when I put it in the integral calculator, the result shows it diverges. I also tried $\int_{0}^{\infty}\frac{\sin^2(x)}{x}\,dx$, it diverges. $\int _{0}^{\infty}\frac{\sin^2(x)}{x^2}\,dx$ converges.$\int_{0}^{\infty}\frac{\sin^2(x)}{x^3}\,dx$ diverges. I thought there is some regularity.

And for the question in the title, how can I prove it? I tried comparison test and limit comparison test. I didn't work it out.

  • $\begingroup$ Here is a link to help you show the result for your latter question. $\endgroup$ – John Smith Mar 28 '17 at 2:01
  • $\begingroup$ The bounds of integration are different in the link I provided, but hopefully you see how one could approach this problem. The image provided by robjon is pretty useful. $\endgroup$ – John Smith Mar 28 '17 at 2:03
  • $\begingroup$ There are two regions you need to consider here: $x\to 0$ and $x\to \infty$. The integral of $\frac{\sin^2(x)}{x^3}$ diverges because of it's behavior close to $x=0$ (for example $\int_1^\infty \frac{\sin^2(x)}{x^3}$ is finite). $\endgroup$ – Winther Mar 28 '17 at 2:04

It is enough to understand if the integrand function is integrable in a right neighbourhood of the origin and in a left neighbourhood of $+\infty$. $\sin(x)^2$ behaves like $x^2$ in a right neighbourhood of the origin and it is a non-negative function with mean value $\frac{1}{2}$, hence $$ \int_{0}^{+\infty}\frac{\sin^2(x)}{x^\alpha}\,dx $$ converges as soon as $1<\alpha<3$. In such a case it equals $$-\frac{\pi\, 2^{\alpha-3}}{\Gamma(\alpha)\cos\left(\frac{\pi\alpha}{2}\right)}$$ by Euler's Beta function and the reflection formula for the $\Gamma$ function.

  • $\begingroup$ thank you! But what does $\Gamma$ stand for? $\endgroup$ – Userkkr Apr 4 '17 at 18:22
  • $\begingroup$ The Gamma function, the usual extension of the factorial to real or complex numbers, $\Gamma(z+1)=z!$. $\endgroup$ – Jack D'Aurizio Apr 5 '17 at 6:02

For $(\sin^2{x})/x$: Divide the integration range up into intervals $[n\pi,(n+1)\pi]$ for $n=0,1,2,\dotsc$. Then on such an interval, $$ \frac{\sin^2{x}}{x} \geq \frac{\sin^2{x}}{(n+1)\pi}, $$ since $1/x$ is decreasing. Then $\int_{n\pi}^{(n+1)\pi} \sin^2{x} \, dx = \pi/2$, so integrating both sides of the inequality over $[n\pi,(n+1)\pi]$, $$ \int_{n\pi}^{(n+1)\pi} \frac{\sin^2{x}}{x} \, dx \geq \frac{1}{2(n+1)}, $$ and summing up, we find that the integral is bounded below by the harmonic series, which diverges.

For $(\sin^2{x})/x^2$, first check that the integrand is bounded as $ x \downarrow 0 $ (you know $\sin{x}/x \to 1$, right?), and then use the same idea as the first example to bound the integrand above on intervals.

The last one diverges for a different reason: $(\sin^2{x})/x^3 \approx 1/x$ as $x \downarrow 0$, and the integral of $1/x$ diverges.


Note that

$$\begin{align} \int_1^L \frac{\sin^2(x)}{x}\,dx&=\frac12\int_1^L \frac{1-\cos(2x)}{x}\,dx\\\\ &=\frac12\log(L)-\frac12\int_2^{2L}\frac{\cos(x)}{x}\,dx\tag 1 \end{align}$$

From Abel's (Dirichlet's) Test for improper integrals, the integral on the right-hand side of $(1)$ converges. To show this, simply note that $\left|\int_0^L\cos(x)\,dx\right|\le 2$ and $\frac1x$ monotonically decreases.

Inasmuch as $\log(L)\to \infty$, the integral on the right-hand side cannot converge.

And that is that.


In THIS ANSWER, I analyzed convergence (conditional) of the integral $\int_0^\infty \frac{\sin^3(x)}{x}\,dx$ from which it is trivial to analyze the convergence of $\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx$

And in THIS ANSWER, I analyzed the convergence (conditional) of $\int_0^\int \frac{\sin(x)}{x}\,dx$

  • $\begingroup$ Would you please let me know how I can improve my answer? I really want to given you the best answer I can. -Mark $\endgroup$ – Mark Viola Apr 29 '17 at 4:24

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