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  1. #1
    Eater of Worlds
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    cool integral

    Perhaps a number of you are familiar with the solution to this integral. if not, see if you can work it out.

    $\displaystyle \int_{0}^{1}\left[\frac{1}{ln(x)}+\frac{1}{1-x}\right]dx$

    If you do them individually, you just get $\displaystyle {-\infty}, \;\ {\infty}$

    But that is certainly not the solution.

    It has the same solution as:

    $\displaystyle \int_{0}^{1}\frac{1-e^{-x}-e^{\frac{-1}{x}}}{x}dx$
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Perhaps a number of you are familiar with the solution to this integral. if not, see if you can work it out.

    $\displaystyle \int_{0}^{1}\left[\frac{1}{ln(x)}+\frac{1}{1-x}\right]dx$

    If you do them individually, you just get $\displaystyle {-\infty}, \;\ {\infty}$

    But that is certainly not the solution.

    It has the same solution as:

    $\displaystyle \int_{0}^{1}\frac{1-e^{-x}-e^{\frac{-1}{x}}}{x}dx$
    I have worked on this integral for about two weeks and have finally come up with a solution

    First split it into

    $\displaystyle \int_0^1\frac{1-e^{-x}}{x}dx-\int_0^1\frac{e^{\frac{-1}{x}}}{x}dx$

    Then make the sub $\displaystyle x=\frac{1}{u}$ on the second integral turning it into

    $\displaystyle \int_0^1\frac{1-e^{-x}}{x}dx-\int_1^{\infty}\frac{e^{-x}}{x}$

    Integration by parts on both gives

    $\displaystyle =\int_0^1e^{-x}\ln(x)dx+\int_1^{\infty}e^{-x}\ln(x)dx=\int_0^{\infty}e^{-x}\ln(x)dx$

    Now consider the gamma function

    $\displaystyle \Gamma(t)=\int_0^{\infty}x^{t-1}e^{-x}dx$

    Well since the Gamma function is uniformly convergent for all values of x that make it converge we have that

    $\displaystyle \Gamma'(t)=\frac{d}{dx}\int_0^{\infty}x^{t-1}e^{-x}dx=\int_0^{\infty}\frac{\partial}{\partial{t}}x^ {t-1}e^{-x}dx=\int_0^{\infty}x^{t-1}e^{-x}\ln(x)dx$

    $\displaystyle \therefore\quad\Gamma'(1)=\int_0^{\infty}e^{-x}\ln(x)dx=-\gamma$

    The value of Gamma' evaluated at one...I cannot prove..haha
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  3. #3
    Eater of Worlds
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    Wow, I had long forgotten about this. You done fine.
    Here is a tidbit you may find nice:

    The DiGamma function:

    $\displaystyle {\Psi}(1)=\frac{{\Gamma}(1)'}{{\Gamma}(1)}=-1-{\gamma}+\sum_{r=1}^{\infty}\left(\frac{1}{r}-\frac{1}{r+1}\right)=-{\gamma}$
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by galactus View Post
    Wow, I had long forgotten about this. You done fine.
    Here is a tidbit you may find nice:

    The DiGamma function:

    $\displaystyle {\Psi}(1)=\frac{{\Gamma}(1)'}{{\Gamma}(1)}=-1-{\gamma}+\sum_{r=1}^{\infty}\left(\frac{1}{r}-\frac{1}{r+1}\right)=-{\gamma}$
    Thanks Galctus...I have heard of the Digamma function but I was unaware of its series representation...I just knew that $\displaystyle \Gamma'(1)=-\gamma$
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  5. #5
    Super Member PaulRS's Avatar
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    Lemma 1. Let $\displaystyle
    f:\mathbb{R}/\mathbb{R}^ - \to \mathbb{R}
    $ be a function with continuous derivative in its domain.

    Then $\displaystyle
    \sum\limits_{k = 1}^n {f\left( k \right)} = \int_0^n {f\left( x \right)dx} + \int_0^n {\left\{ x \right\} \cdot f'\left( x \right)dx}
    $ where $\displaystyle
    \left\{ x \right\} = x - \left\lfloor x \right\rfloor
    $

    Proof

    $\displaystyle
    \int_0^n {\left\{ x \right\} \cdot f'\left( x \right)dx} = \sum\limits_{k = 0}^{n - 1} {\int_k^{k + 1} {\left\{ x \right\} \cdot f'\left( x \right)dx} }
    $$\displaystyle
    = \sum\limits_{k = 0}^{n - 1} {\int_k^{k + 1} {\left( {x - k} \right) \cdot f'\left( x \right)dx} }
    $$\displaystyle
    = \int_0^n {x \cdot f'\left( x \right)dx} - \sum\limits_{k = 0}^{n - 1} {k \cdot \int_k^{k + 1} {f'\left( x \right)dx} }
    $

    By the Fundamental Theorem of Calculus: $\displaystyle
    \sum\limits_{k = 0}^{n - 1} {k \cdot \int_k^{k + 1} {f'\left( x \right)dx} } = \sum\limits_{k = 0}^{n - 1} {k \cdot \left[ {f\left( {k + 1} \right) - f\left( k \right)} \right]}
    $

    So: $\displaystyle
    \sum\limits_{k = 0}^{n - 1} {k \cdot \left[ {f\left( {k + 1} \right) - f\left( k \right)} \right]} = \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( {k + 1} \right)} - \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( k \right)}
    $$\displaystyle
    = \sum\limits_{k = 0}^{n - 1} {\left( {k + 1 - 1} \right) \cdot f\left( {k + 1} \right)} - \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( k \right)}
    $$\displaystyle
    = \sum\limits_{k = 0}^{n - 1} {\left( {k + 1} \right) \cdot f\left( {k + 1} \right)} - \sum\limits_{k = 0}^{n - 1} {k \cdot f\left( k \right)} - \sum\limits_{k = 0}^{n - 1} {f\left( {k + 1} \right)}
    $$\displaystyle
    = n \cdot f\left( n \right) - \sum\limits_{k = 1}^n {f\left( k \right)}
    $

    Now integrating by parts: $\displaystyle
    \int_0^n {x \cdot f'\left( x \right)dx} = n \cdot f\left( n \right) - \int_0^n {f\left( x \right)dx}
    $ and Lemma 1 follows.$\displaystyle
    \square
    $

    Lemma 2. We have: $\displaystyle
    \mathop {\lim }\limits_{s \to 1^ + } \left( {\zeta \left( s \right) - \tfrac{1}
    {{s - 1}}} \right) = \gamma
    $
    Proof

    First take $\displaystyle
    f\left( x \right) = \tfrac{1}
    {{x + 1}}
    $ in Lemma 1: We get $\displaystyle
    \sum\limits_{k = 1}^n {\tfrac{1}
    {{k + 1}}} = \log \left( {n + 1} \right) - \int_0^n {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^2 }}dx}
    $

    Thus: $\displaystyle
    \sum\limits_{k = 1}^{n + 1} {\tfrac{1}
    {k}} = \log \left( {n + 1} \right) + 1 - \int_0^n {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^2 }}dx}
    $ Now note that the integral in the RHS converges as $\displaystyle
    n \to + \infty
    $ - this in fact shows the existance of the gamma constant- then by definition of the gamma constant: $\displaystyle
    \gamma = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sum\limits_{k = 1}^{n + 1} {\tfrac{1}
    {k}} - \log \left( {n + 1} \right)} \right] = \mathop {\lim }\limits_{n \to + \infty } \left( {1 - \int_0^n {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^2 }}dx} } \right) = 1 - \int_0^\infty {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^2 }}dx}
    $

    Now take $\displaystyle
    f\left( x \right) = \tfrac{1}
    {{\left( {x + 1} \right)^s }}
    $ with $\displaystyle s>1$ in Lemma 1 : $\displaystyle
    \sum\limits_{k = 1}^n {\tfrac{1}
    {{\left( {k + 1} \right)^s }}} = \tfrac{{\left( {n + 1} \right)^{1 - s} }}
    {{1 - s}} - \tfrac{1}
    {{1 - s}} - s \cdot \int_0^n {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^{s + 1} }}dx}
    $

    Thus: $\displaystyle
    \sum\limits_{k = 1}^\infty {\tfrac{1}
    {{\left( {k + 1} \right)^s }}} = \tfrac{1}
    {{s - 1}} - s \cdot \int_0^\infty {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^{s + 1} }}dx}
    $ and then $\displaystyle
    \zeta \left( s \right) = \tfrac{1}
    {{s - 1}} + 1 - s \cdot \int_0^\infty {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^{s + 1} }}dx}
    $

    Therefore: $\displaystyle
    \mathop {\lim }\limits_{s \to 1^ + } \left( {\zeta \left( s \right) - \tfrac{1}
    {{s - 1}}} \right) = 1 - \mathop {\lim }\limits_{s \to 1^ + } s \cdot \int_0^\infty {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^{s + 1} }}dx} = 1 - \int_0^\infty {\tfrac{{\left\{ x \right\}}}
    {{\left( {x + 1} \right)^2 }}dx} = \gamma
    $ $\displaystyle
    \square
    $

    Lemma 3. $\displaystyle
    \zeta \left( s \right) \cdot \Gamma \left( s \right) = \int_0^\infty {\tfrac{{x^{s - 1} }}
    {{e^x - 1}}dx}
    $ : $\displaystyle s>1$
    Proof: $\displaystyle
    \int_0^\infty {\tfrac{{x^{s - 1} }}
    {{e^x - 1}}dx} = \int_0^\infty {x^{s - 1} \cdot \tfrac{{e^{ - x} }}
    {{1 - e^{ - x} }}dx} = \int_0^\infty {x^{s - 1} \cdot \sum\limits_{n = 1}^\infty {e^{ - nx} } dx} = \sum\limits_{n = 1}^\infty {\int_0^\infty {x^{s - 1} \cdot e^{ - nx} dx} }
    $ now since $\displaystyle
    \int_0^\infty {x^{s - 1} \cdot e^{ - nx} dx} = \tfrac{{\Gamma \left( s \right)}}
    {{n^s }}
    $ we are done $\displaystyle
    \square
    $

    Proof : $\displaystyle
    \int_0^1 {\left( {\tfrac{1}
    {{1 - x}} + \tfrac{1}
    {{\log \left( x \right)}}} \right)dx} = \gamma
    $

    Let $\displaystyle I$ be the integral then letting $\displaystyle
    u = - \log \left( x \right)
    $ we get $\displaystyle
    I = \int_0^\infty {e^{ - u} \cdot \left( {\tfrac{1}
    {{1 - e^{ - u} }} - \tfrac{1}
    {u}} \right)du}
    $

    Now note that: $\displaystyle
    \left( {\zeta \left( s \right) - \tfrac{1}
    {{s - 1}}} \right) \cdot \Gamma \left( s \right) = \zeta \left( s \right) \cdot \Gamma \left( s \right) - \tfrac{{\Gamma \left( s \right)}}
    {{s - 1}} = \zeta \left( s \right) \cdot \Gamma \left( s \right) - \tfrac{{\left( {s - 1} \right) \cdot \Gamma \left( {s - 1} \right)}}
    {{s - 1}}
    $ ( $\displaystyle s>1$ )

    Thus: $\displaystyle
    \left( {\zeta \left( s \right) - \tfrac{1}
    {{s - 1}}} \right) \cdot \Gamma \left( s \right) = \zeta \left( s \right) \cdot \Gamma \left( s \right) - \Gamma \left( {s - 1} \right) = \int_0^\infty {\tfrac{{u^{s - 1} }}
    {{e^u - 1}}}du - \int_0^\infty {u^{s - 2} \cdot e^{ - u} du}
    $

    Then: $\displaystyle
    \left( {\zeta \left( s \right) - \tfrac{1}
    {{s - 1}}} \right) \cdot \Gamma \left( s \right) = \int_0^\infty {\left( {\tfrac{{u^{s - 1} }}
    {{e^u - 1}} - u^{s - 2} \cdot e^{ - u} } \right)} du = \int_0^\infty {u^{s - 1} e^{ - u} \cdot \left( {\tfrac{1}
    {{1 - e^{ - u} }} - \tfrac{{1 }}
    {u}} \right)} du
    $

    Applying Lemma 2: $\displaystyle
    \gamma = \mathop {\lim }\limits_{s \to 1^ + } \left( {\zeta \left( s \right) - \tfrac{1}
    {{s - 1}}} \right) \cdot \Gamma \left( s \right) = \mathop {\lim }\limits_{s \to 1^ + } \int_0^\infty {u^{s - 1} e^{ - u} \cdot \left( {\tfrac{1}
    {{1 - e^{ - u} }} - \tfrac{1}
    {u}} \right)} du
    $$\displaystyle
    = \int_0^\infty {e^{ - u} \cdot \left( {\tfrac{1}
    {{1 - e^{ - u} }} - \tfrac{{1 }}
    {u}} \right)} du \square
    $
    Last edited by PaulRS; Apr 25th 2009 at 05:40 AM. Reason: Copy-paste typo at the end
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  6. #6
    Super Member PaulRS's Avatar
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    Lemma. $\displaystyle
    \sum\limits_{k = 2}^\infty {\tfrac{{\zeta \left( k \right)}}
    {k} \cdot \left( { - 1} \right)^k } = \gamma
    $

    Proof

    By definition: $\displaystyle
    \gamma = \mathop {\lim }\limits_{n \to + \infty } \left[ {\sum\limits_{k = 1}^n {\left( {\tfrac{1}
    {k}} \right)} - \log \left( {n + 1} \right)} \right]
    $

    Note that: $\displaystyle
    \log \left( {n + 1} \right) = \sum\limits_{k = 1}^n {\log \left( {1 + \tfrac{1}
    {k}} \right)}
    $ thus: $\displaystyle
    \sum\limits_{k = 1}^n {\left( {\tfrac{1}
    {k}} \right)} - \log \left( {n + 1} \right) = \sum\limits_{k = 1}^n {\left[ {\tfrac{1}
    {k} - \log \left( {1 + \tfrac{1}
    {k}} \right)} \right]}
    $

    We get: $\displaystyle
    \gamma = \sum\limits_{k = 1}^\infty {\left[ {\tfrac{1}
    {k} - \log \left( {1 + \tfrac{1}
    {k}} \right)} \right]}
    $ Consider that: $\displaystyle
    \log \left( {1 + \tfrac{1}
    {k}} \right) = \tfrac{1}
    {k} - \tfrac{1}
    {{2 \cdot k^2 }} \pm ...
    $ then $\displaystyle
    \gamma = \sum\limits_{k = 1}^\infty {\sum\limits_{s = 2}^\infty {\tfrac{{\left( { - 1} \right)^s }}
    {{s \cdot k^s }}} }
    $

    Exchange the summation order ( this can be justified working with the remainder, see here): $\displaystyle
    \gamma = \sum\limits_{s = 2}^\infty {\tfrac{{\zeta \left( s \right)}}
    {s} \cdot \left( { - 1} \right)^s } \blacksquare
    $


    Proof. $\displaystyle \int_{0}^{1}\left(\frac{1}{\log(x)}+\frac{1}{1-x}\right)dx=\gamma
    $

    Letting $\displaystyle
    {u = - \log \left( x \right)}
    $ we have: $\displaystyle
    \int_0^1 {\left( {\tfrac{1}
    {{\log \left( x \right)}} + \tfrac{1}
    {{1 - x}}} \right)dx} = \int_0^{ + \infty } {e^{ - u} \cdot \left( { - \tfrac{1}
    {u} + \tfrac{1}
    {{1 - e^{ - u} }}} \right)du} =
    \int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
    {{u \cdot \left( {e^u - 1} \right)}} \cdot du}

    $

    Now use the Power series expansion of $\displaystyle e^x$: $\displaystyle
    \int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
    {{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \int_0^{ + \infty } {\tfrac{{\tfrac{{u^2 }}
    {{2!}} - \tfrac{{u^3 }}
    {{3!}} + \tfrac{{u^4 }}
    {{4!}} \mp ...}}
    {{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \int_0^{ + \infty } {\tfrac{{\tfrac{u}
    {{2!}} - \tfrac{{u^2 }}
    {{3!}} + \tfrac{{u^3 }}
    {{4!}} \mp ...}}
    {{e^u - 1}} \cdot du}
    $

    $\displaystyle
    \int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
    {{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \tfrac{1}
    {{2!}} \cdot \int_0^{ + \infty } {\tfrac{u}
    {{e^u - 1}} \cdot du} - \tfrac{1}
    {{3!}} \cdot \int_0^{ + \infty } {\tfrac{{u^2 }}
    {{e^u - 1}} \cdot du} \pm ...
    $

    So by Lemma 3of the previous post: $\displaystyle
    \int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
    {{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \tfrac{1}
    {2} \cdot \zeta \left( 2 \right) - \tfrac{1}
    {3} \cdot \zeta \left( 3 \right) \pm ...
    $ And finally, by the Lemma we've just seen: $\displaystyle
    \int_0^{ + \infty } {\tfrac{{e^{ - u} - 1 + u}}
    {{u \cdot \left( {e^u - 1} \right)}} \cdot du} = \gamma \blacksquare
    $
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