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Thread: hard integral

  1. #1
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    hard integral

    compute

    $\displaystyle \int_0^{ + \infty } {e^{ - x^2 } \ln xdx} $
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  2. #2
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    Wolfram Mathematica Online Integrator

    That doesn't look like a fun time!

    Hey, how did you get a picture by your name?
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  3. #3
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    A classic...

    Define $\displaystyle f\left( s \right) = \int\limits_0^\infty {{e^{ - {x^2}}}{x^s}dx} \Rightarrow f'\left( 0 \right) = \int\limits_0^\infty {{e^{ - {x^2}}}\ln xdx}
    $

    Then

    $\displaystyle f\left( s \right) = \int\limits_0^\infty {{e^{ - {x^2}}}{x^s}dx} \underbrace = _{u = {x^2}}\frac{1}
    {2}\int\limits_0^\infty {{e^{ - u}}{u^{\frac{{s - 1}}
    {2}}}dx} = \frac{1}
    {2}\Gamma \left( {\frac{{s + 1}}
    {2}} \right)
    $

    And you have to

    $\displaystyle f'\left( s \right) = \frac{1}
    {4}\Gamma '\left( {\frac{{s + 1}}
    {2}} \right) = \frac{1}
    {4}\psi \left( {\frac{{s + 1}}
    {2}} \right)\Gamma \left( {\frac{{s + 1}}
    {2}} \right) \Rightarrow f'\left( 0 \right) = \frac{1}
    {4}\psi \left( {\frac{1}
    {2}} \right)\Gamma \left( {\frac{1}
    {2}} \right)
    $

    Finally, knowing that $\displaystyle \psi \left( {\frac{1}
    {2}} \right) = - 2\ln 2 - \gamma \wedge \Gamma \left( {\frac{1}
    {2}} \right) = \sqrt \pi
    $

    We conclude that

    $\displaystyle f'\left( 0 \right) = - \frac{{\sqrt \pi }}
    {4}\left( {2\ln 2 + \gamma } \right)
    $

    Saludos Vivanca :P
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  5. #5
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    Quote Originally Posted by vivancos View Post
    Did you post this question because you need help, or did you post this question as a challenge question to other members.
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  6. #6
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    Quote Originally Posted by vivancos View Post
    compute

    $\displaystyle \int_0^{ + \infty } {e^{ - x^2 } \ln xdx} $
    See Moderator comments here: http://www.mathhelpforum.com/math-he...ntegral-d.html
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