= Solution
{c}
\b[The <function> is <Lebesgue integrable>.] The inequalities $|\sin u|\leq\min\{u,1\}$ for $u>0$ give bounds that handle the two ends of the $x$-axis differently. For $0<x<1$,
$$
\int_0^\infty e^{-2xy}|\sin x\sin y|\,dy
\leq x\int_0^\infty e^{-2xy}\,dy=\frac12.
$$
For $x\geq1$,
$$
\int_0^\infty e^{-2xy}|\sin x\sin y|\,dy
\leq\int_0^\infty y e^{-2xy}\,dy=\frac1{4x^2}.
$$
<Tonelli's theorem> therefore proves
$$
\boxed{\iint_{(0,\infty)^2}e^{-2xy}|\sin x\sin y|\,dx\,dy
\leq\frac12+\int_1^\infty\frac{dx}{4x^2}=\frac34<\infty.}
$$
The factors of <sine> suppress the near-axis contributions that caused divergence in part <2/c/i>.
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