3

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.

InsideOut
  • 6,883
blazej24
  • 31

1 Answers1

2

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.

enter image description here