Limit of functions and improper integral

Question

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 $a<b$ and $b\in\mathbb{R}\cup\{\infty\}$.

The improper integral exists iff $f(t)$ is Riemann integrable on $[a,\beta]$ for each $\beta$ with $a<\beta<b$ 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.

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

Answer 1

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.

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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,

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