Question $10.28$, Chapter $10$, in Apostol's "Mathematical Analysis" ($2^{\text{nd}}$ edition):
a) If $s \gt 0$ and $a > 0$, show that the series $$\sum_{n=1}^{\infty}\frac{1}{n}\int_{a}^{\infty}\frac{\sin 2n\pi x}{x^{s}} dx$$ converges and prove that $$ \lim_{a \to \infty}\sum_{n=1}^{\infty}\frac{1}{n}\int_{a}^{\infty}\frac{\sin 2n\pi x}{x^{s}} dx = 0$$ b) Let $f(x) = \sum_{n=1}^{\infty} \frac{\sin 2n\pi x}{n}.$ Show that $$ \int_{0}^{\infty} \frac{f(x)}{x^{s}} dx = (2\pi)^{s-1}\zeta(2-s)\int_{0}^{\infty}\frac{\sin t}{t^{s}} dt, \quad \text{if } 0 \lt s \lt 1,$$ where $\zeta$ denotes the Reimann zeta function.
I've managed to prove part a); the difficulties, for me, lie in part b). In order to show part b), I think the integral needs to be split into two parts: $\int_{0}^{a}$ and $\int_{a}^{\infty}$ for an $a \gt 0$. Beyond that, I see that if I were to manage to justify integrating the left-hand side term by term, then the equation would follow after the appropriate substitution. But how do I justify this? I've tried, unsuccessfully, to apply the Levi Monotone convergence theorem and the Lebesgue Dominated convergence theorem. I'm now out of ideas, and would like some help.
Since the partial sums of $f(x)$ are uniformly bounded, $f(x)$ is dominated by an integrable function on $[0,a]$ for every $a<\infty$, so we can interchange summation and integration over $[0,a]$ by the dominated convergence theorem, i.e.$$\int_{0}^{a}\sum_{n=1}^{\infty}\dfrac{\sin(2n\pi x)}{nx^{s}}\,dx=\sum_{n=1}^{\infty}\int_{0}^{a}\dfrac{\sin(2n\pi x)}{nx^{s}}\,dx.$$Then use the result you have proved,$$\begin{aligned} \int_{0}^{\infty}\sum_{n=1}^{\infty}\dfrac{\sin(2n\pi x)}{nx^{s}}\,dx&=\lim_{a\rightarrow\infty}\int_{0}^{a}\sum_{n=1}^{\infty}\dfrac{\sin(2n\pi x)}{nx^{s}}\,dx=\lim_{a\rightarrow\infty}\sum_{n=1}^{\infty}\int_{0}^{a}\dfrac{\sin(2n\pi x)}{nx^{s}}\,dx\\ &=\lim_{a\rightarrow\infty}\sum_{n=1}^{\infty}\int_{0}^{a}\dfrac{\sin(2n\pi x)}{nx^{s}}\,dx+\lim_{a\rightarrow\infty}\sum_{n=1}^{\infty}\int_{a}^{\infty}\dfrac{\sin(2n\pi x)}{nx^{s}}\,dx\\ &=\sum_{n=1}^{\infty}\int_{0}^{\infty}\dfrac{\sin(2n\pi x)}{nx^{s}}\,dx.\\ \end{aligned}$$