# Evaluating $\int\limits_{1}^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx$

I thought the improper integral $$\int\limits_1^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx$$ converge, while the textbook says it's not.
Of course, $$\lfloor{x}\rfloor$$ is the greatest integer function. Here is my solution:

For any $$n\in \mathbb N$$,$$\lfloor{x}\rfloor=n\;\; \Leftrightarrow \;\; n\leq x
Thus if $$\; N\leq t \begin{align}&\int_{1}^{N}\frac{1}{\lfloor{x}\rfloor!}dx\leq\int_{1}^{t}\frac{1}{\lfloor{x}\rfloor!}dx\leq\int_{1}^{N+1}\frac{1}{\lfloor{x}\rfloor!}dx\\ &\Rightarrow \;\sum_{n=1}^{N-1}\frac{1}{n!}\leq\int_{1}^{t}\frac{1}{\lfloor{x}\rfloor!}dx \leq\sum_{n=1}^{N}\frac{1}{n!}\end{align} Taking $$N\rightarrow\infty$$ both sides also gives $$t\rightarrow\infty$$, and we have $$e-1\leq \int_1^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx \leq e-1$$ Therefore, $$\int_1^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx=e-1$$. □

I've tried a sort of times to find some mistakes in what I wrote. But I didn't get anything till now.
Can somebody point out what I missed? Thank you.

• Yes, it converges. Are you sure your textbook says that it does not ? Feb 23, 2021 at 9:21
• Your work is very correct $\to +1$ Feb 23, 2021 at 9:33
• So: either a mistake in the the textbook, or the OP miscopied the problem here. Since the identity of the textbook is secret, no one here can verify this. Feb 23, 2021 at 12:54
• Thank you guys all! Later I will call the author to verify it. Feb 23, 2021 at 15:42

Setting $$I=[1,\infty), \quad I_n=[n,n+1) \quad \forall n\ge1,$$ we have $$I=\bigcup_{n=1}^{\infty}I_n, \quad I_j\cap I_k=\varnothing \quad \forall j\ne k.$$ Since $$\lfloor x \rfloor =n \quad \forall x \in I_n, \quad n=1,2,3, \ldots$$ it follows that $$\begin{eqnarray} \int_1^{\infty}\frac{dx}{\lfloor x\rfloor !} &=&\int_I\frac{dx}{\lfloor x \rfloor ! }\cr &=&\sum_{n=1}^{\infty}\int_{I_n}\frac{dx}{\lfloor x \rfloor !}\cr &=&\sum_{n=1}^{\infty}\int_{I_n}\frac{dx}{n!}\cr &=&\sum_{n=1}^{\infty}\frac{n+1-n}{n!}\cr &=&\sum_{n=1}^{\infty}\frac{1}{n!}\cr &=&-1+\sum_{n=0}^{\infty}\frac{1}{n!}\cr &=& e-1. \end{eqnarray}$$
\begin{aligned} \int_{1}^{∞} \frac{1}{\left \lfloor x \right \rfloor !} dx&=\lim_{N \to ∞}\lim_{λ_{max} \to 0} \sum_{i=1}^{N} \frac{λ_{i}}{\left \lfloor x_{i} \right \rfloor !} \\ &=\lim_{N \to ∞} \sum_{n=1}^{N} \frac{1}{n!} \\ &=e-1. \end{aligned}