# Integration of x^log(x) ?

1. Apr 25, 2010

### 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?

Last edited: Apr 25, 2010
2. Apr 25, 2010

### Cyosis

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

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