I'm reading The Theory of Riemann zeta function, in the process of proving the analytic continuation we proved using Fourrier Series :
$$ [x] - x + \frac{1}{2} = \sum_{n = 1}^{\infty} \frac{\sin2n \pi x}{n\pi}$$
We write $s=\sigma + it $, for $-1<\sigma<0$ , we have also proved by Abel summation that : $$ \zeta(s) = s \int_{0}^{\infty} \frac{[x] - x + \frac{1}{2}}{x^{s + 1}}dx $$ By replacing the Fourrier Serie in the integral, we had : $$ \zeta(s) = s \int_{0}^{\infty} \frac{\sum_{n = 1}^{\infty} \frac{\sin2n \pi x}{n\pi}}{x^{s + 1}}dx = \frac{s}{\pi} \sum_{n = 1}^{\infty} \frac{1}{n} \int_{0}^{\infty} \frac{\sin 2n\pi x}{x^{s + 1}}dx $$ My Problem : I don't understand why we can permute between the integral and the sum. The book says
"We have still to justify the term-by-term integration. Since the series ( it referred to the Fourrier Serie above) is boundedly convergent, term-by-term integration over any finite range is permissible. It is therefore sufficient to prove that
$$\lim_{\lambda \to \infty } \sum_{n = 1}^{\infty} \frac{1}{n} \int_{\lambda}^{\infty} \frac{\sin (2n\pi x)}{x^{s + 1}} = 0$$
I read A Course In Calculus and Real Analysis, I couldn't find a justification. Instead I tried to apply a result in the book where the conditions where $ \sum_{n=1}^{\infty} f_k $ is uniformly convergent, neither the fact $ \mid \sum_{k=1}^{n} f_k \mid \leq G $ where G is a function that it proper integral is convergent. And would yield to : $\sum_{n = 1}^{\infty} \int_{\lambda}^{\infty} f_n = \int_{\lambda}^{\infty} \sum_{n = 1}^{\infty} f_n $. If i'm not mistaken $f_k = \frac{\sin 2k\pi x}{x^{s+1}}$.
Thank You For Reading .