Similar problem to Gaussian integral

  • Thread starter pierce15
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  • #1
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We all know about the famous equation: [itex]\int_{-\infty}^\infty e^{-x^2} dx=\sqrt{\pi}[/itex].

How about [itex]\int_{-\infty}^\infty e^{-x^4} dx[/itex]?

Or, in general, can we calculate any integral in the form [itex]\int_{-\infty}^\infty e^{-x^n} dx[/itex], where n is an even counting number?
 

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  • #2
Mute
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There's no closed form for every even ##n##. A simple change of variables will turn the integral into a Gamma function integral, which typically doesn't have closed form expressions. It so happens that for the case n = 2 there is a closed form. I don't know if there is a nice, simple expression for any other n, but it doesn't appear to be the case for n =4.
 
  • #3
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There's no closed form for every even ##n##. A simple change of variables will turn the integral into a Gamma function integral, which typically doesn't have closed form expressions. It so happens that for the case n = 2 there is a closed form. I don't know if there is a nice, simple expression for any other n, but it doesn't appear to be the case for n =4.
OK. While you're here, can you take a look at this integral too?

$$\int_{0}^\infty x^{-x} dx$$
 
  • #4
Mute
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OK. While you're here, can you take a look at this integral too?

$$\int_{0}^\infty x^{-x} dx$$
There's no closed form expression for the antiderivative of ##x^{\pm x}##, so the integral typically has to be done numerically.

The definite integral you're asking about doesn't appear to have a closed form result, but there is a nice identity for the integral from 0 to 1. See Sophomore's Dream on wikipedia.

(The integral still has to be calculated numerically, but it has a nice alternate expression in terms of sums).
 
  • #5
lurflurf
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[tex]\int_0^\infty e^{-x^{2n}} \mathop{\text{dx}}=\Gamma \left( 1+\frac{1}{2n} \right)=\left( \frac{1}{2n}\right)!=\Pi\left( \frac{1}{2n}\right)=\int_0^\infty x^{2n} \mathop{e^{-t}}\mathop{\text{dx}}[/tex]

Numerical values can be computed with software or looked up in tables.
 
  • #6
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[tex]\int_0^\infty e^{-x^{2n}} \mathop{\text{dx}}=\Gamma \left( 1+\frac{1}{2n} \right)[/tex]
How would you show this?
 
  • #7
lurflurf
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that should have been
[tex]\int_0^\infty e^{-x^{2n}} \mathop{\text{dx}}=\Gamma \left( 1+\frac{1}{2n} \right)=\left( \frac{1}{2n}\right)!=\Pi\left( \frac{1}{2n}\right)=\int_0^\infty t^{1/(2n)} \mathop{e^{-t}}\mathop{\text{dt}}=\int_0^\infty t^{1+1/(2n)} \mathop{e^{-t}}\mathop{\dfrac{\text{dt}}{t}}[/tex]

The change of variable u=x^(2n) will transform the fist integral into the second, the third just shifts one x and is sometimes taken as a definition of the gamma function.
 
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