Sine integral with an exponential denominator
= Sine integral with an exponential denominator
{title2=$\int_0^\infty\frac{\sin(ax)}{e^x-1}\,dx=\sum_{n\ge1}\frac a{a^2+n^2}$}
For real $a$, the geometric-series expansion of $1/(e^x-1)$ gives an integrable <series> after <multiplication> by $\sin(ax)$: each absolute <integral> is bounded by $|a|\int_0^\infty xe^{-nx}dx=|a|/n^2$. <Termwise integration> is therefore justified and gives the <convergent series> $\sum_{n\ge1}a/(a^2+n^2)$.