hard integral

**vivancos** · March 17th 2010, 02:30 PM

compute

$\int_0^{ + \infty } {e^{ - x^2 } \ln xdx}$

**Anonymous1** · March 17th 2010, 07:33 PM

Wolfram Mathematica Online Integrator

That doesn't look like a fun time!

Hey, how did you get a picture by your name?

**danielomalmsteen** · March 19th 2010, 04:07 AM

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} $

Then

$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

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

We conclude that

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

Saludos Vivanca :P

**vivancos** · March 19th 2010, 09:46 AM

**mr fantastic** · March 20th 2010, 03:30 AM

Originally Posted by vivancos

Did you post this question because you need help, or did you post this question as a challenge question to other members.

**mr fantastic** · March 20th 2010, 01:46 PM

Originally Posted by vivancos

compute

$\int_0^{ + \infty } {e^{ - x^2 } \ln xdx}$

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