1. ## integration

$\int_0^{\infty} [ne^{-x}]dx$
Note:[] implies greatest integer function.

2. Originally Posted by mathkeep
$\int_0^{\infty} [ne^{-x}]dx$
Note:[] implies greatest integer function.

Note that $[ne^{-x}]=0 \mbox{ for }\,ne^{-x}<1\Longleftrightarrow\,e^x>n\Longleftrightarrow\ ,x>\ln n$, and then the integral is actually a finite one: $\int\limits_0^{\ln n}[ne^{-x}]dx$.

Now: $[ne^{-x}]=n\Longleftrightarrow\,x=0$ , and $[ne^{-x}]=n-1\Longleftrightarrow\,n-1\leq ne^{-x} $\Longleftrightarrow 1 $\Longleftrightarrow\,0.

Likewise, $[ne^{-x}]=n-2 \Longleftrightarrow\,\ln\frac{n}{n-1} , and etc., so we get:

$\int\limits_0^{\ln n}[ne^{-x}]dx=\sum\limits_{k=1}^{n-1}\,\,\int\limits_{\ln\frac{n}{n-(k-1)}}^{\ln\frac{n}{n-k}}\!\!\!\!(n-k)dx$ ....

Tonio

3. The function to be integrate is of 'staircase type' so the integral it the sum of rectangular's areas. In this case is...

$\int_{0}^{\infty} [n\cdot e^{-x}]\cdot dx= \sum_{k=1}^{n-1} (n-k)\cdot (\ln \frac{n}{n-k} - \ln \frac{n}{n-k+1}) =$

$= n\cdot \ln n -\sum_{k=1}^{n-1} \ln (n-k+1) = \ln \frac{n^{n}}{n!}$ (1)

Kind regards

$\chi$ $\sigma$

4. Originally Posted by chisigma
The function to be integrate is of 'staircase type' so the integral it the sum of rectangular's areas. In this case is...

$\int_{0}^{\infty} [n\cdot e^{-x}]\cdot dx= \sum_{k=1}^{n-1} (n-k)\cdot (\ln \frac{n}{n-k} - \ln \frac{n}{n-k+1}) =$

$= n\cdot \ln n -\sum_{k=1}^{n-1} \ln (n-k+1) = \ln \frac{n^{n}}{n!}$ (1)

Kind regards

$\chi$ $\sigma$
hey thnks tonio nd chisigma...fr ur help