= Solution
For fixed $n>0$, Tonelli and the allowed exponential integral give
$$
\int_0^n\int_0^\infty|\sin x|e^{-tx}\,dt\,dx=\int_0^n\frac{|\sin x|}x\,dx\leq n<\infty.
$$
The point $x=0$ is irrelevant to the integral. Fubini therefore permits reversing the signed integrals. Evaluating the elementary inner integral yields
$$
\int_0^n\frac{\sin x}x\,dx
=\int_0^\infty\frac{1-e^{-nt}(t\sin n+\cos n)}{1+t^2}\,dt.
$$
The first term has integral $\pi/2$. The absolute value of the remainder is at most
$$
\int_0^\infty e^{-nt}(t+1)\,dt=\frac1{n^2}+\frac1n\to0.
$$
Consequently \b[the improper integral tends to $\pi/2$]. The interchange was justified on each finite strip, without asserting absolute integrability on the infinite quadrant.
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