Htykuut

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.

Convergence of the improper integral.

This is a standard result. A change of variable gives the equivalent integral $$frac12int_0^inftyfrac{cos u}{sqrt{u}} du.$$

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

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 $int_{0}^{+infty}frac{left|cos uright|}{sqrt{u}},du$ is ensured by Kronecker's lemma, since $left|cos uright|$ is a non-negative function with mean value $frac{2}{pi}$. In particular $int_{0}^{M}frac{left|cos uright|}{sqrt{u}},du sim int_{0}^{M}frac{2,du}{pisqrt{u}}=frac{4}{pi}sqrt{M}$ as $Mto +infty$.