Mathematical Reflections 2016, Issue 1 - Problem U365

Problem:
Let $n$ be a positive integer. Evaluate

(a) $\displaystyle \int_{0}^n e^{\lfloor x \rfloor} \ dx$

(b) $\displaystyle \int_{0}^n \lfloor e^x \rfloor \ dx$.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
(a) If $k \leq x < k+1$, where $k \in \mathbb{Z}$, then $\lfloor x \rfloor =k$. So, $$\displaystyle \int_{0}^n e^{\lfloor x \rfloor} \ dx=\sum_{k=0}^{n-1} \int_{k}^{k+1} e^{\lfloor x \rfloor} dx=\sum_{k=0}^{n-1} e^k=\dfrac{e^n-1}{e-1}.$$

(b) If $k \leq e^x < k+1$, where $k \in \mathbb{N}^*$, then $\lfloor e^x \rfloor=k$. This implies that if $\log k \leq x < \log (k+1)$, then $\lfloor e^x \rfloor=k$. Let $m$ be the greatest natural number such that $\log m \leq n$.
So, $$\renewcommand{\arraystretch}{2}\begin{array}{lll}\displaystyle \int_{0}^n \lfloor e^x \rfloor \ dx&=& \displaystyle \sum_{k=1}^{m-1} \int_{\log k}^{\log k+1} k \ dx+\int_{\log m}^n m \ dx\\&=&\displaystyle \sum_{k=1}^{m-1} [(k+1)\log(k+1)-k\log k]-\sum_{k=1}^{m-1} \log(k+1)+m(n-\log m)\\&=&\displaystyle m\log m-\log m!+mn-m\log m\\&=&mn-\log m!\\&=&\lfloor e^n \rfloor n-\log (\lfloor e^n \rfloor!). \end{array}$$