Question

Prove that ${\int }_0^{\infty }\left[n{e}^{-x}\right]dx=ln\left(\frac{n^n}{n!}\right)$, where n is a natural number greater than $1$ and $[\cdot ]$ denotes the greatest integer function.

Answer 1

Given : ${\int }_0^{\infty }\left[n{e}^{-x}\right]dx$

${\int }_0^{\infty }\left[n{e}^{-x}\right]dx=-[{\int }_{\infty }^{\ln n}\left[n{e}^{-x}\right]dx+{\int }_{\ln n}^{\ln \frac{n}{2}}\left[n{e}^{-x}\right]dx+..........{\int }_{\ln \frac{n}{n-1}}^{\ln \frac{n}{n}}\left[n{e}^{-x}\right]dx]=-[{\int }_{\infty }^{\ln n}\left(0\right)dx+1.{\int }_{\ln n}^{\ln \frac{n}{2}}1.dx+.........+(n-1\left){\int }_{\ln \frac{n}{n-1}}^{\ln \frac{n}{n}}dx\right]=-[(\ln \left(\frac{n}{2}\right)-\ln \left(n\right))+2(\ln \left(\frac{n}{3}\right)-\ln \left(\frac{n}{2}\right))+.........(n-1\left)(\ln \left(\frac{n}{n}\right)-\ln \left(\frac{n}{n-1}\right))\right]=[(\ln \left(n\right)-\ln \left(\frac{n}{2}\right))+2\left(\ln \left(\frac{n}{2}\right)-\ln \left(\frac{n}{3}\right)\right)+.........(n-1\left)(\ln \left(\frac{n}{n-1}\right)-\ln \left(\frac{n}{n}\right))\right]=(n-1)\ln \left(n\right)+[1.\ln 2+2.\ln 3+\ln 4+4\ln 5....]-[1.\ln 1+2.\ln 2+3.\ln 3+4\ln 4+....]=(n-1)\ln \left(n\right)+[\ln 1+\ln 2+\ln 3+\ln 4+\ln 5+........]=(n-1)\ln \left(n\right)+\left[\ln \left[\mathrm{1.2.3.4.......}\right(n-1\left)\right]\right]=(n-1)\ln \left(n\right)+\ln (n-1)!=\ln \left(n^{(n-1)}\right)-\ln (n-1)!=\ln \left|\frac{n^{(n-1)}}{(n-1)!}\right|=\ln \left|\frac{n^n}{n(n-1)!}\right|=\ln \left|\frac{n^n}{n!}\right|$

Hence the correct answer is $\ln \left|\frac{n^n}{n!}\right|$