$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx$$
For $x\in\left(\frac1{n+1},\frac1n\right)$, $\left\lfloor\frac1x\right\rfloor=n$, and $\frac1n-\frac1{n+1}=\frac1{n(n+1)}$. So the integral is equal to the Riemann sum
$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx=\lim_{n\to\infty}\sum_{k=1}^n\frac{(-1)^k}{k(k+1)}$$
Now, for $|x|<1$, we have
$$\frac x{1-x}=\sum_{k=1}^\infty x^k\\
\implies -x-\ln(1-x)=C_1+\sum_{k=1}^\infty\frac{x^{k+1}}{k+1}\\
\implies x-\frac{x^2}2+(1-x)\ln(1-x)=C_2+C_1x+\sum_{k=1}^\infty \frac{x^{k+2}}{(k+1)(k+2)}$$
If $x=0$, then we find $C_1=C_2=0$ so that
$$x-\frac{x^2}2+(1-x)\ln(1-x)=\sum_{k=1}^\infty \frac{x^{k+2}}{(k+1)(k+2)}\\
\implies 1-\frac x2+\frac{1-x}x\ln(1-x) = \sum_{k=2}^\infty \frac{x^k}{k(k+1)}\\
\implies 1+\frac{1-x}x\ln(1-x) = \sum_{k=1}^\infty \frac{x^k}{k(k+1)}$$
and when $x=-1$, we have
$$\int_0^1(-1)^{\left\lfloor\frac1x\right\rfloor}\,\mathrm dx=\sum_{k=1}^\infty\frac{(-1)^k}{k(k+1)}=\boxed{1-2\ln2}$$
Is this solution correct?
Best Answer
According to this answer, we get:
$$\mathscr{S}_{-1}:=\int_0^1\left(-1\right)^{\left\lfloor\frac{1}{x}\right\rfloor}\space\text{d}x=\frac{-1+\ln\left(1-\left(-1\right)\right)-\left(-1\right)\ln\left(1-\left(-1\right)\right)}{-1}=$$ $$1-2 \ln (2)\approx-0.386294\tag1$$