The Stair Integral

Consider the integral

$\mathfrak{I}=\displaystyle\int_{0}^{\infty} \left \lfloor{ne^{-x}}\right \rfloor \text{ }\text{d}x$

Substituting $ne^{-x}=t$ we get:

$\mathfrak{I}=\displaystyle\int_{0}^{n} \frac{\left \lfloor{t}\right \rfloor}{t}\text{d}t$

$=\displaystyle\sum_{r=0}^{n-1}\displaystyle\int_{r}^{r+1} \frac{r}{t}\text{d}t$

$=\displaystyle\sum_{r=0}^{n-1} r \ln \bigg( \frac{r+1}{r}\bigg)$

$=\displaystyle\sum_{r=0}^{n-1} (r+1)\ln (r+1)-r\ln (r)-\displaystyle\sum_{r=0}^{n-1} \ln (r+1)$

The first sum is a telescopic sum. Evaluating the above summations we get:

$\mathfrak{I}=\ln \bigg( \dfrac{n^{n}}{n!}\bigg)$

Substituting $n=7$ we get $\left \lfloor{\mathfrak{I}}\right \rfloor=\boxed{5}$

When x=0, $7e^{-x}$ =7, and it keeps decreasing.

Therefore, its graph would be a staircase with six stairs.

Horizontally cut the stairs, you end up with six rectangles, each with height being 1 and the top one being the smallest.

Just work out the width of each rectangle and sum them, the expression is $-\ln\frac67-\ln\frac57-\ln\frac47-\ln\frac37-\ln\frac27-\ln\frac17=5.1...$ .