Similar problem to Gaussian integral

[pierce15]

We all know about the famous equation: $\int_{-\infty}^\infty e^{-x^2} dx=\sqrt{\pi}$.

How about $\int_{-\infty}^\infty e^{-x^4} dx$?

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

 

[Mute]

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.

 

[pierce15]

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?

$$\int_{0}^\infty x^{-x} dx$$

 

[Mute]

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).

 

[lurflurf]

$$\int_0^\infty e^{-x^{2n}} \mathop{\text{dx}}=\Gamma \left( 1+\frac{1}{2n} \right)=\left( \frac{1}{2n}\right)!=\Pi\left( \frac{1}{2n}\right)=\int_0^\infty x^{2n} \mathop{e^{-t}}\mathop{\text{dx}}$$

Numerical values can be computed with software or looked up in tables.

 

[pierce15]

$$\int_0^\infty e^{-x^{2n}} \mathop{\text{dx}}=\Gamma \left( 1+\frac{1}{2n} \right)$$

How would you show this?

 

[lurflurf]

that should have been
$$\int_0^\infty e^{-x^{2n}} \mathop{\text{dx}}=\Gamma \left( 1+\frac{1}{2n} \right)=\left( \frac{1}{2n}\right)!=\Pi\left( \frac{1}{2n}\right)=\int_0^\infty t^{1/(2n)} \mathop{e^{-t}}\mathop{\text{dt}}=\int_0^\infty t^{1+1/(2n)} \mathop{e^{-t}}\mathop{\dfrac{\text{dt}}{t}}$$

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.

 

[JJacquelin]

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

Hi !
Have a look at "The Sophomores Dream Function", by the link :
http://www.scribd.com/JJacquelin/documents