In a question Agno suggested an interesting way to compute $\{x\}$ and $\zeta(s)$.
Define $$ F(x) = \{x\} = x - \lfloor x \rfloor = \frac{i \, \log\left(-e^{\left(-2 i \, \pi x\right)}\right)}{2 \, \pi} + \frac{1}{2}$$ $$ I(x) = \lfloor x \rfloor = x - \frac{i \, \log\left(-e^{\left(-2 i \, \pi x\right)}\right)}{2 \, \pi} - \frac{1}{2} $$
Where $\log$ means the principal branch of the logarithm.
In the other question we couldn't compute $\zeta$ via Agno's definite integral and discussion in comments suggested the main problem is taking the correct branch of $\log$.
Tom Dickens suggested "The integral would have to be broken up into a series of ranges $x=n$ to $x=n+1$, which just reproduces the sum for the zeta function."
Suppose one have to compute $\int_a^b f(\{x\}) dx$.
Q1. When $\int_a^b f(\{x\}) dx = \int_a^b f(F(x)) dx$ symbolically?
Using this method, Maple agrees with Wolfram Alpha on several integrals of the form $\int_{1/2}^2 \cos(x)\{ x \} dx$ while sage 5.10 gives different result (usually off by a rational constant).
Q1a Assuming finite values, is computing purely numerically $\int_a^b f(F(x)) dx$ always correct?
For all branches of $\log$ we take in $I,F$, the result is correct $\mod 1$
Q1b Is $\int_a^b f(F(x)) dx$ always correct $\pmod 1$?
If $\log$ weren't multivalued, $ \sum_a^b f(x) = \int_a^b f(I(x)) dx$, which would transform a sum into a definite integral.
Q2 When $ \sum_a^b f(x) = \int_a^b f(I(x)) dx$ symbolically?
Maple 13 appears to give correct result for $f$ polynomial and can't find closed form for more complicated cases (modulo my errors).
Numerically mpmath gave correct value for $\zeta(3)$ using $I(x)$ as expected.
Added The only testcase so far which doesn't work is the zeta integral in the linked question.
Somewhat surprised complicated sums work:
$$\sum_{n=1}^\infty\sin(\pi n)=\int_1^\infty \sin(\pi I(x)) dx = 0$$.
Q3 Is this trivial and well known?
Part of sage's failures might be explained by $I'(x)=0, F'(x)=1$.
