compute
$\displaystyle \int_0^{ + \infty } {e^{ - x^2 } \ln xdx} $
Wolfram Mathematica Online Integrator
That doesn't look like a fun time!
Hey, how did you get a picture by your name?
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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