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Math Help - About Euler's constant (1)...

  1. #1
    MHF Contributor chisigma's Avatar
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    About Euler's constant (1)...

    It is not a secret for anyone that I'm a 'fan' of Leonhard Euler . One of his most beautiful 'discoveries' is the complex of properties of the constant that brings his name...

    \gamma = \lim_{k \rightarrow \infty} 1 + \frac{1}{2} + \dots + \frac{1}{k} - \ln k = .57721566490153 \dots (1)

    The first problem concening Euler's constant I will propose to You is : starting from definition (1) demonstrate that is...

    \int_{0}^{\infty} \ln t\cdot e^{-t}\cdot dt = - \gamma (2)

    Kind regards

    \chi \sigma
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  2. #2
    Super Member PaulRS's Avatar
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    See my 2 posts here.
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    Here is another way to obtain this identity :
     \int_{0}^{\infty} \ln t\cdot e^{-t}\cdot dt = - \gamma

    Consider

     \int_0^{\infty} e^{-t} \ln(t) ~dt

     = \lim_{n\to\infty} \int_0^{n} ( 1 - \frac{t}{n} )^n \ln(t) ~dt

    Sub  1 - \frac{t}{n} = y ,  n \to \infty

     dt = -ndy

    the integral becomes

     \lim_{n\to\infty} n \int_0^1 y^n [ \ln(n)  + \ln(1-y)]~dy

     = \lim_{n\to\infty} n \left( \ln(n) \cdot \frac{1}{n+1} -  \sum_{k=1}^{\infty} \frac{ 1 }{k( n + k + 1)} \right)

     = \lim_{n\to\infty} \left (\ln(n)- [ \frac{1}{1}  + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n+1} ] \right )

    ps : There is a proof which states that  \int_0^{\infty} e^{-t} \ln(t) ~dt = \lim_{n\to\infty} \int_0^n ( 1- \frac{t}{n} )^n \ln(t) ~dt  but i don't think it is important really ...
    Last edited by simplependulum; November 2nd 2009 at 11:36 PM.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Congratulations for Paul and SP!...

    A similar but more 'general purpose' way to arrive at the result is the use of the following 'nice lilttle formula'. From the well known definition of [negative] exponential function...

    e^{-t} = \lim_{k \rightarrow \infty} (1-\frac{t}{k})^{k} (1)

    ... derives 'spontaneously' the formula...

    \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} \int_{0}^{k} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt (2)

    ... that with the substitution u=\frac{t}{k} becomes...

    \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} k \cdot \int_{0}^{1} f(ku)\cdot (1-u)^{k}\cdot du (3)

    ... and with the substitution u=1-\frac{t}{k} becomes...

    \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} k \cdot \int_{0}^{1} f \{k(1-u)\}\cdot u^{k}\cdot du (4)

    Now if in (4) we set f(t)= \ln t we obtain...

    \int_{0}^{\infty}\ln t\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} k\cdot \ln k \cdot \int_{0}^{1}u^{k}\cdot du + k\cdot \int_{0}^{1} \ln (1-u)\cdot u^{k}\cdot du=

    = \lim_{k \rightarrow \infty} \frac{k}{k+1}\cdot \ln k - k \cdot \sum_{n=1}^{\infty} \frac{1}{n}\cdot \int_{0}^{1} u^{k+n}\cdot du =

    = \lim_{k \rightarrow \infty} \ln k - k\cdot \sum_{n=1}^{\infty} \frac{1}{n\cdot (k+n)} (5)

    If now we remember the identity...

     \sum_{n=1}^{\infty} \frac{1}{n\cdot (k+n)} = \frac{1}{k} \sum_{n=1}^{k} \frac{1}{n} (6)

    ... we arrivwe at the final result...

     \int_{0}^{\infty} \ln t\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} (\ln k - \sum_{n=1}^{k}\frac{1}{n}) = -\gamma (7)

    Kind regards

    \chi \sigma
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chisigma View Post
    \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt= \lim_{k \rightarrow \infty} \int_{0}^{k} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt (2)
    Isn't the above kind of a deep statement? Wouldn't you need to at least supply some reasoning for why it's true?
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  6. #6
    MHF Contributor chisigma's Avatar
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    e^{-t}=\lim_{k \rightarrow \infty} (1-\frac{t}{k})^{k} \rightarrow

    \rightarrow \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} \int_{0}^{\infty} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt \rightarrow

    \rightarrow \int_{0}^{\infty} f(t)\cdot e^{-t}\cdot dt = \lim_{k \rightarrow \infty} \int_{0}^{k} f(t)\cdot (1-\frac{t}{k})^{k}\cdot dt

    Kind regards

    \chi \sigma
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Why can you pull the limit out of the integral sign? I also believe you can but I'm not quite sure I know how to justify it.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Why can you pull the limit out of the integral sign? I also believe you can but I'm not quite sure I know how to justify it.
    That was my point. When it is permissable for extracting a limit from an integrand to the exterior is nontrivial.

    And saying that \lim_{n\to\infty}\int_0^{\infty}f_n(x)=\lim_{n\to\  infty}\int_0^{n}f_n(x) is also highly suspect.

    Take for example f_n(x)=\frac{1}{n}

    Then \lim_{n\to\infty}\int_0^{\infty}\frac{dx}{n}=\inft  y

    but \lim_{n\to\infty}\int_0^{n}\frac{dx}{n}=1
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Drexel28 View Post
    That was my point. When it is permissable for extracting a limit from an integrand to the exterior is nontrivial.

    And saying that \lim_{n\to\infty}\int_0^{\infty}f_n(x)=\lim_{n\to\  infty}\int_0^{n}f_n(x) is also highly suspect.

    Take for example f_n(x)=\frac{1}{n}

    Then \lim_{n\to\infty}\int_0^{\infty}\frac{dx}{n}=\inft  y

    but \lim_{n\to\infty}\int_0^{n}\frac{dx}{n}=1
    The correct procedure requires to remember that by definition is...

    a_{n} = \int_{0}^{\infty} f_{n} (x)\cdot dx = \lim_{t \rightarrow \infty} \int_{0}^{t} f_{n}(x)\cdot dx (1)

    ... so that the limit...

    \lim_{n \rightarrow \infty} a_{n} (2)

    ... exists if \forall n the limit (1) exists. In your example is...

    a_{n} = \lim_{t \rightarrow \infty} \int_{0}^{t} \frac{1}{n}\cdot dx (3)

    ... and none of the a_{n} exists so that the expression...

    \lim_{n\to\infty}\int_0^{\infty}\frac{dx}{n}=\inft  y (4)

    ... is a nonsense. Anyway your observation is probably correct so that I propose to overcame any difficulty with the following formal statement...

    \lim_{t \rightarrow \infty} \int_{0}^{t} f(x)\cdot e^{-x}\cdot dx = \lim_{k \rightarrow \infty} \int_{0}^{k} f(x)\cdot (1- \frac{x}{k})^{k}\cdot dx (5)

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 4th 2009 at 11:07 PM.
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  10. #10
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    Let  f_n(z) = \int_0^n \left ( 1 - \frac{t}{n} \right )^n t^{z-1} ~dt

    and we have

     \Gamma(z) - f_n(z) = \int_0^n \left ( e^{-t} - ( 1 - \frac{t}{n} )^n \right ) t^{z-1}~dt + \int_n^{\infty} e^{-t} t^{z-1}~dt

    now , we prove this inequality

     0 \leq e^{-t} - \left ( 1 - \frac{t}{n} \right )^n \leq \frac{t^2}{n}e^{-t}

    consider

     1 + y \leq \ e^{y} \leq \frac{1}{1 - y }

    Replace  y with  \frac{t}{n}

     \left ( 1 - \frac{t}{n} \right )^n \leq e^{-t} \leq \left ( 1 +  \frac{t}{n} \right )^{-n}



    Therefore ,

     0 \leq e^{-t} - \left ( 1 - \frac{t}{n} \right )^n \leq e^{-t} \left ( 1 - ( 1 - \frac{t^2}{n^2} )^n \right )


    but  ( 1 - x )^n \geq 1 - nx

    therefore

     1 - \left ( 1 - \frac{t^2}{n^2} \right )^n \leq  \frac{t^2}{n^2}


     0 \leq e^{-t} - \left ( 1 - \frac{t}{n} \right )^n \leq \frac{t^2}{n}e^{-t}




    After proving the inequality , Take limit  n \to \infty

    we can obtain  \lim_{n\to\infty} f_n(z) = \Gamma(z)
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  11. #11
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    Quote Originally Posted by chisigma View Post
    Anyway your observation is probably correct so that I propose to overcame any difficulty with the following formal statement...

    \lim_{t \rightarrow \infty} \int_{0}^{t} f(x)\cdot e^{-x}\cdot dx = \lim_{k \rightarrow \infty} \int_{0}^{k} f(x)\cdot (1- \frac{x}{k})^{k}\cdot dx (5)
    This is in not really more self evident than the previous assertion. However, simplependulum provides us with a proof. Indeed,

    Quote Originally Posted by simplependulum View Post
    now , we prove this inequality

     0 \leq e^{-t} - \left ( 1 - \frac{t}{n} \right )^n \leq \frac{t^2}{n}e^{-t}
    (this also holds for non-integer values of n as soon as they're greater than 1, cf. the proof)

    Hence, from this inequality, for t>1,

    \left|\int_{0}^{t} f(x)\cdot e^{-x}\cdot dx - \int_{0}^{t} f(x)\cdot (1- \frac{x}{t})^t\cdot dx\right| \leq \frac{1}{t}\int_0^t f(x)x^2 e^{-x} dx\leq \frac{1}{t}\int_0^\infty f(x) x^2e^{-x}dx\to_{t\to\infty} 0.

    (Provided the integral is finite, so under mild hypotheses on f)
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    I copied this proof from a book which talks about elements of complex analysis , almost the last chapter .


    Perhaps this excellent proof is completed by some GREAT Mathematicians , if i have enough time , i will complete the proof here . We can see that some parts of the original proof are missing ( e.g. proof of   ( 1 - x )^n \geq 1 - nx ) .
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  13. #13
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    Hence, from this inequality, for t>1,

    \left|\int_{0}^{t} f(x)\cdot e^{-x}\cdot dx - \int_{0}^{t} f(x)\cdot (1- \frac{x}{t})^t\cdot dx\right| \leq \frac{1}{t}\int_0^t f(x)x^2 e^{-x} dx\leq \frac{1}{t}\int_0^\infty f(x) x^2e^{-x}dx\to_{t\to\infty} 0.


    (Provided the integral is finite, so under mild hypotheses on f)

    This inequality appears at the end of the proof in my book ! This is the main point , when n tends to infinity , their difference (  f_n(z) - \Gamma(z) ) tends to zero .

    Great !
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  14. #14
    Moo
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    Quote Originally Posted by Bruno J. View Post
    Why can you pull the limit out of the integral sign? I also believe you can but I'm not quite sure I know how to justify it.
    Dominated convergence theorem - Wikipedia, the free encyclopedia
    or Monotone convergence theorem - Wikipedia, the free encyclopedia
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