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Integral of exp(-x^n)

  1. Oct 17, 2011 #1
    Hello,

    How do I prove or find a proof for [itex] \int e^{-(x_1^2+\dots+x_\nu^2)^{k/2}} \mathrm d x = \frac{\pi^{\nu/2} \Gamma \left( \frac{\nu}{k}+1 \right)}{\Gamma \left( \frac{\nu}{2} +1\right)}[/itex]?

    EDIT: I think I have it, and as I can't delete this thread, I will sketch the solution:
    first of all, notice that proving it for the case of [itex]\nu = 1[/itex] is enough, because we can always go back to that specific case by using polar coordinates (replacing [itex]x_1^2 + \dots + x_\nu^2 \mapsto r^2[/itex]); and then:

    [itex] \int_{-\infty}^{+\infty} e^{-|x|^k} \mathrm dx = 2 \int_{0}^{+\infty} e^{-x^k} \mathrm dx[/itex] and if we substitute with y=x^k => [itex]\mathrm d y = k y^{(k-1)/k} \mathrm d x[/itex], this is equal to [itex]2\int_0^{+\infty} e^{-y} y^{\frac{1}{k}-1} \frac{1}{k} \mathrm d y = \frac{2}{k} \Gamma \left(\frac{1}{k} \right)[/itex] and using [itex]x \Gamma(x) = \Gamma(x+1)[/itex] we get:
    [itex] \int_{-\infty}^{+\infty} e^{-|x|^k} \mathrm dx = 2 \Gamma \left( \frac{1}{k} + 1 \right) = \frac{\pi^{1/2} \Gamma \left( \frac{1}{k}+1 \right)}{\Gamma \left( \frac{1}{2} +1\right)}[/itex]
     
    Last edited: Oct 17, 2011
  2. jcsd
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