Proving that $\int_{1}^{\infty} \frac{\sin^{2}(3x)}{x} dx$ diverges

Question

I have to prove that $$\int_{1}^{\infty} \frac{\sin^{2}(3x)}{x}\,dx$$ diverges.

Can anyone give a hand? I'm totally stuck.

Answer 1

We have, by integration by parts:

$$ \int_{1}^{M}\frac{\sin^2(3x)}{x}\,dx = \left.\left(\frac{1}{2}-\frac{\sin(6x)}{12 x}\right)\right|_{1}^{M}+\int_{1}^{M}\left(1-\frac{\sin(6x)}{6x}\right)\,\frac{dx}{2x}$$ but $1-\frac{\sin 6x}{6x}$ is trivially greater than $\frac{5}{6}$ for any $x\geq 1$ and $$\frac{5}{12}\cdot\lim_{M\to +\infty}\int_{1}^{M}\frac{dx}{x} = +\infty.$$

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Actually the integral diverges like $\frac{1}{2}\log(M)>\frac{5}{12}\log(M)$, since $\int_{1}^{+\infty}\frac{\cos(6x)}{x}\,dx$ is convergent by Dirichlet's test (integral version), but no matter the constant, the main term of the asymptotics is $\log(M)$ as expected.

Answer 2

Note that the integrand is positive, so that we might restrict the domain of integration and we know that the original integral will be at least as big. Now, $$\sin^2(3x)\geq \frac{1}{2}\iff |\sin x |\ge\frac{\sqrt2}{2}\iff x\in [\pi/4+k\pi,\frac{3\pi}{4}+k\pi],\quad k\in\mathbb{Z}.$$ Thus $$\int_1^\infty\frac{\sin^2(3x)}{x}\geq\frac{1}{2}\sum_{k=0}^{\infty}\int_{[\frac{\pi}{4}+k\pi,\frac{3\pi}{4}+k\pi]}\frac{1}x,$$ which can be easily shown to diverge as follows: \begin{align*}\sum_{k=0}^{N}\int_{[\frac{\pi}{4}+k\pi,\frac{3\pi}{4}+k\pi]}\frac{1}xdx& -\sum_{k=0}^{N-1}\int_{[\frac{\pi}{4}+k\pi,\frac{3\pi}{4}+k\pi]}\frac{1}xdx\\& = \int_{[\frac{\pi}{4}+N\pi,\frac{3\pi}{4}+N\pi]}\frac{1}xdx\\& \geq \underbrace{(\frac{3\pi}{4}+N\pi+1-(\frac{\pi}{4}+N\pi))}_{=\frac{\pi}{2}}\cdot\min_{x\in [\pi/4+N\pi,\frac{\pi}{4}+N\pi+1]}\frac{1}{x}\\ & = \frac{\pi}{2}\frac{1}{\frac{\pi}{4}+N\pi+1},\end{align*} which diverges like the harmonic series.

Answer 3

Write it as $$\int_3^\infty\frac{\sin^2x}x\,dx$$

And observe that $$\int_3^\infty\frac{\sin ^2x}x\,dx\ge\sum_{n\pi\ge 3}\int_{n\pi}^{(n+1)\pi}\frac{\sin^2x}{x}\,dx\ge\\\ge\sum_{n\pi\ge 3}\int_{n\pi}^{(n+1)\pi}\frac{\sin^2x}{(n+1)\pi}\,dx=\sum_{n\pi\ge 3}\frac{\int_0^\pi\sin^2x\,dx}{(n+1)\pi}=\infty$$

Answer 4

By a change of variable, $x=u-{\pi\over2}$, and a couple of trig identities,

$$I=\int_1^\infty{\sin^23x\over x}dx=\int_{1+{\pi\over2}}^\infty{\sin^23(u-{\pi\over2})\over u-{\pi\over2}}du=\int_{1+{\pi\over2}}^\infty{\cos^23u\over u-{\pi\over2}}du=\int_{1+{\pi\over2}}^\infty{1-\sin^23x\over x-{\pi\over2}}dx$$

Consequently

$$2I=2\int_1^{1+{\pi\over2}}{\sin^23x\over x}dx-{\pi\over2}\int_{1+{\pi\over2}}^\infty{\sin^23x\over x(x-{\pi\over2})}dx+\int_{1+{\pi\over2}}^\infty{1\over x-{\pi\over2}}dx$$

The first two integrals on the right are convergent, the third is divergent.