Integral with the floor function

Observe that $1 - \frac{\lfloor x \rfloor}{x} = \frac{x - \lfloor x \rfloor}{x} > \frac{x - \lfloor x \rfloor}{\lfloor x \rfloor + 1}$ for all $x \ge 1$ , since $x < \lfloor x \rfloor + 1$ for all $x$ .

The sought area is $\displaystyle\int_1^\infty \frac{x - \lfloor x \rfloor}{x} \, dx$ . Due to the above observation, this is greater than $\displaystyle\int_1^\infty \frac{x - \lfloor x \rfloor}{\lfloor x \rfloor + 1} \, dx$ . Break this integral on unit intervals:

$\displaystyle\int_1^\infty \frac{x - \lfloor x \rfloor}{\lfloor x \rfloor + 1} \, dx$

$= \displaystyle\sum_{k=1}^\infty \int_k^{k+1} \frac{x - \lfloor x \rfloor}{\lfloor x \rfloor + 1} \, dx$

Since $k$ is an integer, for $k \le x < k+1$ we haave $\lfloor x \rfloor = k$ , so we can simplify the integral further:

$= \displaystyle\sum_{k=1}^\infty \int_k^{k+1} \frac{x - k}{k + 1} \, dx$

$= \displaystyle\sum_{k=1}^\infty \left. \dfrac{\frac{1}{2}x^2 - kx}{k+1} \right|_{x=k}^{k+1}$

$= \displaystyle\sum_{k=1}^\infty \dfrac{\frac{1}{2}(k+1)^2 - k(k+1)}{k+1} - \dfrac{\frac{1}{2}k^2 - k^2}{k+1}$

$= \displaystyle\sum_{k=1}^\infty \dfrac{\frac{1}{2}(2k+1) - k}{k+1}$

$= \displaystyle\sum_{k=1}^\infty \dfrac{1}{2(k+1)}$

$= \dfrac{1}{2} \cdot \displaystyle\sum_{k=2}^\infty \dfrac{1}{k}$

But the latter factor is infinite: it is the divergent series $\sum_{k=1}^\infty \frac{1}{k}$ , subtracted by the finite quantity $\frac{1}{1}$ . Multiplying by a positive constant $\frac{1}{2}$ leaves it infinity.

Our integral is greater than this result, but this result is infinity, so our integral must be infinity itself.