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  1. #1
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    integration

    \int_0^{\infty} [ne^{-x}]dx
    Note:[] implies greatest integer function.
    Last edited by mathkeep; November 19th 2009 at 06:38 AM.
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  2. #2
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    Quote Originally Posted by mathkeep View Post
    \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}<n\Longleftrightarrow\,\frac{n-1}{n}\leq e^{-x}<1 \Longleftrightarrow 1<e^x\leq \frac{n}{n-1} \Longleftrightarrow\,0<x\leq \ln\frac{n}{n-1}.

    Likewise, [ne^{-x}]=n-2 \Longleftrightarrow\,\ln\frac{n}{n-1}<x\leq \ln \frac{n}{n-2} , 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
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  3. #3
    MHF Contributor chisigma's Avatar
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    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
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  4. #4
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    Quote Originally Posted by chisigma View Post
    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
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