Let $\Sigma_n a_n(xt)^n$ be the standard Taylor series expansion of $sin(xt)$. I want to show that $\int_0^{\infty}\frac{\Sigma_n a_n(xt)^n}{e^t-1}dt=\Sigma_n\int_0^{\infty}\frac{a_n(xt)^n}{e^t-1}dt$. I think the dominated convergence theorem will do the trick, by making $|\Sigma_{n=1}^k a_n(xt)^n|$ dominated by some integrable function $g$ for all $k$. But what dominator do I use?
The obvious choice of dominator is $\frac{1}{e^t-1}$, but $\int_0^{\infty}\frac{dt}{e^t-1}$ does not converge. So what other dominator can I use?
If all you are interested in is whether the integral converges, I think we can argue as follows to say that it does: Since $\frac{\sin xt}{e^{t}-1}\to x$ as $t\to 0^+$ is suffices to consider $\int_{1}^{\infty}\frac{\sin xt}{e^{t}-1}dt,$ in which case we have
$\int_{1}^{\infty}\frac{\sin xt}{e^{t}-1}dt\le \int_{1}^{\infty}\frac{|\sin xt|}{(t+\frac{t^{2}}{2}+\frac{t^{3}}{6}+O(t^{4}))}dt=\int_{1}^{\infty}\frac{|\sin xt|}{t(1+\frac{t}{2}+\frac{t^{2}}{6}+O(t^{3}))}dt\le \int_{1}^{\infty}\frac{dt}{1+\frac{t}{2}+\frac{t^{2}}{6}}\le 6\int^{\infty}_1\frac{dt}{t^{2}}$