$\sum_{a\lt n\le b}\phi (n)=\int_a^b \phi (x)\, dx+\int_a^b (x-[x]-\frac{1}{2})\phi '(x)\, dx+(a-[a]-\frac{1}{2})\phi (a)-(b-[b]-\frac{1}{2})\phi (b)$ From Titchmarsh's The Theory of the Riemann Zeta-Function, page 13:

Let $\phi (x)$ be any function with a continuous derivative in the interval $[a,b]$. Then, if $[x]$ denotes the greatest integer not exceeding $x$,
$$\sum_{a\lt n\le b}\phi (n)=\int_a^b \phi (x)\, dx+\int_a^b \left(x-[x]-\frac{1}{2}\right)\phi '(x)\, dx+\left(a-[a]-\frac{1}{2}\right)\phi (a)-\left(b-[b]-\frac{1}{2}\right)\phi (b).$$

There's no proof of this in the book and I don't know what's the 'name' of this theorem. I'd like to understand this theorem but don't know where to start at all.
 A: Let $\rho(t)=\frac12 -(t-[t])=\frac{1}{2} - \{t\}$, where $\{t\}$ is the fractional part of $t$.
Sketch of proof:
I leave the details to you. Here is one way to approach this identity.

*

*First, notice that $\rho$ is a $1$-periodic function, and that $\rho'(t)=-1$ for $x\in [k,k-1)$, $k\in\mathbb{Z}$. For $k\leq \alpha<b\leq k+1$, use integration by parts twice (once with $u=f(t)$ and $dv=\rho'(t)\,dt$; and another with
$u=f'(t)$ and $dv=\sigma'(t)\,dt=\rho(t)\,dt$) to get

$$
\begin{align}
-\int^\beta_\alpha f(t)\,dt &= \int^\beta_\alpha f(t)\rho'(t)\,dt\\
&=\rho(\beta-)f(\beta)-\rho(\alpha)f(\alpha)-\int^\beta_\alpha \rho(t)\,f'(t)\,dt
\end{align}
$$
You can now add over integer intervals $[k,k+1]\subset(a,b]$ and then over potentially fractional intervals $(a,[a]+1]$, $[[b],b]$ to obtain the desired result.

Edit: A more general and elegant proof can be obtained by integration by parts:
Lemma: Let  $F$ and $G$ be right-continuous functions of locally finite variation on
$I$, and let $\mu_G$, $\mu_F$ are the signed measures induced by $G$ and $F$ respectively.  Then,  for any compact interval $[a,b]\subset I$,
$$
\begin{align}
\int_{(a,b]} F(t)\,\mu_G(dt)=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t-)\,\mu_F(dt)
\end{align}
$$
where $G(t-)=\lim_{s\nearrow t}G(s)$.
For the OP,
Consider the counting measure  $\mu(dt)=\sum_{n\in\mathbb{Z}}\delta_{n}$ and the Lebesgue measure $\lambda$, both defined on $(\mathbb{R}\mathscr{B}(\mathbb{R}))$.
Let$\phi(dt)=(\lambda-\mu)(dt)$. Notice that $\Phi(t):=\phi((0,t])=t-[t]=\{t\}$.
$$
\begin{align}
\sum_{a< n\leq b}f(n)-\int^b_af(t)\,dt &=-\int^b_af(t)\,(\mu(dt)-\lambda(dt))=-\int^b_af(t)\phi(dt)
\end{align}
$$
Applying the Lemma above with $f$ in place of $F$ and $\Phi$ in place of $G$, we have that $\mu_f(dt)=f'(t)\,dt$ and $\mu_{\Phi}(dt)=\phi(dt)$ and so,
$$
\begin{align}
\int^b_af(t)\phi(dt) &= f(t)\Phi(t)|^b_a -\int^b_a\Phi(t-)\, f'(t)\,dt\\
&=f(b)\{b\}-f(a)\{a\}-\int^b_a\Phi(t)\,f'(t)\,dt\\
&= f(b)(b-[b])-f(a)(a-[a)] -\int^b_a(t-[t])\,f'(t)\,dt
\end{align}
$$
where the change from $\Phi(t-)$ to $\Phi(t)$ follows from the fact that $\Phi(t-)=\Phi(t)$ $\lambda$-a.s.
The conclusion follows by adding and subtracting $\frac12$ in the last integral.
A: This can be obtained by Abel-Summation: $$\sum_{a<n\leq b} f(n) = f(b) \sum_{a<n\leq b} 1 - \int_a^b \sum_{a<n\leq t} 1 \cdot f'(t) \, {\rm d}t \\ = f(b) \left( \lfloor b \rfloor - \lfloor a \rfloor \right) - \int_a^b \left( \lfloor t \rfloor - \lfloor a \rfloor \right) f'(t) \, {\rm d}t \\ = f(b) \lfloor b \rfloor - f(a) \lfloor a \rfloor + \int_a^b \left(t - \lfloor t \rfloor - \frac{1}{2} + \frac{1}{2} - t \right) f'(t) \, {\rm d}t \\ = f(a) \left( a - \lfloor a \rfloor - \frac{1}{2} \right) - f(b) \left( b - \lfloor b \rfloor - \frac{1}{2} \right) + \int_a^b f(t) \, {\rm d}t + \int_a^b \left(t - \lfloor t \rfloor - \frac{1}{2} \right) f'(t) \, {\rm d}t \\ = f(a) \, B_1\left( a - \lfloor a \rfloor \right) - f(b) \, B_1\left( b - \lfloor b \rfloor \right) + \int_a^b f(t) \, {\rm d}t + \int_a^b B_1\left( t - \lfloor t \rfloor \right) f'(t) \, {\rm d}t \, ,$$
where $B_1(x)$ is the first Bernoulli polynomial. As mentioned before, the $1/2$-terms are redundant.
By successively integrating by parts using $\int B_n(x) \, {\rm d}x = \frac{B_{n+1}(x)}{n+1}$, you'll obtain Euler-Maclaurin formula if $a,b$ are integers.
