Definite Integral $\int_0^1 \left \{\frac{(-1)^{\lfloor 1/x \rfloor}}{x} \right\}\, dx$ The curly brackets mean 'FractionalPart' which, I believe, is defined as {${x}$}$=x-\lfloor x \rfloor$ where $x \in \mathbb{R}$.
The conjecture I have found but can not prove is that the definite integral has a closed form of $A+B*ln(C/\pi)$ for positive integers A,B and C. The graph of the integrand gets 'very wild' near x=0 and I have come up with values from -.14478 to 3.928 - which means I can't get a handle on it at all. Any help or guidance to material that could teach me new techniques would also be appreciated. Values of A=B=C=1 does give me a result close to my negative estimate.
Also, using telescoping sum tricks, $\displaystyle \int_0^1(-1)^{\lfloor 1/x \rfloor} dx=1-2\ln(2)$ , so the conjecture seems to have a similar format. The denominator of  'x'  in the integrand, this time, has me really puzzled.
 A: We have that $\left\lfloor\frac{1}{x}\right\rfloor = n$ for any $x\in \left(\frac{1}{n+1},\frac{1}{n}\right]$. The integral of $\frac{1}{x}$ over such interval is $\log\left(1+\frac{1}{n}\right)$,
hence:

$$ \int_{0}^{1}(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{dx}{x}=\sum_{n\geq 1}(-1)^n \log\left(1+\frac{1}{n}\right)=\color{red}{\log\frac{2}{\pi}} $$

since the middle term is just the logarithm of Wallis' product. If we add a fractional part to the integrand function, things become a little more complicated, but not that much.
$$\begin{eqnarray*} \int_{0}^{1}\left\{(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{1}{x}\right\}\,dx &=& \sum_{n\geq 1}\int_{\frac{1}{n+1}}^{\frac{1}{n}}\left\{\frac{(-1)^n}{x}\right\}\,dx\\
&=&\int_{\frac{1}{2}}^{1}\left(2-\frac{1}{x}\right)+\int_{\frac{1}{3}}^{\frac{1}{2}}\left(\frac{1}{x}-2\right)\,dx+\ldots\end{eqnarray*}$$
hence:

$$\begin{eqnarray*} \int_{0}^{1}\left\{(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}\frac{1}{x}\right\}\,dx &=& \log\frac{2}{\pi}+\sum_{k=1}^{+\infty}\left(\frac{2k}{2k(2k-1)}-\frac{2k}{2k(2k+1)}\right)\\&=&\log\frac{2}{\pi}+\sum_{k\geq 1}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\\&=&\color{red}{1+\log\frac{2}{\pi}}.\end{eqnarray*}$$

