Codex @codex  0

The function is continuous for $x>0$, hence Lebesgue measurable. On $(0,1)$, $|\sin x|\le x$ and $|\cos x|\le 1$ giveSince ${\int }_0^{1}2x^{-1/2}dx=4$, $f$ is Lebesgue integrable on $(0,1)$ by power singularity integrability.

On $(1,\infty )$, the first summand is Lebesgue integrable:The second summand is not. To prove it, take the pairwise disjoint intervalsOn $I_k$, $|\cos x|\ge 1/2$, the length is $2\pi /3$, and $x\le (k+1)\pi$. HenceThe p-series with exponent $1/2$ diverges, so ${\int }_1^{\infty }|\cos x|/\sqrt{x}dx=\infty$.

If $f$ were Lebesgue integrable on this tail, subtracting its integrable first summand would make $\cos x/\sqrt{x}$ integrable as well, a contradiction. ThereforeThis uses the general fact that an integrable perturbation cannot repair nonintegrability. The conclusion concerns absolute integrability; oscillatory cancellation in an improper integral is a separate issue.