Reciprocal log floor

evaluate

\int_0^1 \frac{1}{\lfloor\log_{2018}{x}\rfloor}dx
\begin{align*} \int_0^1 \frac{1}{\lfloor\log_{2018}{x}\rfloor}dx &= \sum_{k=0}^\infty (2018^{-k}-2018^{-(k+1)})\frac{1}{-(k+1)} \\&= \sum_{k=0}^\infty 2018^{-k}(1-2018^{-1})\frac{-1}{k+1} \\&= \frac{2017}{2018}\sum_{k=0}^\infty \frac{(-1)^{2k-1}}{k+1}(-1)^k(2018^{-1})^k \\&= 2017\sum_{k=1}^\infty \frac{(-1)^{k-1}(-1)^k}{k}(2018^{-1})^{k-1}2018^{-1} \\&= 2017\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}(-2018^{-1})^{k} \\&=2017\ln(1-2018^{-1}) \\&= \ln(2017^{2017})-\ln(2018^{2017}) \end{align*}