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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!

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  • $\begingroup$ Using Mathematica says it doesn't converge though! $\endgroup$ – Euler_Salter Dec 19 '16 at 2:15
  • $\begingroup$ try by substitution of $u=x^2$ then you'll probably need by parts $\endgroup$ – Euler_Salter Dec 19 '16 at 2:16
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    $\begingroup$ Since you can’t get a closed-form antiderivative of $\sin x^2$, I find it highly unlikely that this can be done by methods of Calculus I. $\endgroup$ – Lubin Dec 19 '16 at 2:20
  • $\begingroup$ @Lubin Indeed. This is was on a previous final real analysis exam question and it was asking to prove if it converged or diverged. $\endgroup$ – Nikitau Dec 19 '16 at 2:22
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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.

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    $\begingroup$ Ah, thank you so much! I cannot believe I didn't think of a u-sub. $\endgroup$ – Nikitau Dec 19 '16 at 2:26
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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 $t\to+\infty$.

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  • $\begingroup$ Just what I was thinking. If the integrand had been $\sin x^2$, that would have been a much more interesting problem. $\endgroup$ – Lubin Dec 19 '16 at 2:21
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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.

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$$I=\int_0^\infty\sin^2(x^2)\,dx = \int_0^\infty \frac{\sin^2(z)} {2\sqrt{z}} \, dz = \int_0^\infty \frac1 {2\sqrt{z}} \,dz + \underbrace{\Re\int_0^\infty \frac{e^{2 iz}}{2\sqrt{z}} \, dz}_{\text{finite}}$$

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 )

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  • $\begingroup$ Similar to mine. So (+1). And Happy Holidays! $\endgroup$ – Mark Viola Dec 19 '16 at 2:25
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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.

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