How to evaluate this integral: $$\int_{0}^{\infty} \biggl\lfloor{\frac{n}{e^{x}}\biggr\rfloor} \ dx, $$where $n \in \mathbb{N}$.
The same integral when asked to evaluate for $n=2$ (say) i can do it by splitting the limits from $x = 0$ to $x = \log{2}$ where $e^{x}$ takes the value 1, and then from $x= \log{2}$ to $x = \infty$. But how to do this for the general case $n$. I thought of two ways:
Using induction on $n$. This will not work!
Splitting the limits from $x =0$, to $x = \log{n}$ also seems to cause some problems for me.