TITLE: How to integrate $\int\limits_0^1 \left(-1\right)^{^{\left\lfloor\frac{1}{x}\right\rfloor}} dx$?
QUESTION [11 upvotes]: 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?

REPLY [4 votes]: 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,

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