Integral of fractional

The given function vanishes at integral values of $x$ since fractional part of $x$ is zero at these values.

We know that for non-integral values of $x$ :

$\lceil x \rceil = \lfloor x \rfloor + 1$

The given integral now reduces to:

$\displaystyle \int_{0}^{1} \frac{(-1)^{\lfloor x \rfloor} \cdot 2016 \{x\}}{2^{\lceil x \rceil}} dx + \displaystyle \int_{1}^{2} \frac{(-1)^{\lfloor x \rfloor} \cdot 2016 \{x\}}{2^{\lceil x \rceil}} dx + \ldots$

$\displaystyle \frac{(-1)^{0}\cdot 2016}{2^{1}} \displaystyle \int_{0}^{1} \{x\} dx + \frac{(-1)^{1}\cdot 2016}{2^{2}} \displaystyle \int_{1}^{2} \{x\} dx + \ldots$

Using the property:

$\displaystyle \int_{r}^{r+1} \{x\} dx = \displaystyle \int_{0}^{1} \{x\} dx = \frac{1}{2} ; \forall r \in \mathbb{Z}$

We get

$\bigg( \frac{(-1)^{0}}{2^{1}} + \frac{(-1)^1}{2^2} + \ldots \bigg) \cdot 2016 \cdot \displaystyle \int_{0}^{1} \{x\} dx$

$\displaystyle \Rightarrow \bigg( \displaystyle \sum_{r=0}^{\infty} \frac{(-1)^{r}}{2^{r+1}} \bigg) \cdot 2016 \cdot \frac{1}{2}$

$\displaystyle \Rightarrow 1008 \cdot \frac{1}{2} \cdot \displaystyle \sum_{r=0}^{\infty} \big( \frac{-1}{2} \big)^{r}$

This is an infinite geometric progression. The sum yields

$\displaystyle \Rightarrow 504 \times \frac{1}{1-\frac{-1}{2}}$

$\displaystyle \Rightarrow \boxed{336}$