The theory of riemann zeta function titchmarsh page 15 question in the proof of the functional equation I am currently reading Titchmarsh's book about the Riemann Zeta function and came across a problem in a proof of the functional equation that I cannot solve.
To be precise, I am referring to this book
here at page 15.
I have two problems there:
First:
It states that $$[x]-x+\frac{1}{2}=\sum_{n=1}^{\infty}\frac{sin(2n\pi x)}{n\pi}$$ where $[x]$ is the greatest integer less or equal to $x$.
I know that for $0< y <2\pi $ holds  $$\frac{\pi-y}{2}=\sum_{n=1}^{\infty}\frac{sin(ny)}{n}$$ and when I take $y=2\pi x$, I get 
$$\sum_{n=1}^{\infty}\frac{sin(2\pi nx)}{n\pi}=\frac{1}{\pi} \cdot \frac{\pi -2\pi x}{2}= \frac{1}{2}-x$$
My question is: what happened to $[x]$? Why does equallity above still hold?
My second question is why we can change series and integral in
\begin{align*}
\zeta(s)&= s\int_0^{\infty} \frac{[x]-x+\frac{1}{2}}{x^{s+1}}dx\\
 &= s\int_0^{\infty} \frac{\sum_{n=1}^{\infty}\frac{sin(2\pi nx)}{n\pi}}{x^{s+1}}dx\\
&= \frac{s}{\pi}\int_0^{\infty}\sum_{n=1}^{\infty} \frac{sin(2\pi nx)}{n x^{s+1}}dx\\
&=\frac{s}{\pi}\sum_{n=1}^{\infty}\int_0^{\infty} \frac{sin(2\pi nx)}{n x^{s+1}}dx
\end{align*}
Titchmarsh states that to justify the term-by-term integration it suffices to show that $$\lim\limits_{\lambda \to \infty} \sum_{n=1}^{\infty} \frac{1}{n}\int_{\lambda}^{\infty} \frac{sin(2\pi nx)}{ x^{s+1}}dx=0 $$ 
I understand the proof why this limit equals $0$ but I do not understand why this proves that we can integrate term-by-term. I know that if the series converges uniformly or absolute we could do that but I know that it is at least not absolute convergent.
I also don't understand why $$\sum_{n=1}^{\infty}\frac{sin(2n\pi x)}{n\pi} $$ is boundedly convergent.
Thank you very much in advance, i am having a rough time recently because of these questions.
 A: in general, for proving from scratch that you can exchange $\sum$ and $\int$  (or any two limits), start with finite bounds. clearly when $\lambda,A,B$ are finite :
$$ \int_0^\lambda \sum_{n=A}^B\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx =  \sum_{n=A}^B \int_0^\lambda\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = 
\sum_{n=A}^B \int_0^{n\lambda} \frac{\sin(2 \pi  y)}{\pi n} (y/n)^{-s-1} \frac{dy}{n} $$ $$ = \sum_{n=A}^B n^{s-1} \int_0^{n\lambda}\frac{\sin(2 \pi  y)}{\pi } y^{-s-1}dy$$
now all you need is noticing that when $Re(s) \in ]-1,0[$ : $\displaystyle\sum_{n} n^{s-1}$
is absolutely convergent, and 
$\displaystyle\left|\int_0^{n\lambda}\frac{\sin(2 \pi  y)}{\pi } y^{-s-1}dy\right| <C \left|\int_0^\infty\frac{\sin(2 \pi  y)}{\pi } y^{-s-1}dy\right|$
(you can use integration by part for proving that the integral converges)
hence, when $Re(s) \in ]-1,0[$ the series $$\sum_{n= A}^\infty n^{s-1} \int_0^{n\lambda}\frac{\sin(2 \pi  y)}{\pi } y^{-s-1}dy$$ converges absolutely (and hence it $\to 0$ when $A \to \infty$)
finally, prove that when $A \to \infty$ :
$$ \int_0^\lambda \sum_{n=A}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx \to 0$$
(again you can use integration by part)
and from all this you get that :
$$ \int_0^\lambda \sum_{n=1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx  = \int_0^\lambda \sum_{n=1}^A\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx  + \int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx $$
$$ = \sum_{n=1}^A \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx + \int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx $$
$$= \sum_{n=1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx - 
\sum_{n=A+1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx+\int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx$$
$$ = \sum_{n=1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx\qquad\qquad(\text{ letting  } \ A \to \infty)$$
(since the two other terms $\to 0$)
and from what Titchmarsh prove, that 
$$\lim\limits_{\lambda \to \infty} \sum_{n=1}^{\infty} \frac{1}{n}\int_{\lambda}^{\infty} \frac{sin(2\pi nx)}{ x^{s+1}}dx=0$$
you get that when $Re(s) \in ]-1,0[$ :
$$ \int_0^\infty \sum_{n=1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx =  \sum_{n=1}^\infty\int_0^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx$$
which (with the functional equations for the $\Gamma$ function) proves the functional equation for $\zeta(s)$.
