Test for divergence of $\int_{0}^{\infty} \frac{\sin^2(x)}{x}dx$ without evaluating the integral

I would like to prove that $$\int_{0}^{\infty} \frac{\sin^2(x)}{x}dx$$ diverges without actually evaluating the integral. Is there a convergence test from calculus or real analysis that can show that this integral diverges?

Thanks.

Edit: Someone pointed out that this is a possible duplicate. However, the question put forth as a possible duplicate asks about $\sin(x^2)$, not about $\sin^2(x)$.

It is a divergent integral by Kronecker's lemma, since $\sin^2(x)$ is a non-negative function with mean value $\frac{1}{2}$. In more explicit terms, by integration by parts we have

$$\int_{\pi}^{N\pi}\frac{\sin^2(x)}{x}\,dx =\color{blue}{\left[\frac{1}{2}-\frac{\sin(2x)}{4x}\right]_{\pi}^{N\pi}}+\color{red}{\frac{1}{2}\int_{\pi}^{N\pi}\frac{dx}{x}}+\color{blue}{O(1)}$$ where the blue terms are bounded, but the red term equals $\frac{1}{2}\log N$.

• I don't understand why you choose $\pi$ and $N\pi$ as the limits of integration. – user484604 Dec 4 '17 at 19:34
• @Taliant: aesthetics only, but any other choice equally does the job. – Jack D'Aurizio Dec 4 '17 at 19:40
• So let's say $N\pi = \infty$. Can we conclude from that that the integral in red diverges, and so $\frac{\sin^2(x)}{x}$ diverges? – user484604 Dec 4 '17 at 21:36
• @Taliant: the red integral is unbounded as $N\to +\infty$, hence $\int_{0}^{+\infty}\frac{\sin^2(x)}{x}\,dx = +\infty$. – Jack D'Aurizio Dec 4 '17 at 21:43

We see there is no problem around $0$ so the problem lies in the convergence of: \begin{align} \int^\infty_M \frac{\sin^2(x)} {x} dx \end{align} For $M>0$, let's take it very large (you'll see the reason in the next line).

We prove that it diverges by a (badass) contradiction. Assume it converges. Then we know that: \begin{align} \int^\infty_{M} \frac{\sin^2(x+\pi/2)}{x+\pi/2}dx \end{align} converges too. Hence by comparison the following converges too: \begin{align} \int^\infty_{M} \frac{\sin^2(x+\pi/2)}{x}dx = \int^\infty_{M} \frac{\cos^2(x)}{x}dx \end{align} But then we get that the next one is also convergent since sum of convergent integrals is convergent: \begin{align} \int^\infty_{M} \frac{\sin^2(x)+\cos^2(x) }{x}dx =\int^\infty_M\frac{1}{x} dx \end{align} I think this is a beautiful contradiction. Hence the integral that we was considering was not convergent in the first place.

• Very clever trick ! – Gabriel Romon Dec 5 '17 at 12:12

\begin{align} \int_0^\infty\frac{\sin^2(x)}{x}\,\mathrm{d}x &=\sum_{k=1}^\infty\int_{(k-1)\pi}^{k\pi}\frac{\sin^2(x)}{x}\,\mathrm{d}x\\ &\ge\sum_{k=1}^\infty\frac1{k\pi}\int_{(k-1)\pi}^{k\pi}\sin^2(x)\,\mathrm{d}x\\ &=\sum_{k=1}^\infty\frac1{2k} \end{align}

Integrate from $x=n\pi$ to $x=(n+1)\pi$. With the squared sine being nonnegative the integrand is greater than $(\sin^2x)/((n+1)\pi)$. Conclude that the definite integral from $x=n\pi$ to $x=(n+1)\pi$ is greater than $\pi/(2(n+1))$ and then compare the full integral from $x=0$ to $x=\infty$ to the harmonic series.