Integrating (floor(x) e^-x) from 0 to inf

Question

How does one obtain the following result:

\begin{equation} \int_0^\infty \text{floor}(x) e^{-x} dx= \frac{1}{e-1} \end{equation}

where $\text{floor}(x)$ is the greatest integer not exceeding x (from practice booklet).

Answer 1

Because of the nature of the floor function, the integral becomes a sum:

$$\sum_{k=0}^{\infty} k \int_{k}^{k+1} dx \, e^{-x} = \sum_{k=0}^{\infty} k \left (e^{-k}-e^{-(k+1)}\right ) $$

$$ \sum_{k=0}^{\infty} k \, e^{-k}= -\left [\frac{d}{da} \sum_{k=0}^{\infty} e^{-a k} \right ]_{a=1}= -\left [\frac{d}{da} \frac1{1-e^{-a}}\right ]_{a=1} = \frac{e}{(e-1)^2}$$

Therefore, the integral is

$$\frac{e}{(e-1)^2} - \frac1{e} \frac{e}{(e-1)^2} = \frac1{e-1} $$

Answer 2

Hint 1: Note that on $[n,n+1)$, the expression $\lfloor x\rfloor e^{-x}$ equals $ne^{-x}$.

Hint 2: You can also write $\displaystyle{\int_0^Nf(x)\;dx = \sum_{n=0}^{N-1}\int_n^{n+1}f(x)\;dx}$