For the non integrability in the Lebesgue sense, one can use the intervals $$
I_n=\left(\dfrac1{n\pi+3\pi/4},\dfrac1{n\pi+\pi/4}\right).
$$
On each $I_n$, $|\sin(1/t)|\geqslant1/\sqrt2$ and $t\leqslant1/(n\pi)$ hence $|\sin(1/t)/t|\geqslant n\pi/\sqrt2$. The length of $I_n$ is of order $1/n^2$ hence the integral of $|t\sin(1/t)|$ on $I_n$ is at least of order $1/n$. Summing these shows the Lebesgue integral diverges.
Proofs that $\displaystyle\int_1^x\frac{\sin(t)}t\mathrm dt$ converges when $x\to+\infty$ are all over the site.
To show that the integral $I=\int_0^\infty \frac{\cos(x)}{\sqrt{x}}\,dx$ exists as an improper Riemann integral, we proceed as follows. We express the integral of interest as the sum
$$\int_0^L \frac{\cos(x)}{\sqrt x}\,dx=\int_0^1 \frac{\cos(x)}{\sqrt x}\,dx+\int_1^L \frac{\cos(x)}{\sqrt x}\,dx\tag 1$$
The first integral on the right-hand side of $(1)$ is absolutely integrable such that $2\cos(1)\le \int_0^1 \frac{\cos(x)}{\sqrt x}\,dx\le 2$.
For the second integral on the right-hand side of $(1)$, we integrate by parts with $u=x^{-1/2}$ and $v=\sin(x)$ to reveal
$$\int_1^L \frac{\cos(x)}{\sqrt x}\,dx=\left(\frac{\sin(L)}{\sqrt L}-\sin(1)\right)+\frac12\int_1^L \frac{\sin(x)}{x^{3/2}}\,dx\tag 2$$
Inasmuch as the second integral on the right-hand side of $(2)$ absolutely converges as $L\to \infty$, the integral on the left-hand side converges (it does not absolutely converge).
Putting it together, we have shown that the integral in $(1)$ converges and exists as an improper Riemann integral.
Best Answer
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 $$ \int_{0}^\infty\left\vert\frac{\cos(u)}{u}\right\vert\ du=\infty $$