Results 1 to 6 of 6

Math Help - cool integral

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    cool integral

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

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

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

    But that is certainly not the solution.

    It has the same solution as:

    \int_{0}^{1}\frac{1-e^{-x}-e^{\frac{-1}{x}}}{x}dx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    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.

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

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

    But that is certainly not the solution.

    It has the same solution as:

    \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

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

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

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

    Integration by parts on both gives

    =\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

    \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

    \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

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

    The value of Gamma' evaluated at one...I cannot prove..haha
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Wow, I had long forgotten about this. You done fine.
    Here is a tidbit you may find nice:

    The DiGamma function:

    {\Psi}(1)=\frac{{\Gamma}(1)'}{{\Gamma}(1)}=-1-{\gamma}+\sum_{r=1}^{\infty}\left(\frac{1}{r}-\frac{1}{r+1}\right)=-{\gamma}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    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:

    {\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 \Gamma'(1)=-\gamma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Lemma 1. Let <br />
f:\mathbb{R}/\mathbb{R}^ -   \to \mathbb{R}<br />
be a function with continuous derivative in its domain.

    Then <br />
\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} <br />
where <br />
\left\{ x \right\} = x - \left\lfloor x \right\rfloor <br />

    Proof

    <br />
\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} } <br />
<br />
 = \sum\limits_{k = 0}^{n - 1} {\int_k^{k + 1} {\left( {x - k} \right) \cdot f'\left( x \right)dx} } <br />
<br />
 = \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} } <br />

    By the Fundamental Theorem of Calculus: <br />
\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]} <br />

    So: <br />
\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)} <br />
<br />
 = \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)} <br />
<br />
 = \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)} <br />
<br />
 = n \cdot f\left( n \right) - \sum\limits_{k = 1}^n {f\left( k \right)} <br />

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

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

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

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

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

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

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

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

    Proof : <br />
\int_0^1 {\left( {\tfrac{1}<br />
{{1 - x}} + \tfrac{1}<br />
{{\log \left( x \right)}}} \right)dx}  = \gamma <br />

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

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

    Thus: <br />
\left( {\zeta \left( s \right) - \tfrac{1}<br />
{{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} }}<br />
{{e^u  - 1}}}du  - \int_0^\infty  {u^{s - 2}  \cdot e^{ - u} du} <br />

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

    Applying Lemma 2: <br />
\gamma  = \mathop {\lim }\limits_{s \to 1^ +  } \left( {\zeta \left( s \right) - \tfrac{1}<br />
{{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}<br />
{{1 - e^{ - u} }} - \tfrac{1}<br />
{u}} \right)} du<br />
<br />
 = \int_0^\infty  {e^{ - u}  \cdot \left( {\tfrac{1}<br />
{{1 - e^{ - u} }} - \tfrac{{1 }}<br />
{u}} \right)} du \square<br />
    Last edited by PaulRS; April 25th 2009 at 06:40 AM. Reason: Copy-paste typo at the end
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    Lemma. <br />
\sum\limits_{k = 2}^\infty  {\tfrac{{\zeta \left( k \right)}}<br />
{k} \cdot \left( { - 1} \right)^k }  = \gamma <br />

    Proof

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

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

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

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


    Proof. \int_{0}^{1}\left(\frac{1}{\log(x)}+\frac{1}{1-x}\right)dx=\gamma<br />

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

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

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

    So by Lemma 3of the previous post: <br />
\int_0^{ + \infty } {\tfrac{{e^{ - u}  - 1 + u}}<br />
{{u \cdot \left( {e^u  - 1} \right)}} \cdot du}  = \tfrac{1}<br />
{2} \cdot \zeta \left( 2 \right) - \tfrac{1}<br />
{3} \cdot \zeta \left( 3 \right) \pm ...<br />
And finally, by the Lemma we've just seen: <br />
\int_0^{ + \infty } {\tfrac{{e^{ - u}  - 1 + u}}<br />
{{u \cdot \left( {e^u  - 1} \right)}} \cdot du}  = \gamma \blacksquare<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cool forum! I need a little help....
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 20th 2010, 02:57 PM
  2. Very cool
    Posted in the Calculus Forum
    Replies: 11
    Last Post: December 21st 2009, 11:32 AM
  3. A few cool ones here
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: March 26th 2009, 06:00 PM
  4. cool series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 4th 2008, 06:53 AM
  5. cool theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 14th 2008, 04:50 PM

Search Tags


/mathhelpforum @mathhelpforum