Sine integral with an exponential denominator (${\int }_0^{\infty }\frac{\sin (ax)}{e^x-1}dx={\sum }_{n\ge 1}\frac{a}{a^2+n^2}$)

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 $|a|{\int }_0^{\infty }x{e}^{-nx}dx=|a|/n^2$. Termwise integration is therefore justified and gives the convergent series ${\sum }_{n\ge 1}a/(a^2+n^2)$.