Term by term integration fourier series 
$s= \sigma +it$, for $-1<\sigma<0$, we have 
  $$ \int_{0}^\infty \sum_{n=1}^\infty \frac{ \sin 2 n \pi x}{n x^{s+1}} \, dx = \sum_{n=1}^\infty \frac{1}{n} \int_0^{\infty}\frac{\sin  2n \pi x }{ x^{s+1}} \, dx  $$ 

For the justification, the author writes, 

As $\sum_{n=1}^\infty \frac{\sin 2n\pi x }{ n \pi } $ is boundedly convergent, term by term integration over any finite range is permissible. It suffices to show that 
  $$ 
\lim_{\lambda \rightarrow \infty} \sum_{n=1}^\infty \frac{1}{n} \int_\lambda^\infty \frac{ \sin 2 n \pi x }{ x^{s+1}} \, ds = 0 $$

I know the series is uniformly bounded, but what is meant in the bolded sentences? Why is term by term integration permissible over finite range? Also why is this sufficient? 

What I think: I believe there are two steps here.


*

*Let $f_n(x) = \frac{ \sin 2 n \pi x }{ nx^{s+1} } $. We first prove it for "finite range", 
$$ \int_0^{\lambda} \sum f_n \, dx  = \sum \int_0^{\lambda} f_n \, dx  $$

*If we show it for the finite case, it "suffices" to prove 
$$ \lim_{\lambda \rightarrow 0} \sum \int_0^{\lambda}  f_n - \sum \int_0^{\infty} f_n   = \lim_{\lambda \rightarrow \infty} \sum \int_{\lambda}^\infty f_n = 0 $$ 

EDIT: Proof 1:

Let $g(x)=\frac{1}{x^{s+1}}, s_m(x) = \sum_1^m \frac{ \sin 2 n \pi x }{ n \pi }$ and $s(x) = \sum_1^\infty \frac{ \sin 2 n \pi x }{ n \pi }$ .   It suffices to show equality for a fixed $s$, with $-1 < \sigma < 0$. Also, as $\lambda<\infty$, it suffices to show it holds for closed unit interval $[ k , k +1 ] $ for $k \in \mathbb{N}$. Consider 
  $$   \int_0^{1}  g s  \, dx  - \lim _{ m \rightarrow \infty} \int_0^{1}  gs_m  \, dx = \lim _{ m \rightarrow \infty}  \int_{\delta}^{1 - \delta} + \lim _{ m \rightarrow \infty} \Big \{ \int_0^{\delta} + \int_{\delta}^{1 - \delta} g [s - s_m] \, dx  \Big \} $$
  For the first term, $g$ is bounded on $[ \delta, 1-\delta]$, $s_m$ converges uniformly on this interval, the term limits to $0$ as $m \rightarrow \infty$. For the second term, as $s_m$ is boundedly convergent in $\mathbb{R}$,  and 
  $$ \int_0^{\delta} + \int_{1- \delta} ^{1} \frac{1}{x^{\sigma +1 } }\, dx \le \frac{-x^{\sigma} }{\sigma} \Big|_0^{\delta} + \Big| _{1 - \delta }^{1 } \rightarrow 0 $$
  as $\delta \rightarrow 0$. The integral also limits to $0$ as $m \rightarrow \infty$. Hence, we have equality for finite range. 

EDIT: Proof 2(Using DCT) 

Let $h(x) = \sum \frac{2 n\pi x}{ n} $, which is convergent to $\pi (  [x] -x + \frac{1}{2} )$. We apply Dominated Convergence Theorem. We consider the series, $s_m(x) = \sum_1^m \frac{2n \pi x}{nx^{x+1} }$. So for any finite interval $[0 , \lambda] $, we have 
  $$ |s_m(x) | \le M \frac{1}{x^{s+1}}  $$ 
  Further, we have, from noting that 
\begin{align*}  \int_0^{\lambda} \frac{1}{x^{s+1}}  \,ds &= \frac{-1}{s x^s} \Big|_1^{\lambda} +  \pi \int_0^1 \frac{\frac{1}{2}- x }{x^{s+1}}  \\ 
&= C_1 + C_2 x^{-s} \Big|_0^1 + C_3 x^{-s+1} \Big|_0^1  < \infty
\end{align*}
  as $-1 < \sigma < 0 $. Thus, we have a sequence of functions $s_m(x) \rightarrow s(x)$ such that it is dominated by an integrable function. By Dominated Convergence Theorem, we have 
  $$ \int_0^{\lambda} \sum_1^\infty \frac{ \sin 2 n \pi x }{n \pi x^{s+1}} \, dx = \sum_1^\infty \frac{1}{n \pi }\int_0^{\lambda} \frac{\sin 2 n \pi x }{x^{s+1} } \, dx $$ 

 A: The series is not uniformly convergent. $\sum \frac{\sin (2n\pi x)}{n}$ is the Fourier series of the sawtooth function $h(x) = \pi\bigl(\frac{1}{2} - \lbrace x\rbrace\bigr)$ where $\lbrace x\rbrace = x - \lfloor x\rfloor$ is the fractional part of $x$ (to have pointwise convergence of the Fourier series everywhere, set $h(0) = 0$ rather than $\frac{\pi}{2}$ as the formula would), so your function is $h(x)\cdot x^{-(s+1)}$, which has jump discontinuities at all integers. Since every term of the series is continuous, it follows that the convergence is not uniform.
I don't know what the author exactly means with the phrase "boundedly convergent". If they mean that the partial sums are uniformly bounded, that is not quite correct. For $-1 < \sigma < 0$, the factor $x^{-(s+1)}$ is unbounded near $0$, so $h(x) x^{-(s+1)}$ is unbounded near $0$. Thus the partial sums cannot be uniformly bounded. They are uniformly bounded on $[\varepsilon, +\infty)$ for every $\varepsilon > 0$, though.
In any case, since the partial sums of the Fourier series of $h$ are uniformly bounded, the series in question is dominated by an integrable function on $[0,\lambda]$ for every $\lambda < +\infty$, so we can interchange summation and integration over $[0,\lambda]$ by the dominated convergence theorem.
In the framework of the Riemann integral, where we don't have a dominated convergence theorem, we can argue with the uniform convergence of the Fourier series of $h$ on $[\delta, 1-\delta]$ plus the overall boundedness of these partial sums (at positive integers, $x^{-(s+1)}$ is bounded, at $0$ this factor is integrable, so we get an overall small bound for the integral over $[0,\delta]$ choosing $\delta$ appropriately small) to justify the interchange of summation and integration over $[0,\lambda]$.
Then it remains to show that we can take the limit as $\lambda \to \infty$. Knowing the function $h$, it is not difficult to see that
$$\lim_{\lambda \to \infty} \int_{\lambda}^{\infty} \frac{h(x)}{x^{s+1}}\,dx = 0.$$
Then showing
$$\lim_{\lambda \to \infty} \sum_{n = 1}^{\infty} \frac{1}{n}\Biggl\lvert\int_{\lambda}^{\infty} \frac{\sin (2n\pi x)}{x^{s+1}}\,dx\Biggr\rvert = 0\tag{$\ast$}$$
completes the justification. Note that I took the absolute value of the integrals. Without doing that, we'd need another justification for
$$\lim_{\lambda \to \infty} \sum_{n = 1}^{\infty} \frac{1}{n}\int_0^{\lambda} \frac{\sin (2n\pi x)}{x^{s+1}}\,dx = \sum_{n = 1}^{\infty} \frac{1}{n} \lim_{\lambda \to \infty} \int_0^{\lambda} \frac{\sin (2n\pi x)}{x^{s+1}}\,dx.$$
With the absolute value, that is straightforward.
To see $(\ast)$, we integrate by parts,
\begin{align}
\Biggl\lvert\int_{\lambda}^{\infty} \frac{\sin (2n\pi x)}{x^{s+1}}\,dx \Biggr\rvert
&= \Biggl\lvert\biggl[-\frac{\cos (2n\pi x)}{2n x^{s+1}}\biggr]_{\lambda}^{\infty} - \frac{s+1}{2n\pi} \int_{\lambda}^{\infty} \frac{\cos (2n\pi x)}{x^{s+2}}\,dx\Biggr\rvert \\
&= \Biggl\lvert\frac{\cos (2n\pi\lambda)}{2n\pi \lambda^{s+1}} - \frac{s+1}{2n\pi}\int_{\lambda}^{\infty} \frac{\cos (2n\pi x)}{x^{s+2}}\,dx\Biggr\rvert \\
&\leqslant \frac{1}{2n\pi \lambda^{\sigma + 1}} + \frac{\lvert s+1\rvert}{2n\pi} \int_{\lambda}^{\infty} \frac{dx}{x^{\sigma + 2}} \\
&= \frac{1}{2n\pi \lambda^{\sigma + 1}}\biggl( 1 + \frac{\lvert s+1\rvert}{\sigma + 1}\biggr).
\end{align}
Thus we have
$$\sum_{n = 1}^{\infty} \frac{1}{n}\Biggl\lvert\int_{\lambda}^{\infty} \frac{\sin (2n\pi x)}{x^{s+1}}\,dx\Biggr\rvert \leqslant \frac{\pi}{12}\biggl(1 + \frac{\lvert s+1\rvert}{\sigma + 1}\biggr)\cdot \frac{1}{\lambda^{\sigma + 1}}.$$
Since $\sigma + 1 > 0$, $(\ast)$ follows.
A: Read again Titchmarsh, he wrote something else. Given what you know, I would sketch a proof of the functional equation :
For $\Re(s) \in (0,1)$ 
$$\int_0^\infty \sin(nx) x^{-s-1}dx= n^{s}\int_0^\infty \sin(x) x^{-s-1}dx$$
Thus
$$\eta(1-s) \int_0^\infty \sin(x) x^{-s-1}dx = \lim_{N \to \infty} \sum_{n=1}^N (-1)^{n+1} n^{s-1}\int_0^\infty \sin(x) x^{-s-1}dx \\=\lim_{N \to \infty} \int_0^\infty \left(\sum_{n=1}^N \frac{(-1)^{n+1}}{n}\sin(nx)\right) x^{-s-1}dx$$ 
Also by absolute convergence and convergence of the Fourier series in $L^2(0,2\pi)$ it is clear that 
$$\lim_{N \to \infty} \int_1^\infty \left(\sum_{n=1}^N \frac{(-1)^{n+1}}{n}\sin(nx)\right) x^{-s-1}dx = \int_1^\infty \left(\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin(nx)\right) x^{-s-1}dx\\ = (2\pi)^{s} s(2^s-1) \zeta(s) \tag{1}$$
Thus we are left with the limit as $N \to \infty$ of
$$\sum_{n=1}^N \frac{(-1)^{n+1}}{n} \int_0^1 \sin(nx) x^{-s-1}dx= \sum_{n=1}^N (-1)^{n+1}n^{s-1} \int_0^{n} \sin(x) x^{-s-1}dx \tag{2}$$
Since $\sum_{n=1}^\infty (2n-1)^{s-1}-(2n)^{s-1}$ converges absolutely as well as $\int_0^\infty |\sin(x) x^{-s-1}|dx$, we get that $(2)$ converges to $C$. 
Finally since $\int_0^1 \left(\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin(nx)\right) x^{-s-1}dx=0$ we get that $C=0$ and hence 
$$\eta(1-s)\int_0^\infty \sin(x) x^{-s-1}dx =  (2\pi)^{s} s(2^s-1) \zeta(s) \qquad \Re(s) \in (0,1)$$
