I have seen many questions about almost the same integral; the only difference is the absolute value. How can I check if this integral converges (as real integral)?
$\displaystyle\int_0^\infty |\sin (x^2)| \; dx$
EDIT: I've read this answer in topic you linked, and I don't fully understand the steps in:
Now we observe that $|\frac{\cos(x)}{2 \sqrt x}|\geq \frac{\cos^2(x)}{2 \sqrt x}=\frac{1}{4\sqrt x}+\frac{\cos(2x)}{4\sqrt x}$.
And this, which arguments?
By the same arguments above $\displaystyle\int_{0}^{\infty}\frac{\cos(2x)}{4\sqrt x}dx$ converges.
Consider the inequation
$$|\sin x^2|\ge\frac1{\sqrt 2},$$
which is true in the intervals
$$\left[\sqrt{k\frac\pi2-\frac\pi4},\sqrt{k\frac\pi2+\frac\pi4}\right]$$ for odd $k$.
This establishes the bounding
$$\int_0^\infty|\sin x^2|dx\ge{\frac{\sqrt\pi}2}\sum_{\text{odd }k }\left(\sqrt{k+\frac12}-\sqrt{k-\frac12}\right).$$
Then
$$\sqrt{k+\frac12}-\sqrt{k-\frac12}=\frac1{\sqrt{k+\dfrac12}+\sqrt{k-\dfrac12}}$$ decays like $\dfrac1{\sqrt k}$, making a divergent sum.
The plot below shows the area that diverges.