# Improper integral $\int_0^\infty \cos(x^2)$ exists but $\cos(x^2)$ is not Lebesgue integrable

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.

• Lebesgue integrability requires $\int_0^\infty|\cos(x^2)|\,dx\lt\infty$. – Barry Cipra Nov 22 '18 at 15:14
• @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. – D. Brito Nov 22 '18 at 16:00
• 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. – Barry Cipra Nov 22 '18 at 16:05
• 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 – Masacroso Nov 22 '18 at 16:22

This is a standard result. A change of variable gives the equivalent integral $$\frac12\int_0^\infty\frac{\cos u}{\sqrt{u}}\ du.$$
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 $$\int_{0}^\infty\left\vert\frac{\cos(u)}{u}\right\vert\ du=\infty$$
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 $$M\to +\infty$$.