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