Similar problem to Gaussian integral

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
Homework Helper
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.

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

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

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

lurflurf
Homework Helper
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.

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