Gaussian integral .

can some one tell me how to go about solving the gaussian integral
e^(-x^2) I know it has no elementary integral . but i was told the improper integral from -inf to positive inf can be solved and some said that i haft to do it complex numbers or something and help would be great , this is not a homework question.

rock.freak667
Homework Helper
$$\int_{- \infty} ^{\infty} e^{-x^2} dx = 2 \int_0 ^{\infty} e^{-x^2} dx$$

Write I as ∫e-x2 dx

If you change the variable x for y and then multiply the two I's you will get

$$I^2 = (\int_0 ^{\infty} e^{-x^2} dx)(\int_0 ^{\infty} e^{-y^2} dy)$$

this is the same as

$$I^2= \int_0 ^{\infty} e^{-x^2-y^2}dxdy$$

Convert to polar coordinates now.

can some one tell me how to go about solving the gaussian integral
e^(-x^2) I know it has no elementary integral . but i was told the improper integral from -inf to positive inf can be solved and some said that i haft to do it complex numbers or something and help would be great , this is not a homework question.

http://en.wikipedia.org/wiki/Gaussian_integral

Why don't you search by yourself before asking?

Mute
Homework Helper
The Gaussian integral is not one that you can do with contour integrals (complex numbers)*. Typically to evaluate it you consider

$$I = \int_{-\infty}^\infty dx~e^{-x^2};$$
then,

$$I^2 = \int_{-\infty}^\infty dx~e^{-x^2} \int_{-\infty}^\infty dy~e^{-y^2} = \int_{0}^{2\pi} d\theta \int_{0}^\infty dr r e^{-r^2}$$
and evaluating.

An intro calculus textbook should treat this example more carefully and rigorously than I've written (identifying $(x,y) \in (-\infty,\infty) \mbox{U} (-\infty,\infty)$ with $(r,\theta) \in [0,\infty) \mbox{U} [0,2\pi)$ technically takes some care to do).

*I once read an account of how one could derive the result using a contour integral, but it was not at all a simple (or obvious) integrand that one needed to use in order to get the result.

An intro calculus textbook should treat this example more carefully and rigorously than I've written (identifying $(x,y) \in (-\infty,\infty) \mbox{U} (-\infty,\infty)$ with $(r,\theta) \in [0,\infty) \mbox{U} [0,2\pi)$ technically takes some care to do).

Why when i convert the integral to polar form, do the intervals become $(r,\theta) \in [0,\infty) \mbox{U} [0,2\pi)$. Do you prove to me?Thanks.

Redbelly98
Staff Emeritus
Why when i convert the integral to polar form, do the intervals become $(r,\theta) \in [0,\infty) \mbox{U} [0,2\pi)$. Do you prove to me?Thanks.