- #1

- 315

- 2

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?

- Thread starter pierce15
- Start date

- #1

- 315

- 2

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?

- #2

Mute

Homework Helper

- 1,388

- 10

- #3

- 315

- 2

OK. While you're here, can you take a look at this integral too?

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

- #4

Mute

Homework Helper

- 1,388

- 10

There's no closed form expression for the antiderivative of ##x^{\pm x}##, so the integral typically has to be done numerically.OK. While you're here, can you take a look at this integral too?

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

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

Homework Helper

- 2,432

- 132

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

- #6

- 315

- 2

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

- #7

lurflurf

Homework Helper

- 2,432

- 132

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.

[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.

Last edited:

- #8

- 798

- 34

Hi !OK. While you're here, can you take a look at this integral too?

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

Have a look at "The Sophomores Dream Function", by the link :

http://www.scribd.com/JJacquelin/documents

- Last Post

- Replies
- 4

- Views
- 4K

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 900

- Last Post

- Replies
- 8

- Views
- 3K

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K