Integral with floor function

Question

Find the integral $$\int_1^{1000}\frac{dx}{x+⌊\log_{10}(x)⌋}$$ The logarithm is creating some problems along with floor function.

Answer 1

We have that $\lfloor \log_{10}(x)\rfloor $ equals $0$ for $x\in(0,10)$, $1$ for $x\in[10,100)$ and $2$ for $x\in[100,1000)$, hence the value of the integral is just: $$ \int_{1}^{10}\frac{dx}{x}+\int_{10}^{100}\frac{dx}{x+1}+\int_{100}^{1000}\frac{dx}{x+2}=\log(10)+\log\frac{101}{11}+\log\frac{1002}{102}$$ or: $$ \color{red}{\log\frac{168670}{187}}.$$

Answer 2

HINT:

$$n\le\log_{10}(10x)<n+1\iff10^n\le(10x)<10^{n+1}$$

For example, $n=1,10\le(10x)<100\iff1\le x<10$