= Solution
At each $x>0$, eventually $x<n$, and the stipulated power limit gives $f_n(x)\to e^{-x}\sin x$. At nonpositive $x$ the functions vanish. On their positive support,
$$
|f_n(x)|\le(1-x/n)^n\le e^{-x}.
$$
The supplied integrable exponential therefore dominates the sequence. Dominated convergence gives
$$
\lim_n\int f_n=\int_0^\infty e^{-x}\sin x\,dx=\operatorname{Im}\frac1{1-i}=\boxed{\frac12}.
$$
The final improper integral is absolutely convergent, so agrees with its <Lebesgue integral>.
Back to article page
