Solve Gaussian Integral: e^(-x^2)

[cragar]

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]

[tex]\int_{- \infty} ^{\infty} e^{-x^2} dx = 2 \int_0 ^{\infty} e^{-x^2} dx[/tex]


Write I as ∫e^-x2 dx

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


[tex]I^2 = (\int_0 ^{\infty} e^{-x^2} dx)(\int_0 ^{\infty} e^{-y^2} dy)[/tex]

this is the same as

[tex]I^2= \int_0 ^{\infty} e^{-x^2-y^2}dxdy[/tex]

Convert to polar coordinates now.

 

[netheril96]

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]

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

[tex]I = \int_{-\infty}^\infty dx~e^{-x^2};[/tex]
then,

[tex]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}[/tex]
and evaluating.

An intro calculus textbook should treat this example more carefully and rigorously than I've written (identifying [itex](x,y) \in (-\infty,\infty) \mbox{U} (-\infty,\infty)[/itex] with [itex](r,\theta) \in [0,\infty) \mbox{U} [0,2\pi)[/itex] 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.

 

[cragar]

Thanks for your answers

 

[coki2000]

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

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

 

[Redbelly98]

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

If you are integrating over "all space", then r and θ must take on all possible values. Hence those integration limits, 0≤r<∞ and 0≤θ≤2π

 

[Gib Z]

http://www.math.uconn.edu/~kconrad/blurbs/analysis/probint.pdf

I find the fourth proof particularly nice.

 

[Count Iblis]

Another way to do this is to consider the integral of sin^n(x) or cos^n(x) and use that to derive Stirling's approximation by first deriving the Wallis product.

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

Since the asymptotics of n! can also be obtained directly from its integral representation and then using the saddle point method, this yields a derivation of the Gaussian integral.