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Does the following integral converge?
$$\int_1^\infty \sin^2 (x^2) \, dx$$
I tried $$\int_1^\infty \sin^2(x^2) \, dx=\int_1^\infty \frac{1-\cos(2x^2)}{2} \, dx = \frac{\sqrt{2}}{4} \int_1^\infty\frac{1-\cos(u)}{2\sqrt{u}} \, du$$
The idea is that, I want to compare the original integral to a divergent $p$-integral. But I am not sure how to proceed from here.
No, it does not. We have, due to $\sin^2(z)=\frac{1-\cos(2z)}{2}$,
$$ \int_{1}^{M}\sin^2(x^2)\,dx \stackrel{x^2\to z}{=} \int_{1}^{M^2}\frac{\sin^2(z)}{2\sqrt{z}}\,dz =\frac{M-1}{2}-\int_{1}^{M^2}\frac{\cos(2z)}{4\sqrt{z}}\,dz$$
where $\int_{1}^{+\infty}\frac{\cos(2z)}{4\sqrt{z}}\,dz$ is finite by Dirichlet's test/integration by parts, but $\frac{M-1}{2}$ grows unbounded.
In particular,
$$\int_{1}^{M^2}\frac{\cos(2z)}{4\sqrt{z}}\,dz = \left.\frac{\sin(2z)}{8\sqrt{z}}\right|_{1}^{M^2}+\int_{1}^{M^2}\frac{\sin(2z)}{16 z\sqrt{z}}\,dz$$
is bounded in absolute value by $\frac{3}{8}$ (by the trivial inequality $\sin\leq 1$), hence
$$\boxed{ \int_{1}^{M}\sin^2(x^2)\,dx \color{red}{\geq \frac{M}{2}-\frac{7}{8}}. }$$
We know that $\sin^2(x)\geq\dfrac{1}{2}$ $\text{whenever}\,\,\dfrac{\pi}{4}+n\pi\leq x\leq\dfrac{3\pi}{4}+n\pi, n\in\mathbb{N}$. Hence $$\sqrt{\dfrac{\pi}{4}+n\pi}\leq x\leq\sqrt{\dfrac{3\pi}{4}+n\pi}\Longrightarrow \sin^2(x^2)\geq\dfrac{1}{2}$$ Let $I_n=\left[\sqrt{\dfrac{\pi}{4}+n\pi},\sqrt{\dfrac{3\pi}{4}+n\pi}\right]$. Then the length of $I_n$, $n\geq 1$, is more than $\dfrac{C}{\sqrt{n}}$ for some constant $C>0$ independent of $n$. Hence: $$\int_1^{\infty}\sin^2(x^2)dx\geq\sum_{n\geq 1}^{\infty}\int_{I_n}\sin^2(x^2)dx\geq \sum_{n\geq 1}^{\infty}\dfrac{1}{2}\dfrac{C}{\sqrt{n}}$$ which is infinite.
$$\int\sin^2x^2\,dx=\frac12\int dx-\frac12\int\cos2x^2\,dx.$$ The second term is a Fresnel cosine integral, which is known to converge.
