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Math Help - Nice identity with natural log. and floor function

  1. #1
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    Nice identity with natural log. and floor function

    Prove that for each natural n\ge2 it's \ln n!=\int_{1}^{n}{\frac{n-\left\lfloor x \right\rfloor }{x}\,dx}.
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  2. #2
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    Induction

    Use proof by induction
    (I): Let n=2
    Then \int_1^{n}\frac{n-\left\lfloor{x}\right\rfloor}{x}\,dx=\int_1^2\frac  {2-\left\lfloor{x}\right\rfloor}{x}\,dx
    and on 1<x<2, \left\lfloor{x}\right\rfloor=1, so the above integral is
    \int_1^2\frac{2-1}{x}\,dx=\int_1^2\frac{dx}{x}=\ln{2}=\ln{2!}.
    (II): Assume that \int_1^{n}\frac{n-\left\lfloor{x}\right\rfloor}{x}\,dx=\ln{n!} for n.
    Now,
    \int_1^{n+1}\frac{n+1-\left\lfloor{x}\right\rfloor}{x}\,dx=\int_1^{n}\fr  ac{n+1-\left\lfloor{x}\right\rfloor}{x}\,dx+\int_n^{n+1}\  frac{n+1-\left\lfloor{x}\right\rfloor}{x}\,dx
    =\int_1^{n}\frac{n-\left\lfloor{x}\right\rfloor}{x}\,dx+\int_1^{n}\fr  ac{1}{x}\,dx+\int_n^{n+1}\frac{n+1-\left\lfloor{x}\right\rfloor}{x}\,dx
    =\ln{n!}+\int_1^{n}\frac{dx}{x}+\int_n^{n+1}\frac{  n+1-\left\lfloor{x}\right\rfloor}{x}\,dx.
    Using the fact that \left\lfloor{x}\right\rfloor=n for n<x<n+1 in the rightmost integral,
    \int_1^{n+1}\frac{n+1-\left\lfloor{x}\right\rfloor}{x}\,dx=\ln{n!}+\int_  1^{n}\frac{dx}{x}+\int_n^{n+1}\frac{n+1-n}{x}\,dx
    =\ln{n!}+\int_1^{n}\frac{dx}{x}+\int_n^{n+1}\frac{  dx}{x}
    =\ln{n!}+\int_1^{n+1}\frac{dx}{x}
    =\ln{n!}+\ln(n+1)
    =\ln\left(n!(n+1)\right)
    =\ln(n+1)!
    And we have our proof.

    --Kevin C.
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  3. #3
    MHF Contributor kalagota's Avatar
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    i was about to give the same proof.. but i can't fix the latex.. haha
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  4. #4
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    Okay!

    We have \ln n!=\sum\limits_{k=2}^{n}{\ln k}, hence \sum\limits_{k=2}^{n}{\int_{1}^{k}{\frac{dx}{x}}}=  \sum\limits_{k=2}^{n}{\sum\limits_{j=1}^{k-1}{\int_{j}^{j+1}{\frac{dx}{x}}}}=\sum\limits_{j=1  }^{n-1}{\sum\limits_{k=j+1}^{n}{\int_{j}^{j+1}{\frac{dx  }{x}}}}, finally \sum\limits_{j=1}^{n-1}{\int_{j}^{j+1}{\frac{n-j}{x}\,dx}}=\sum\limits_{j=1}^{n-1}{\int_{j}^{j+1}{\frac{n-\left\lfloor x \right\rfloor }{x}\,dx}}=\int_{1}^{n}{\frac{n-\left\lfloor x \right\rfloor }{x}\,dx}.\quad\blacksquare

    And we're done!
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  5. #5
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    brilliant solution!!
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