Does this intergal $\int_0^\infty\frac{ \sin^2 x} x\,dx$ converge? I'm going to make use of the series $\displaystyle \sum_{n=0}^\infty \frac 1{n+1}$.
and that $\displaystyle \int_0^\infty \frac{ \sin^2 x} x \, dx = \sum_{n=0}^\infty\int_{n\pi}^{(n+1)\pi} \frac{ \sin^2 x} x \,dx$
If I use variable substitution $t=x-n\pi$ it gives
$$\tag 1 \sum_{n=0}^\infty\int_0^\pi  \frac{\sin^2 t }{n\pi+t}dt  $$
gives  
$$\tag 2\frac 1 \pi\sum_{n=0}^\infty\int_0^\pi \frac{\sin^2 t }{n+1} \, dt$$
$$\tag 3 \frac 1 \pi \int_{n=0}^\pi \sin^2 t \;dt\cdot \sum_{n=0}^\infty \frac 1 {1+n}$$
I don't really know how to explain this or what i have done. If someone knows how to solve this.
 A: First note that in the interval $[n \pi, (n+1) \pi)$, where $n > 0$, we have that $\dfrac{\sin^2(x)}x \geq \dfrac{\sin^2(x)}{(n+1) \pi}$.
This is because $\dfrac1x$ is a decreasing function and hence in the interval $[n \pi, (n+1) \pi)$, we have $$\dfrac1x > \dfrac1{(n+1) \pi}$$
Hence,
$$\int_{n \pi}^{(n+1) \pi} \dfrac{\sin^2(x)}x dx > \int_{n \pi}^{(n+1) \pi} \dfrac{\sin^2(x)}{(n+1) \pi} dx = \dfrac1{n \pi} \dfrac{\pi}2 = \dfrac1{2(n+1)}$$
Hence, $$\int_{0}^{\infty} \dfrac{\sin^2(x)}x dx = \int_{0}^{\pi} \dfrac{\sin^2(x)}x dx + \int_{\pi}^{2\pi} \dfrac{\sin^2(x)}x dx + \int_{2\pi}^{3\pi} \dfrac{\sin^2(x)}x dx + \cdots$$
Hence,
$$\int_{0}^{\infty} \dfrac{\sin^2(x)}x dx = \lim_{k \to \infty} \sum_{n=0}^{k} \int_{n \pi}^{(n+1) \pi} \dfrac{\sin^2(x)}x dx$$
We showed at the beginning of the post that
$$\int_{n \pi}^{(n+1) \pi} \dfrac{\sin^2(x)}x dx > \dfrac1{2(n+1)}$$
Hence, putting these two together, we get that
$$\int_{0}^{\infty} \dfrac{\sin^2(x)}x dx > \dfrac12\lim_{k \to \infty} \sum_{n=0}^{k} \dfrac1{n+1}$$
The series on the right side is called the harmonic series and look here to figure out why it diverges.
