1. ## hard integral

compute

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

2. Wolfram Mathematica Online Integrator

That doesn't look like a fun time!

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

3. 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

4. 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.

5. Originally Posted by vivancos
compute

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