= Solution
Use $|\sin(\alpha x)|\le|\alpha|x$ for $x\ge0$. For the source's functions,
$$
\int_{\mathbb R}|f_n(x)|\,dx
\le|\alpha|\int_0^\infty xe^{-nx}\,dx=\frac{|\alpha|}{n^2}.
$$
The last <integral> follows by the substitution $u=nx$ and <integration by parts>. Since $\sum_n n^{-2}$ converges, (ii) permits summing the <integrals> and identifies an integrable almost-everywhere sum.
For every $x>0$, the <geometric series> gives
$$
\sum_{n=1}^\infty e^{-nx}\sin(\alpha x)
=\frac{\sin(\alpha x)}{e^x-1}.
$$
The functions vanish on $x\le0$, and the single endpoint does not change the <integral>. Use the supplied <integral> of $f_n$ to obtain
$$
\boxed{\int_0^\infty\frac{\sin(\alpha x)}{e^x-1}\,dx
=\sum_{n=1}^\infty\frac{\alpha}{\alpha^2+n^2}.}
$$
The estimates include negative $\alpha$ through its absolute value, and the case $\alpha=0$ gives zero on both sides.
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