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Math Help - hard integral

  1. #1
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    hard integral

    compute

    \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 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} <br />

    Then

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

    And you have to

    f'\left( s \right) = \frac{1}<br />
{4}\Gamma '\left( {\frac{{s + 1}}<br />
{2}} \right) = \frac{1}<br />
{4}\psi \left( {\frac{{s + 1}}<br />
{2}} \right)\Gamma \left( {\frac{{s + 1}}<br />
{2}} \right) \Rightarrow f'\left( 0 \right) = \frac{1}<br />
{4}\psi \left( {\frac{1}<br />
{2}} \right)\Gamma \left( {\frac{1}<br />
{2}} \right)<br />

    Finally, knowing that \psi \left( {\frac{1}<br />
{2}} \right) =  - 2\ln 2 - \gamma  \wedge \Gamma \left( {\frac{1}<br />
{2}} \right) = \sqrt \pi  <br />

    We conclude that

    f'\left( 0 \right) =  - \frac{{\sqrt \pi  }}<br />
{4}\left( {2\ln 2 + \gamma } \right)<br />

    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

    \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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