$\displaystyle\int_{0}^{\infty}\left[ne^{-x}\right]dx=\log\frac{n^{n-1}}{(n-1)!}$

Question

$\displaystyle\int_{0}^{\infty}\left[ne^{-x}\right]dx=\log\frac{n^{n-1}}{(n-1)!}, \ $ where $[\,\cdot\,]$ is the greatest integer function.

Put $e^{-x}=t$

$$ \begin{aligned} \int_{0}^{\infty}\left[ne^{-x}\right]dx & =\int_{1}^{0}\left(\left[n t\right]\cdot\frac{-1}{t}\right)dt =\int_{0}^{1}\left(\left[n t\right]\times\frac{1}{t}\right)dt \\ & =\int_{0}^{\frac{1}{n}}0\,dt+\int_{\frac{1}{n}}^{\frac{2}{n}}\frac{1}{t}\,dt+\int_{\frac{2}{n}}^{\frac{3}{n}}\frac{2}{t}\,dt+\int_{\frac{3}{n}}^{\frac{4}{n}}\frac{3}{t}\,dt+ \dots +\int_{\frac{n-1}{n}}^{1}\frac{n-1}{t}\,dt \\ &=\log 2+2\log\frac{3}{2}+3\log\frac{4}{3}+ \dots +(n-1)\log\frac{n}{n-1} \end{aligned} $$

However I did not the result in the form given in answer.

Answer 1

$$ \log \frac{2^1 \cdot 3^2 \cdot 4^3 \cdots n^{n-1}}{1^1 \cdot 2^2 \cdot 3^3 \cdots (n-1)^{n-1}}=\log \frac{n^{n-1}}{1 \cdot 2 \cdot 3 \cdots (n-1)}. $$

Answer 2

You have your answer now you need to use that $$ n\log(x)=\log(x^n) \qquad \textrm{ and }\qquad \log(ab)=log(a)+\log(b) $$