KingOfActing
lukewarm mess
Re: MX2 2016 Integration Marathon
![](https://latex.codecogs.com/png.latex?\bg_white $We split the sum into each integer $1 \leq m \leq n $ to obtain: $\\\int_0^{\infty} \lfloor{ne^{-x}}\rfloor\,dx = \sum_{m=1}^{n-1} \int_{\ln{\left(\frac{n}{m+1} \right )}}^{\ln{\left(\frac{n}{m} \right )}}m\,dx=\sum_{m=1}^{n-1}m\ln{\left(1+\frac{1}{m} \right )})
As far as I know, you can't simplify that sum (unless you want to turn it into a logarithm of a product).
As far as I know, you can't simplify that sum (unless you want to turn it into a logarithm of a product).