$\int_{0}^{\infty}\frac{\sin^2(x)}{x}\,dx$ how to analyze it?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

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