Jump to content
Science Forums

Integral of e^(x^2)*dx?


Edge

Recommended Posts

Gosh, it's been a long since I last visited.

Anyway, I was making a math study the other day on integrals and things like that. You know, you get to the point where you study complex ways of integrating functions...

However, I wonder... is there an integral for the function mentioned above...

Can we integrate every function no matter how weird is it?

I tried integrating that equation:

e^(x^2)dx ... and man, that was frustating... you only end up repeating the same process again and again...



so, I ask... is there a solution to this one? If not, why not?

 

http://www.physicsforums.com

Link to comment
Share on other sites

If you are looking for a Primitive to my knowledge there is none but the defined integral has values if you integrate over an ininite interval:

A way to see this is to start from the integral squared (change from x to -x, bounds do not change because you get one minus sign from the variable change and one for leaving the bounds as they are) and then to pass in polar coordinates:

[math](\int_{-\infty}^{\infty}e^{-x^2}dx)^2=(\int_{-\infty}^{\infty}e^{-x^2}dx)\cdot (\int_{-\infty}^{\infty}e^{-y^2}dy)=[/math]

[math](\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy)= \int_0^{2\pi}d\phi\int_0^\infty dr r e^{-r^2}=-\pi e^{-r^2}\vert^\infty_0=\pi[/math]

hence the integral is [imath]\sqrt{\pi}[/imath]

Link to comment
Share on other sites

It is of an interesting note that in general the vast majority of functions cannot be easily integrated into a closed form, by this, I mean NOT a power series expansion, which is how we define the "imaginary error function" as noted by lambus. As noted by Sanctus, it is certainly possible to integrate this function over the entire real line, or any infinite interval. Interestingly enough, you can integrate a function of the form f(x) = x^5 * e^(x^2). Before ripping your hair out, make a rationalizing substiution and don't be afraid to do make another substiution and apply parts more than once. It's a fun integral, and interesting, I might say. I have actually meant to take time to look at functions of the form f(x) = x^m * e^(x^k), but I have yet to get around to that.

 

And the answer to your question is a resouding no, no we cannot integrate any function. The function e^(x^2) is of a class a functions who antiderivatives are defined as transcendental (neither elementary nor algebraic) and are not expressed in terms of a "normal" function. And the reason for there be no explicit formulae (as opposed to an infinite chain of polynomials) for certain integrals is rather complicated one, and I think I need some time to brood over that one (when it's not six in the morning). But interesting question, you'd never believe how many people get their doctorates in mathematics and never ask such simple questions.

 

And Sanctus, I've seen the derivation of this integral, but the one thing I never understood is why the integral squared is equal to the integral times the integral with respect to y. This is probably due to my unfamiliarity with multivariable calculus.

Link to comment
Share on other sites

The way I see it Nootropic is: the defined integral is just a number (including infinities, so maybe it gets abit more complicated) and a the root of a number squared is just a number. About y, again in the definite integral it is just a kind of summation variable (there was an expression for these variables in english, but I don't remember now, maybe "dummy variable"?) so you can call it whatever you want.

Also maybe the derivation of the integral the way I showed is complete only if you first show that the the definte integral not squared converges.

Hope you understand what I want to say.

Link to comment
Share on other sites

Also maybe the derivation of the integral the way I showed is complete only if you first show that the the definte integral not squared converges.
Actually, I'd say the computation of the square shows the integral to be convergent. It finds the integral in r to be, and a quantity that's neither defined nor finite couldn't have a finite defined square.
Link to comment
Share on other sites

The gamma function is in the class of functions defined by integrals. It is relatively easy to show that gamma(x) = (n-1)! using integration by parts. It would not be a far stretch to say the gamma function is an extension to all real numbers not including the negative integers (to extend it to all but the negative integers note that the gamma functions satisfies the functional equation f(x+1) = xf(x) and rearrange to f(x+1)/x = f(x), which will change the domain of the gamma function).

Link to comment
Share on other sites

  • 2 years later...
  • 2 weeks later...

hence the integral is [imath]\sqrt{\pi}[/imath]

 

Hi!

 

You have made a nice proof. Later I have found similar proof here: Gaussian integral - Wikipedia, the free encyclopedia

 

But what about constructive proof of Normal Distribution probability density function:

 

???

I am assured all of us know about its properties and practical implementation.

But how did Carl Friedrich Gauss obtained this formulae?

Do You know any books or papers where this problem explained?

Link to comment
Share on other sites

what do you mean by constructive proof?

 

Sanctus, I mean a proof like the proof stated here: Gauss_formulae_constructive_proof.djvu (7 pages)

Text is written in russian but You can easily understand it looking at figures & formulas.

The explanation of DJVU format is here: How to open .djvu files..........

 

Common idea of proof is usage of target shooting model...

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
×
×
  • Create New...