Prove that $\int_{0}^{\infty} |\sin x^2|\,dx$ doesn't converge. I want to prove that $I=\int_{0}^{\infty} |\sin x^2|\,dx$ doesn't converge. It is easy to see that $I=\sum_{i=0}^{\infty} a_n$, where $a_n=\int_{\sqrt{n\pi}}^{\sqrt{(n+1)\pi}} |\sin x^2|\,dx>\frac{\sqrt{(n+1)\pi}-\sqrt{n\pi}}{2}=b_n$(the area of the triangle with base=$(\sqrt{(n+1)\pi}-\sqrt{n\pi})$ and $h=1$.
Then $\sum a_n > \sum b_n$, but $\sum b_n$ converges, so I don't get any conclusion. Does anyone have other idea? Thank you.
 A: For any $k\in\mathbb{N}^*$ we have:
$$\begin{eqnarray*}\int_{0}^{\sqrt{\pi k}}|\sin x^2|\,dx &=& \frac{1}{2}\int_{0}^{\pi k}\frac{|\sin x|}{\sqrt{x}}\,dx\\ &=&\frac{1}{2}\int_{0}^{\pi}|\sin x|\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x+\pi}}+\ldots+\frac{1}{\sqrt{x+(k-1)\pi}}\right)\,dx\\&\geq&\frac{1}{\pi}\int_{0}^{\pi/2}\frac{kx}{\sqrt{x+(k-1)\pi}}\,dx\geq\frac{k}{\pi\sqrt{(k-1/2)\pi}}\cdot\frac{\pi^2}{8}\\&\geq&\frac{\sqrt{(k-1/2)\pi}}{8},\end{eqnarray*}$$
ensuring divergence.
A: Here is the idea that you asked for: Re-evaluate your opinion that the sum of bn converges. Trivially, bn is roughly some constant divided by the square root of n. And adding bn from 1 to N is roughly some constant times the square root of N. 
Instead of integrating |$\sin x^2$|, integrate the constant 1. Subdivide the interval [0, inf] in exactly the same way as you did. Each sub-integral is equal to $2 b_n$ instead of $b_n$. That should make it obvious that your idea that the sum of $b_n$ converges cannot be right.
