Simple question, simple answer

Function $\{x\}$ is same on every interval $(n,n+1)$ , where $n$ is an integer, and function $\lfloor x\rfloor$ is constant and equal to $n$ on every interval $(n,n+1)$ where $n$ in an integer.It's easy to see that on interval $(0,1)$ , $x=\{x\}$ . Because of this we can write the integral as:

$\int_0^\infty (-\{ x \} )^{\lfloor x \rfloor}=\sum_{r=0}^{\infty}{\int_{0}^{1}{(-x)^r dx}}$

Now by evaluating this easy polynomial integral we get:

$\int_0^\infty (-\{ x \} )^{\lfloor x \rfloor}=\sum_{r=0}^{\infty}{\frac{(-1)^r}{r+1}}$

Now we evaluate the sum by writing it as:

$\sum_{r=0}^{\infty}{\frac{(-1)^r}{r+1}}=\lim_{n \to \infty}{(\sum_{k=1}^{n} ( \frac{1}{2k-1}-\frac{1}{2k} ) )}=\lim_{n \to \infty}{ \sum_{k=1}^{n} (( \frac{1}{2k-1}+\frac{1}{2k} ) - 2 \sum_{k=1}^{n} \frac{1}{2k}} )=$

$=\lim_{n \to \infty}{ (\sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k})} = \lim_{n \to \infty}{\sum_{k=n+1}^{2n} \frac{1}{k}}$

This sum can be covered to an integral:

$\lim_{n \to \infty}{\sum_{k=n+1}^{2n} \frac{1}{k}}=\lim_{n \to \infty}{\sum_{k=1}^{n} \frac{1}{n+k}}=\lim_{n \to \infty}{\sum_{k=1}^{n} \frac{1}{1+\frac{k}{n}}}\frac{1}{n}=\int_{0}^{1}{\frac{dx}{x+1}}=\ln2$

And finally we have:

$\displaystyle {\int_0^\infty (-\{ x \} )^{\lfloor x \rfloor}=\ln2}$

opening in intervals of 1 and puting limit we get te very known series of log2