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!
Best Answer
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.