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Show that the improper integral $\int_0^\infty \cos(x^2)$ exists but $\cos(x^2)$ is not Lebesgue integrable.

I'm asked to prove the above statement. I know that the integral is a special one, but I've not yet found a proof of its existence. And as for proving that it is not Lebesgue integrable, I don't have any idea. All tips appreciated.

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    $\begingroup$ Lebesgue integrability requires $\int_0^\infty|\cos(x^2)|\,dx\lt\infty$. $\endgroup$ – Barry Cipra Nov 22 '18 at 15:14
  • $\begingroup$ @BarryCipra Okay, any ideas how I would go about proving that it tends to infinity? I thought about 'splitting' the integral into $\sum_{n=0}^{\infty} \int_{\sqrt{n\pi /2}}^{\sqrt{(n+1)\pi /2}} |\cos x^2|$ and then seeing that each term maybe was greater than some divergent series but I do not see it. $\endgroup$ – D. Brito Nov 22 '18 at 16:00
  • $\begingroup$ The answer given by user587192 effectively does what you describe. Alternatively, see if you can show that the set on which $|\cos(x^2)|\ge{1\over2}$ has infinite Lebesgue measure. $\endgroup$ – Barry Cipra Nov 22 '18 at 16:05
  • $\begingroup$ take a look here for a generalized case about the existence of the improper integral. By the other side the comment of @Barry conclude the other part $\endgroup$ – Masacroso Nov 22 '18 at 16:22
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Convergence of the improper integral.

This is a standard result. A change of variable gives the equivalent integral $$\frac12\int_0^\infty\frac{\cos u}{\sqrt{u}}\ du.$$

See this post for the value of the improper integral: https://en.wikipedia.org/wiki/Fresnel_integral#Limits_as_x_approaches_infinity

See also the following questions:
- Definite integral of $\cos (x)/ \sqrt{x}$?
- A simple proof of the fact that $\int_0^{+\infty} \cos(x)/\sqrt{x} \text{d}x \neq 0$

Lebesgue integrability.

Consider the integrals $$ \int_0^\infty\left\vert\frac{\cos u}{\sqrt{u}}\right\vert\ du=\int_0^{\pi/2}\left\vert\frac{\cos(u)}{u}\right\vert\ du+ \int_{\pi/2}^\infty\left\vert\frac{\cos(u)}{u}\right\vert\ du. $$ For the second one, note that $$ \int_{k\pi+\frac{\pi}{2}}^{(k+1)\pi+\frac{\pi}{2}}|\cos u|\ du = 2, $$ which implies that $$ \frac{2}{a_{k+1}}\leq \int_{k\pi+\frac{\pi}{2}}^{(k+1)\pi+\frac{\pi}{2}}\left\vert\frac{\cos u}{\sqrt{u}}\right\vert\ du \leq\frac{2}{a_k} $$ where $a_k = \sqrt{k\pi+\frac{\pi}{2}}$. But $$ \sum \frac{1}{a_k}=\infty. $$ So one must have $$ \int_{\pi/2}^\infty\left\vert\frac{\cos(u)}{u}\right\vert\ du=\infty. $$ and thus $$ \int_{0}^\infty\left\vert\frac{\cos(u)}{u}\right\vert\ du=\infty $$

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The convergence of $\int_{0}^{+\infty}\frac{\cos u}{\sqrt{u}}\,du$, as an improper Riemann integral, is ensured by Dirichlet's test, since $\cos u$ has a bounded primitive and $\frac{1}{\sqrt{u}}$ is decreasing towards zero. Vice-versa, the divergence of $\int_{0}^{+\infty}\frac{\left|\cos u\right|}{\sqrt{u}}\,du$ is ensured by Kronecker's lemma, since $\left|\cos u\right|$ is a non-negative function with mean value $\frac{2}{\pi}$. In particular $\int_{0}^{M}\frac{\left|\cos u\right|}{\sqrt{u}}\,du \sim \int_{0}^{M}\frac{2\,du}{\pi\sqrt{u}}=\frac{4}{\pi}\sqrt{M}$ as $M\to +\infty$.

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