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Integration of x^log(x) ?

  1. Apr 25, 2010 #1
    Hi,

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

    [tex]\int { e^{\frac{-log(x)^2}{2} } dx }[/tex]

    Through 'simplification' ( [tex] e^{log(x)} = x [/tex] ), I have translated the function to:
    [tex] \int { x^{ \frac{-log(x)}{2} } dx } [/tex] , 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 [tex] \int { x^c } = \frac{x^{c+1}}{c+1}[/tex] 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 [tex] \frac{ 1}{x^{\frac{log(x)}{2}} } [/tex] using [tex] u = \frac{log(x)}{2} \ \ du = \frac {1}{2x} dx [/tex] , which almost works, except that the problem then becomes [tex] \int {2 du^u } [/tex] , which seems quite odd.

    Any hints for how to approach this problem?
     
    Last edited: Apr 25, 2010
  2. jcsd
  3. Apr 25, 2010 #2

    Cyosis

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    Homework Helper

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

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