= Solution
The <function> is continuous for $x>0$, hence <Lebesgue measurable>. On $(0,1)$, $|\sin x|\leq x$ and $|\cos x|\leq1$ give
$$
|f(x)|\leq\frac{x}{x\sqrt x}+\frac1{\sqrt x}=2x^{-1/2}.
$$
Since $\int_0^12x^{-1/2}\,dx=4$, \b[$f$ is <Lebesgue integrable> on $(0,1)$] by <power singularity integrability>.
On $(1,\infty)$, the first summand is <Lebesgue integrable>:
$$
\int_1^\infty\frac{|\sin x|}{x^{3/2}}\,dx\leq\int_1^\infty x^{-3/2}\,dx=2.
$$
The second summand is not. To prove it, take the pairwise <disjoint> <intervals>
$$
I_k=\left[k\pi-\frac\pi3,\ k\pi+\frac\pi3\right],\qquad k\geq1.
$$
On $I_k$, $|\cos x|\geq1/2$, the length is $2\pi/3$, and $x\leq(k+1)\pi$. Hence
$$
\int_{I_k}\frac{|\cos x|}{\sqrt x}\,dx
\geq\frac{\pi}{3\sqrt{(k+1)\pi}}
=\frac{\sqrt\pi}{3\sqrt{k+1}}.
$$
The <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. Therefore
$$
\boxed{f\in L^1((0,1)),\qquad f\notin L^1((1,\infty)).}
$$
This 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.
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