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

I thought the improper integral $$\int_{1}^{\infty} \frac{1}{\lfloor{x}\rfloor!}dx$$ 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 ? – TheSilverDoe Feb 23 at 9:21
• Your work is very correct $\to +1$ – Claude Leibovici Feb 23 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. – GEdgar Feb 23 at 12:54
• Thank you guys all! Later I will call the author to verify it. – JJLEE Feb 23 at 15:42