There's a very simple formula in the book I am currently reading: $$\sum\limits_{a<x\leq b}f(x) = \int\limits_{a}^{b}f(x)\mathrm{d}x + \rho(b)f(b)-\rho(a)f(a) - \int\limits_{a}^{b}\rho(x)f'(x)\mathrm{d}x$$ where $f\in \mathrm{C}^{1}[a,b]$ and $\rho(x) = 1/2-\{x\}$ and $x$ on LHS is an integer (in $(a,\;b]$).
Let's see. Suppose that $[a]+1 = x_{0}<x_{1}<\ldots < x_{N-1} < x_{N} = [b]$ -- all integers within $(a,b]$. $$\int\limits_{x_{j-1}}^{x_{j}}\rho(x)f'(x)\mathrm{d}x = \rho(x)f(x)\Bigg\rvert_{x_{j-1}}^{x_{j}} + \int\limits_{x_{j-1}}^{x_{j}}f(x)\mathrm{d}x = \dfrac{f(x_{j})-f(x_{j-1})}{2} + \int\limits_{x_{j-1}}^{x_{j}}f(x)\mathrm{d}x$$ Summing these from $j=1$ to $N$ we get $$\int\limits_{x_{0}}^{x_{N}}\rho(x)f'(x)\mathrm{d}x = \dfrac{f(x_{N}) - f(x_{0})}{2} + \int\limits_{x_{0}}^{x_{N}}f(x)\mathrm{d}x$$ Finally since $$\int\limits_{a}^{x_{0}}\rho(x)f'(x)\mathrm{d}x = f(x_{0})/2 - \rho(a)f(a) + \int\limits_{a}^{x_{0}}f(x)\mathrm{d}x$$ and similarly $$\int\limits_{x_{N}}^{b}\rho(x)f'(x)\mathrm{d}x = \rho(b)f(b) - f(x_{N})/2 + \int\limits_{x_{N}}^{b}f(x)\mathrm{d}x$$ we obtain $$\int\limits_{a}^{b}\rho(x)f'(x)\mathrm{d}x = \rho(b)f(b) - \rho(a)f(a) + \int\limits_{a}^{b}f(x)\mathrm{d}x$$ which clearly contradicts the formula in the beginning. What am I doing wrong? Must be a stupid question, but I genuinely do not see...
To be clear: $[x]$ denotes the integer part of $x$ and $\{x\}$ the fractional part. Then $x=[x]+\{x\}$ and for $\rho(x)=1/2-\{x\}\,,$ $$ \rho'(x)=\frac{d}{dx}\Big([x]-x\Big)=\sum_{z\in\mathbb Z}\delta_z(x)-1\,. $$ Integration by parts gives \begin{align} &\int_{(a,b]}\rho(x)\,f'(x)\,dx=-\int_{(a,b]}\rho'(x)\,f(x)\,dx+\rho(x)\,f(x)\,\Big|^b_a\\ &=-\sum_{z\in (a,b]}f(z)+\int_{(a,b]}f(x)\,dx+\rho(a)f(a)-\rho(b)f(b) \end{align} from which the formula follows.
Note that because we are dealing with Dirac measures the range over which we integrate is ambiguous when we write $\int_a^b\rho'(x)\,f(x)\,dx\,.$