Improper integral involving floor function

Evaluate the integral:

$$\mathcal{L}_1=\int_{0}^{\infty}\left \lfloor x \right \rfloor e^{-x}\, {\rm d}x$$

Solution



We note that:

$$\begin{aligned}
\int_{0}^{\infty}\left \lfloor x \right \rfloor e^{-x}\,{\rm d}x  &=\int_{0}^{1}0 \, {\rm d}x + \int_{1}^{2}e^{-x}\, {\rm d} x + \int_{2}^{3}2 e^{-x}\, {\rm d}x + \cdots  \\
 &= \sum_{k=1}^{\infty}k\int_{k}^{k+1}e^{-x}\, {\rm d}x\\
 &= \sum_{k=1}^{\infty}k\left [ e^{-k} - e^{-k-1} \right ]\\
&= \frac{e}{\left ( e-1 \right )^2}- \frac{1}{\left ( e-1 \right )^2}= \frac{1}{e-1}
\end{aligned}$$

The sums are easily computed since we are invoking the Taylor series of $\frac{1}{1-x}$ and differentiate it.