Why is it allowed to write this way, $$\int_{0}^{m} \sum_{k=1}^{\infty}\frac{1}{2\pi k}\frac{\sin(2\pi k t)}{t^{s+1}}\,\mathrm{d}t=\sum_{k=1}^{\infty}\frac{1}{2\pi k}\int_{0}^{m} \frac{\sin(2\pi k t)}{t^{s+1}} \,\mathrm{d}t?$$ The author wrote that it follows from Lebesgue's dominated convergence theorem, but I don't really see how it satisfies the theorem.
Note: it is proven that $$\sum_{k=1}^{n}\frac{\sin(k\theta)}{k}$$ is uniformly bounded for all $\theta\in\mathbb{R}$.
Consider the sequence of functions $$f_n(t) = \frac{1}{2\pi t^{s + 1}}\sum \limits_{k = 1}^n \frac{\sin(2\pi kt)}{k}.$$
As you've said, the sum $\sum \limits_{k = 1}^n \frac{\sin(2\pi kt)}{k}$ is uniformly bounded by a constant $M$. This yields $|f_n(t)| \le \frac{M}{2\pi} |t|^{-(\Re(s)+1)}$, which gives an integrable upper bound for $f_n$. Now the dominated convergence theorem yields the assertion.
Edit: More precisely, write $g_k(t) = \frac{\sin(2\pi k t)}{2\pi k t^{s + 1}}$. Since $\frac{M}{2\pi} |t|^{-(\Re(s)+1)}$ is an integrable function on the interval $[0, m]$, the dominated convergence theorem yields $$\begin{align*} \int \limits_0^m \sum \limits_{k = 1}^\infty g_k(t) dt &= \int\limits_0^m \lim \limits_{n \to \infty} \sum \limits_{k = 1}^n g_n(t) dt = \int\limits_0^m \lim \limits_{n \to \infty} f_n(t) dt = \lim \limits_{n \to \infty} \int\limits_0^m f_n(t) dt\\ &= \lim \limits_{n \to \infty} \int_0^m \sum \limits_{k = 1}^n g_k(t) dt \end{align*}$$