= Solution
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 $r\,dr\,d\theta$. Applying the nonnegative <change of variables formula> to the <absolute value> gives
$$
\int_{\mathbb R^2}|k(x,y)|\,dx\,dy
=2\pi\int_0^\infty|\sin(r^2)|r^{1-2\alpha}\,dr.
$$
For $0<r\leq1$, $|\sin(r^2)|\leq r^2$, so the radial <integrand> is bounded by $r^{3-2\alpha}$. This is integrable at zero when $\alpha<2$. For $r\geq1$, $|\sin(r^2)|\leq1$, giving the integrable tail bound $r^{1-2\alpha}$ when $\alpha>1$. In particular,
$$
\begin{aligned}
\|k\|_1
&\leq2\pi\left(\int_0^1r^{3-2\alpha}\,dr+\int_1^\infty r^{1-2\alpha}\,dr\right)\\
&=\pi\left(\frac1{2-\alpha}+\frac1{\alpha-1}\right)<\infty.
\end{aligned}
$$
Therefore
$$
\boxed{k\in L^1(\mathbb R^2)\qquad(1<\alpha<2).}
$$
Equivalently, the <substitution rule> $u=r^2$ reduces its exact <L1 norm> to
$$
\|k\|_1=\pi\int_0^\infty\frac{|\sin u|}{u^\alpha}\,du.
$$
This is <radial sine-power kernel integrability>, reducing a two-dimensional question to the <sine power kernel integrability> thresholds at zero and infinity.
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