$\int_{1}^{\infty}\frac{[3x]}{[x]!}dx$ I saw an interesting definite integral at somewhere and would like to share.
$$\int_{1}^{\infty}\frac{[3x]}{[x]!}dx$$
The solution given is
$$\int_{1}^{\infty}\frac{[3x]}{[x]!}dx=\lim_{n \to \infty}\sum_{k=1}^{n-1}\int_{k}^{k+1}\frac{[3x]}{[x]!}dx$$
$$=\lim_{n\to \infty}\sum_{k=1}^{n-1}\left ( \int_{k}^{k+\frac{1}{3}}\frac{3k}{k!}dx+\int_{k+\frac{1}{3}}^{k+\frac{2}{3}}\frac{3k+1}{k!}dx+\int_{k+\frac{2}{3}}^{k+1}\frac{3k+2}{k!}dx \right )$$
$$=\lim_{n \to \infty}\sum_{k=1}^{n-1}\left ( \frac{1}{3}\left ( \frac{3k}{k!} \right )+\frac{1}{3}\left ( \frac{3k+1}{k!} \right ) +\frac{1}{3}\left ( \frac{3k+2}{k!} \right )\right )=\lim_{n \to \infty}\sum_{k=1}^{n-1}\frac{3k+1}{k!}=3\sum_{k=1}^{\infty}\frac{1}{(k-1)!}+\sum_{k=1}^{\infty}\frac{1}{k!}$$
$$=3e+e-1=4e-1$$
I wonder is there another way to solve this definite integral?

Interesting problem and nice solution.
Doing the same kind of work, if $n$ is an integer, I think that we may conjecture that
$$\int_1^\infty \frac{\lfloor n x\rfloor }{\lfloor x\rfloor !}\,dx=\frac {4e} 3 n-1 $$
I think that
$$\int_1^\infty \frac{\lceil n x\rceil }{\lceil x\rceil !}\,dx=\frac {e} 2 n+\frac 13 $$
Try to prove if I am correct or not with the second.
