Does $\int_0^\infty \sin^2 (x^2)\, dx$ converge or diverge? I'm trying to show determine if $\int_0^\infty \sin^2(x^2)\,dx$ converges. By continuity, we have that $\sin^2(x^2)$ is continuous on $[0,1]$, and therefore (by a theorem) it is Riemann integrable on $[0,1]$. And so, we will have that if $\int_1^\infty \sin^2(x^2) \,dx$ converges then so will $\int_0^\infty \sin^2(x^2) \, dx$.
And so I'm left with $\int_0^\infty \sin^2(x^2) \,dx$ and I have no idea how to integrate this. Hints or help would be very much welcomed!
 A: Like you noted, it is enough to investigate whether $\int_1^{\infty} \sin^2(x^2) \, dx$ converges. Performing the substitution $u = x^2$, we are lead to the following integral:
$$ \frac{1}{2} \int_1^{\infty} \frac{\sin^2(u)}{\sqrt{u}} \, du = \frac{1}{4} \int_1^{\infty} \frac{1 - \cos(2u)}{\sqrt{u}} \, du. $$
Now, the integral
$$ \int_1^{\infty} \frac{\cos(2u)}{\sqrt{u}} \, du $$
converges by Dirichlet's test and since $\int_1^{\infty} \frac{1}{\sqrt{u}} \, du$ diverges, the original integral diverges.
A: Heuristically, when $x^2$ is large, $\sin^2(x^2)>\frac12$ about half of the time, and it is never negative. So for large $t$, $\int_0^t \sin^2(x^2)\,dx$ will grow at an asymptotic rate of at least $+\frac 14$, and therefore it diverges for $t\to+\infty$.
A: $$I=\int_0^\infty\sin^2(x^2)\,dx = \int_0^\infty \frac{\sin^2(z)} {2\sqrt{z}} \, dz = \int_0^\infty \frac1 {2\sqrt{z}} \,dz + \underbrace{\Re\int_0^\infty \frac{e^{2 iz}}{2\sqrt{z}} \, dz}_{\text{finite}}$$
which is clearly divergent (the convergence of the second part after the last equality sign is, for example, an easy application of Cauchy's theorem
)
A: HINT:
First enforce the substitution $u=x^2$.  Then, we have
$$\int_0^L \sin^2(x^2)\,dx=\int_0^{L^2} \frac{\sin^2(u)}{2\sqrt u}\,du=\int_0^{L^2}\frac{1-\cos(2u)}{4\sqrt u}\,du$$
Use the Abel-Dirichlet test to show that $\int_0^{L^2}\frac{\cos(u)}{\sqrt u}\,du$ converges and hence conclude that the integral of interest diverges.
A: Changing variables $x^2 \mapsto t$, the integral becomes
$$
\int_0^\infty \frac{\sin^2 t}{2\sqrt{t}} dt \, . 
$$
The integrand is $\ge \frac{1}{4\sqrt{k\pi}}$ whenever $t \in ((k-3/4)\pi, (k-1/4)\pi)$. Since $\sum_k \frac{1}{\sqrt{k}}$ diverges, so does the integral.   
