As the title says, I'm trying to justify why the (Riemann) integral $\int_1^\infty\frac{\sin(x)}{x^2}dx$ exists. I'm not interested in determining its value. It seems to me that I should use the fact that the integral $\int_1^B\frac{\sin(x)}{x^2}dx$ exists for every $B>0$, because the integrand is continuous, and the estimate $$\left|\int_1^B\frac{\sin(x)}{x^2}dx\right| \le\int_1^B\left|\frac{\sin(x)}{x^2}\right|dx \le\int_1^B\frac 1{x^2}dx<\infty.$$ However, the estimate only says that in the limit $B\rightarrow\infty$ the integral is finite if it exists, but I don't see how I conclude that the limit does exist.

  • $\begingroup$ Notice that $\lim_{B\to\infty}\frac d{dB}\int_1^B\frac{\sin(x)}{x^2}\ dx=0$. What should that tell you? $\endgroup$ – Simply Beautiful Art Apr 14 '17 at 22:08
  • 1
    $\begingroup$ What is your definition of convergence? You've proven it is absolutely convergent and this implies convergence. $\endgroup$ – Mark Viola Apr 14 '17 at 22:15
  • $\begingroup$ Absolute convergence does not imply convergence here, since I am using the Riemann integral, not the Lebesgue integral. $\endgroup$ – cthl Apr 14 '17 at 22:20
  • 2
    $\begingroup$ @cthl No, that is not correct. An improper Riemann integral that is absolutely convergent is convergent! What on earth are you talking about?? $\endgroup$ – Mark Viola Apr 14 '17 at 22:28
  • $\begingroup$ To check that absolute convergence implies convergence for the integral of Riemann it is enough to check the epsilon-delta definition of convergence of $\int|f|$ and compare with the epsilon-delta definition of convergence of $\int f$ $\endgroup$ – Masacroso Apr 14 '17 at 22:36

You should prove the theorem that if $f \colon [a, \infty) \rightarrow \mathbb{R}$ is Riemann integrable on each interval $[a,b]$ for $b > a$ and $\int_a^{\infty} |f(x)| \, dx$ exists (as an improper Riemann integral) then so does $\int_a^{\infty} f(x) \, dx$ (absolute convergence implies regular convergence).

The main idea of the proof is to use the Cauchy criterion of convergence for an improper integral. Namely, the improper integral $\int_a^{\infty} g(x) \, dx$ converges iff for every $\varepsilon > 0$ we can find $M \geq a$ such that for all $y > x > M$ we have

$$ \left| \int_x^{y} g(x) \, dx\right| < \varepsilon. $$

Assuming this criterion, if $\int_a^{\infty} |f(x)| \, dx$ exists then for any $\varepsilon > 0$ we can find $M \geq a$ such that for all $y > x > M$ we have

$$ \left | \int_x^y |f(x)| \, dx \right| = \int_x^y |f(x)| \, dx < \varepsilon $$

but then we also have

$$ \left| \int_x^y f(x) \, dx \right| \leq \int_x^y |f(x)| \, dx < \varepsilon $$

so by the Cauchy criterion, $\int_a^{\infty} f(x) \, dx$ also conveges.

  • $\begingroup$ Very well written. (+1) $\endgroup$ – Mark Viola Apr 14 '17 at 22:30
  • $\begingroup$ Thank you, this took me a while to understand. If I understand this correctly, your first statement implicitly assumes that $f$ is integrable on every interval $\left[a,b\right]$, as is the case for my original function. I didn't see this implicit assumption and was concerned about a function like $f(x)=\frac{g(x)}{x^2}$, where $g(x)=1$ if $x\in\mathbb Q$ and $g(x)=-1$ otherwise, since in this case $\int_1^\infty |f|$ exists, but $\int_1^\infty f$ doesn't. $\endgroup$ – cthl Apr 14 '17 at 22:56
  • $\begingroup$ @cthl: Yeah. When one talks about the improper Riemann integral of $f$ on $[a,\infty)$, one implicitly assumes that $f$ is Riemann integrable on $[a,b]$ for all $b > a$ (otherwise, one cannot even talk about the limit of $\int_a^b f(x) \, dx$ as $b \to \infty$) so that such pathologies cannot occur. I'll edit the answer to be more precise. $\endgroup$ – levap Apr 14 '17 at 23:01

Define $F:[1,+\infty) \to\mathbb R$ by $F(x) = \int_1^x s^{-2}\sin(s)\,ds$. You know $F$ is well defined, so we only have to prove $\lim\limits_{x\to+\infty}F(x)$ exists and is finite.

The estimate $$|F(y)-F(x)|\leq \left|\int_x^ys^{-2}\,ds\right|\leq\int_{\min\{x,y\}}^\infty s^{-2}\,ds=\frac{1}{\min\{x,y\}}$$ shows that if $x,y > M$, then $|F(x)-F(y)|<1/M$. If we take any sequence $(x_n) \to +\infty$, this implies $(F(x_n))$ is Cauchy, and therefore converges to a real number.

Now you can argue that $\lim\limits_{n\to\infty} F(y_n)$ is the same for any $(y_n)\to\infty$ and therefore that $\lim\limits_{x\to+\infty} F(x)$ exists.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.