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Assign any value, for example zero, at $(0,0)$, where the displayed quotient is undefined. A single point is a Lebesgue null set, so this choice does not affect Lebesgue integrability. Elsewhere the function is continuous, hence it is Lebesgue measurable on the plane after this extension.

In polar coordinates, $x^2+y^2=r^2$ and the area element is $rdrd\theta$. Applying the nonnegative change of variables formula to the absolute value givesFor $0<r\le 1$, $|\sin (r^2)|\le r^2$, so the radial integrand is bounded by $r^{3-2\alpha }$. This is integrable at zero when $\alpha <2$. For $r\ge 1$, $|\sin (r^2)|\le 1$, giving the integrable tail bound $r^{1-2\alpha }$ when $\alpha >1$. In particular,ThereforeEquivalently, the substitution rule $u=r^2$ reduces its exact L1 norm toThis is radial sine-power kernel integrability, reducing a two-dimensional question to the sine power kernel integrability thresholds at zero and infinity.

The function is continuous for $x>0$, hence Lebesgue measurable. Near zero, the bound $|\sin x|\le x$ givesThus $f$ is Lebesgue integrable on $(0,1]$ for $q<2$. The convergence is an instance of power singularity integrability.

For the tail, take the pairwise disjoint intervalsOn each $I_k$, $|\sin x|\ge 1/2$, $x\le (k+1)\pi$, and the interval length is $2\pi /3$. ThereforeSumming the nonnegative integrals over these disjoint intervals yieldsby divergence of the p-series. Hence $f$ is not Lebesgue integrable on $[1,\infty )$ for $0<q\le 1$.

The test is absolute integrability: oscillatory sign cancellation does not make the absolute-value integral finite. A conditionally convergent improper integral is a different notion. The sine power kernel integrability criterion separates these endpoint thresholds.

Both functions are continuous on the positive domains in question, so they are measurable. For the first function, the decreasing substitution $t=1/x$ and $|dx/dt|=t^{-2}$ giveFor every integer $k\ge 1$,Summing this lower bound gives a divergent harmonic series. Thus $f$ is not Lebesgue integrable on $(0,1)$.

For the second function, $|\sin x|\le x$ on $(0,1)$ and $|\sin x|\le 1$ everywhere giveTherefore $g$ is Lebesgue integrable on $(0,\infty )$. The sine power kernel integrability threshold is $1<p<2$: one endpoint requires $p<2$, and the other requires $p>1$. The signed improper integral for $f$ does converge by the Dirichlet test for improper integrals, but conditional improper convergence does not give a finite integral of $|f|$.

Assign an arbitrary finite value at the origin. Polar integration and the substitution rule $u=r^2$ reduce the joint L1 norm exactly to $\pi {\int }_0^{\infty }|\sin u|u^{-\alpha }du$. Near zero, $|\sin u|\le u$ gives $u^{1-\alpha }$; at infinity, $|\sin u|\le 1$ gives $u^{-\alpha }$. These are integrable when $\alpha <2$ and $\alpha >1$, respectively. Their explicit integrals give the displayed bound. This is a planar version of sine power kernel integrability, with the radial area element determining the exponents.