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## Integral of e^(x^2)*dx? Rate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://scienceforums.com/index.php?app=forums&module=ajax&section=topics&do=rateTopic&t=11973&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>

### #1Edge

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Posted 19 July 2007 - 08:44 PM

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?
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### #2sanctus

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Posted 19 July 2007 - 10:56 PM

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:
$(\int_{-\infty}^{\infty}e^{-x^2}dx)^2=(\int_{-\infty}^{\infty}e^{-x^2}dx)\cdot (\int_{-\infty}^{\infty}e^{-y^2}dy)=$
$(\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy)= \int_0^{2\pi}d\phi\int_0^\infty dr r e^{-r^2}=-\pi e^{-r^2}\vert^\infty_0=\pi$
hence the integral is $\sqrt{\pi}$

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### #3 Guest_Lambus_*

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Posted 20 July 2007 - 12:50 AM

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### #4Nootropic

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Posted 21 July 2007 - 12:45 AM

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.
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### #5sanctus

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Posted 21 July 2007 - 01:09 AM

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.

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### #6Nootropic

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Posted 22 July 2007 - 12:38 AM

Differential galois theory!
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### #7sanctus

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Posted 22 July 2007 - 12:48 AM

???

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### #8 Guest_Lambus_*

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Posted 22 July 2007 - 01:17 AM

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### #9Nootropic

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Posted 22 July 2007 - 06:32 AM

Guess that works, Lambus. Differential galois theory allows one to look at a function and determine whether or not its antiderivative is elementary. Of course I could have explained this in my post, but who bothers to explain?

Check it out Differential Galois theory - Wikipedia, the free encyclopedia
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### #10 Guest_Lambus_*

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Posted 22 July 2007 - 09:41 PM

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### #11Qfwfq

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Posted 22 July 2007 - 11:45 PM

sanctus said:

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.
Inutil insegnà al mus, si piart timp, in plui si infastidìs la bestie.

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### #12Nootropic

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Posted 23 July 2007 - 12:33 PM

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).
"In heaven all the interesting people are missing."
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### #13takeonme

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Posted 01 February 2010 - 09:13 PM

Nootropic said:

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.

I'm not sure if you ever got round to doing this, but I think it can be integrated quite easily using a reduction formula as long as m is odd.
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### #14Listener

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Posted 15 February 2010 - 08:45 AM

sanctus said:

hence the integral is $\sqrt{\pi}$

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?
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### #15sanctus

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Posted 16 February 2010 - 04:00 AM

Listener, what exactly is your question? I mean what do you mean by constructive proof?

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