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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.


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

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

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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,

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

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  • $\begingroup$ 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? $\endgroup$
    – Philipp
    Commented Jan 15, 2021 at 21:58
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    $\begingroup$ Yes - correct. I added a proof that convergence for one sequence suffices when $f \geqslant 0$. $\endgroup$
    – RRL
    Commented Jan 15, 2021 at 22:17
  • $\begingroup$ 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? $\endgroup$
    – Philipp
    Commented Jan 16, 2021 at 19:58
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    $\begingroup$ 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? $\endgroup$
    – RRL
    Commented Jan 16, 2021 at 20:29
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    $\begingroup$ So please consider only $\beta_n = b - 1/n$ where everything could be OK as you say. $\endgroup$
    – RRL
    Commented Jan 16, 2021 at 20:30

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