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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 $I'(x)=0, F'(x)=1$.

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