamma(2/3)/6$. Aug 9, 2020 at 21:15
• @Gae.S. Thanks, just wondering is there a name for the theorem states that $\limsup\limits_{x\to\infty} f(x)\ge0$ and $\liminf\limits_{x\to\infty}f(x)\le 0$.
– user533661
Aug 9, 2020 at 21:29

If $$f: \mathbb{R} \to \mathbb{R}$$ is nonnegative and uniformly continuous on $$(1, \infty)$$ and $$\int_1^\infty f(x) dx$$ converges, then $$\lim_{x \to \infty} f(x) = 0$$.
Assume for contradiction that $$\lim_{x \to \infty} f(x) \neq 0$$. Then by definition there exists $$\epsilon \gt 0$$ such that: for every $$m \gt 0$$ we have some $$x \gt 0$$ satisfying $$x \gt m$$ and $$f(x) \ge \epsilon$$. Thus we can construct a strictly increasing sequence $$1 \lt x_1 \lt x_2 \lt x_3 \lt ...$$ with $$x_n \to \infty$$ as $$n \to \infty$$ and $$f(x_n) \ge \epsilon$$ for $$n = 1, 2, 3,...$$. Moreover, we can construct the sequence such that $$x_n + 1 \lt x_{n + 1}$$ for $$n = 1, 2, 3,...$$. Now by uniform continuity there is some $$\delta \gt 0$$ with $$\delta \lt \frac{1}{2}$$ such that $$\lvert f(t) - f(x_n) \rvert \le \frac{1}{3}\epsilon$$ for every $$t \in (x_n - \delta, x_n + \delta)$$ and for $$n = 1, 2, 3,...$$. Thus $$f(t) \ge f(x_n) - \lvert f(t) - f(x_n) \rvert \ge \epsilon - \frac{1}{3}\epsilon = \frac{2}{3}\epsilon$$ for every $$t \in (x_n - \delta, x_n + \delta)$$ and for $$n = 1, 2, 3,...$$. Hence $$\int_{x_n - \delta}^{x_n + \delta} f(t) dt \ge \int_{x_n - \delta}^{x_n + \delta} \frac{2}{3}\epsilon dt =\frac{4}{3}\epsilon\delta$$ for $$n = 1, 2, 3,...$$. Hence $$\int_1^\infty f(x) dx \ge \sum_{n = 1}^{\infty} \int_{x_n - \delta}^{x_n + \delta} f(t) dt \ge \sum_{n = 1}^{\infty} (\frac{4}{3}\epsilon\delta)$$ which clearly do diverge, a contradiction. Hence $$\lim_{x \to \infty} f(x) = 0$$ as desired. QED.
Just making the n-th triangle having area $${(\frac{1}{2})}^n$$.