improper integral for a function involving floor function I am trying to study the improper integrability of $\int_{1}^{\infty}1-\frac{\lfloor x\rfloor}{x} d{x}$. I tried the definition of the improper integral as a limit with no success. Any hint? 
 A: Here is a hint:


*

*Write (with $\{x\}$ being the fractional part of $x$):
$$\int_1^n \frac{\{x\}}{x}dx = 
\sum_{k=1}^{n-1} \int_k^{k+1} \frac{\{x\}}{x}dx = 
\sum_{k=1}^{n-1} \int_0^1 \frac{\{x+k\}}{x+k}dx$$

*Check that $\{x+k\} = x$ for $x \in (0,1)$

*Get an explicit formula for the last $\sum\int$, you should almost be there
A: \begin{align}
\int_1^\infty \left(1 - \frac{\lfloor x\rfloor}x\right) dx 
&= \lim_{N\to\infty} \int_1^{N+1} \left(1 - \frac{\lfloor x\rfloor}x\right) dx \\
&= \lim_{N\to\infty} \sum_{k=1}^N \int_k^{k+1} \left(1 - \frac kx\right) dx \\
&= \lim_{N\to\infty} \sum_{k=1}^N \left( k\ln k - k\ln(k+1) + 1\right) \\
&= \lim_{N\to\infty} \left( N(1-\ln(N+1)) + \sum_{k=1}^N \ln k \right),
\end{align}
which is divergent, so the improper integral does not exist.
A: $$\int_{1}^{\infty}1-\frac{\lfloor x\rfloor}{x} d{x}=\int_{1}^{\infty}(\frac{x-\lfloor x\rfloor}{x}) d{x}=\Sigma_{m=1}^\infty \int_0^1 \frac{x}{m+x} d{x}=\Sigma_{m=1}^\infty \int_0^1 (1-\frac{m}{m+x}) d{x}=\Sigma_{m=1}^\infty(x-m \ln(m+x) |_0^1)=\Sigma_{m=1}^\infty(1-m \ln \frac{m+1}{m})=\Sigma_{m=1}^\infty(1-\ln (\frac{m+1}{m})^m)=\Sigma_{m=1}^\infty(\ln e ( \frac{m+1}{m})^{-m})$$
This series diverges.
