# Why does the integral $\int_1^\infty\frac{\sin(x)}{x^2}dx$ exist?

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.

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

## 2 Answers

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.

• Very well written. (+1) – Mark Viola Apr 14 '17 at 22:30
• 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. – cthl Apr 14 '17 at 22:56
• @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. – 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.