# Limit of functions and improper integral

As a criterion for the existence of the limit of a function we have:

$$\lim\limits_{x\to a} f(x)$$ exists iff for each sequence $$(a_n)$$ with $$\lim\limits_{n\to \infty}a_n=a$$ the limit $$\lim\limits_{n\to \infty}f(a_n)$$ exists.

Let's consider an improper integral $$\int\limits_a^{b}f(t)dt$$ where $$a and $$b\in\mathbb{R}\cup\{\infty\}$$.

The improper integral exists iff $$f(t)$$ is Riemann integrable on $$[a,\beta]$$ for each $$\beta$$ with $$a<\beta and the limit $$\lim\limits_{\beta\to b} \int\limits_a^{\beta}f(t)dt$$ exists. In fact, $$\lim\limits_{\beta\to b} \int\limits_a^{\beta}f(t)dt$$ can be interpreted as a limit of a function, let's call it $$F(\beta):=\int\limits_a^{\beta}f(t)$$. Hence, it must be possible to prove the existence of $$\lim\limits_{\beta\to b}F(\beta)$$ by the above criterion. However, in our lecture it says that in the context of improper integrals it is sufficient to find only one sequence such that $$\lim\limits_{n\to \infty} F(\beta_n)=\lim\limits_{n\to \infty} \int\limits_a^{\beta_n}f(t)dt$$ exists.

Why is it sufficient to show it only for one sequence?

• Maybe you're working under the hypothesis $f\ge0$.
– user239203
Jan 15, 2021 at 17:16
• @Gae.S. No, there are no restrictions imposed on $f$. Jan 15, 2021 at 17:33

For a counterexample, take $$f(x) = \cos x$$, $$a = 0$$, $$b = \infty$$, and $$\beta_n = n\pi$$.

We have $$\beta_n \to \infty$$, and

$$\lim_{n \to \infty}\int_a^{\beta_n}f(x) \, dx = \lim_{n \to \infty}\int_0^{n\pi} \cos x\,dx = \lim_{n \to \infty}(\sin n\pi - \sin 0) = 0$$

However, the improper integral $$\displaystyle\int_0^\infty \cos x \, dx$$ does not converge, since $$\underset{\beta \to \infty}\lim \sin \beta$$ does not exist.

If $$f \geqslant 0$$, then $$F(\beta)$$ is nonnegative and nondecreasing. Suppose there exists a sequence $$\beta_n$$ where $$\beta_n \to \infty$$ and

$$\lim_{n \to \infty} F(\beta_n) = \lim_{n \to \infty}\int_a^{\beta_n} f(x) \, dx = I$$

For any $$\epsilon > 0$$, there exists $$N$$ such that $$|F(\beta_n) - I| < \epsilon$$ for all $$n \geqslant N$$. If $$\beta > \beta_N$$, then since $$\beta_n \to \infty$$, there exists $$m > N$$ such that $$F(\beta_N) \leqslant F(\beta) \leqslant F(\beta_m),$$ and

$$-\epsilon < F(\beta_N) - I < F(\beta)-I < F(\beta_m)- I < \epsilon$$

Thus,

$$\lim_{\beta \to \infty} F(\beta) = \lim_{\beta \to \infty}\int_a^{\beta} f(x) \, dx = I$$

• Ah ok, then there is a mistake in the lecture notes. There should be imposed an additional property on $f$, e.g. $f\geq 0$ as @Gae. S pointed out. Right? Jan 15, 2021 at 21:58
• Yes - correct. I added a proof that convergence for one sequence suffices when $f \geqslant 0$.
– RRL
Jan 15, 2021 at 22:17
• It feels like that it is necessary that the sequence $(\beta_n)$ must go to $\infty$. Why is that so? Why is it not possible that $b\in\mathbb{R}$? Or is it? Jan 16, 2021 at 19:58
• By definition $\int_a^b f(x) \, dx$ is the limit from the left, that is $\int_a^b f(x) \, dx = \lim_{\beta \to b-}\int_a^\beta f(x) \, dx$ along with the obvious requirement that $f$ is R-integrable on $[a,\beta]$ for all $a < \beta < b$. How do you define it as $\lim_{\beta \to b+} \int_a^\beta f(x) \, dx$ where the interval $[a,\beta]$ includes $b$ where there is a singularity?
– RRL
Jan 16, 2021 at 20:29
• So please consider only $\beta_n = b - 1/n$ where everything could be OK as you say.
– RRL
Jan 16, 2021 at 20:30