Solving the Integral of x^log(x): How to Approach?

[foo_daemon]

Hi,

I need to solve the following integral from 0 to \infty :
Please note that my professor has defined log(x) = ln(x) , i.e. 10 is not the default base.

\int { e^{\frac{-log(x)^2}{2} } dx }

Through 'simplification' ( e^{log(x)} = x ), I have translated the function to:
\int { x^{ \frac{-log(x)}{2} } dx } , which appears easier to integrate.

However, I am uncertain where to go from here (or if this is even the right direction). Obviously the formula \int { x^c } = \frac{x^{c+1}}{c+1} doesn't work here, as c is not a constant.

I have tried using mathematica to see if I could reverse-engineer the integral, but the resulting 'error' function seems significantly complex and doesn't help.

I have tried substitution with it in the form \frac{ 1}{x^{\frac{log(x)}{2}} } using u = \frac{log(x)}{2} \ \ du = \frac {1}{2x} dx , which almost works, except that the problem then becomes \int {2 du^u } , which seems quite odd.

Any hints for how to approach this problem?

 

[Cyosis]

The exponential form is definitely the easier one. Substitute u= \log x/ \sqrt{2}. Then complete the square and use that

\int_{-\infty}^\infty e^{-ax^2}=\sqrt{\pi/a}.