How to integrate $\int\limits_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx$? As my title says, I need help integrating with floor functions,
$$\int\limits_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx$$
What does this even mean exactly? How would approach this? 
 A: One may write 
\begin{align*}
\displaystyle \int_{0}^{1} \left(-1\right)^{\large ^{\left\lfloor\frac{1}{x}\right\rfloor}}  \mathrm{d}x  
&= \sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}}  \left(-1\right)^{\large ^{\left\lfloor\frac{1}{x}\right\rfloor}}  \mathrm{d}x  \\
&= \sum_{k=1}^{\infty} \int_{k}^{k+1} \left(-1\right)^{\large ^{\left\lfloor u\right\rfloor}} \: \frac{\mathrm{d} u}{u^{2}} \\
&= \sum_{k=1}^{\infty} \int_{k}^{k+1} \left(-1\right)^{{k}} \: \frac{\mathrm{d} u}{u^{2}} \\
&= \sum_{k=1}^{\infty}\frac{\left(-1\right)^{{k}}}{k (k+1)} \\
&=  \sum_{k=1}^{\infty}\frac{\left(-1\right)^k}{k}+\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k+1}\\
&=-\log 2-\log 2+1
\end{align*} where we have used the standard identity
$$
\log(1+x)=-\sum_{k=1}^{\infty}\frac{\left(-1\right)^k}{k}x^k, \quad |x|<1,
$$ when  $x \to 1^-$ (via Abel's theorem).
Finally,

$$\int_{0}^{1} \left(-1\right)^{\large ^{\left\lfloor\frac{1}{x}\right\rfloor}}  \mathrm{d}x  
= 1-2 \log 2.
$$

A: Essentially, you have this:
$$\int_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx=\int_\frac{1}{2}^1\left(-1\right)^1dx+\int_\frac{1}{3}^\frac{1}{2}\left(-1\right)^2dx+\int_\frac{1}{4}^\frac{1}{3}\left(-1\right)^3dx+\int_\frac{1}{5}^\frac{1}{4}\left(-1\right)^4dx+\ldots$$
$$=\left(-1+\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+ \left(-\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots$$
$$=-1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\ldots\right)$$
$$=-1+2(1-\log2)=1-\log 4$$
