Can anyone show that:
log(N-1)! $\displaystyle \leq$ $\displaystyle \int_1^N \! log(x) \, dx$ $\displaystyle \leq$ log(N)!
$\displaystyle \log (n!) = \log n(n-1)(n-2)\cdots2\cdot1 = \log n + \log (n-1) + \cdots + \log 2$. This is a Riemann sum for the integral $\displaystyle \int \log x$. Can you show that the Riemann sum on the left of your inequality is less than the integral?
What Maddas is doing is setting up a Riemann sum for $\displaystyle \int_1^N log(x)dx$ using the integers as the "break points"- that is $\displaystyle \delta x= 1$. Since log(x) is an increasing function, you get the "lower sum" using the left endpoints of each interval and the "upper sum" using the right endpoints.