The integration of e^(x^2)


by Owais Iftikhar
Tags: integration
Owais Iftikhar
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#1
Dec11-05, 03:34 PM
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i am tryaing to find the integral of "e" raise to power "x" square. by conventional method by dividing e^x^2 with derivative of the power of "e" i.e. "2x". But i am not confident on my answer. Please tell me am i right???
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arildno
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#2
Dec11-05, 03:37 PM
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You won't be able to find a nice anti-derivative to the function [itex]e^{x^{2}}[/itex]
HallsofIvy
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#3
Dec11-05, 04:12 PM
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I don't know how many times we have to say this- it's given in any good calculus book. The anti-derivative of [itex]e^{x^2}[/itex] is not an "elementary" function. Because [itex]e^{-x^2}[/itex] is used so much in statistics, the "error" function, erf(x), is defined as an anti-derivative of [itex]e^{-x^2}[/itex] but I don't believe anyone has ever bothered to give a name to the anti-derivative of [itex]e^{x^2}[/itex].

That "conventional" method you talk about, only works when the derivative of the power is a constant.

D H
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#4
Dec12-05, 05:53 AM
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The integration of e^(x^2)


No need to give the integral of [itex]e^{x^2}[/itex] a new name, as it already has one: erf(ix).
Since
[tex]\int_0^x{\mathrm e}^{-t^2}{\mathrm d}t =
\sqrt{\frac{\pi}2}\operatorname{erf}(x)[/tex]

[tex]\int_0^x{\mathrm e}^{t^2}{\mathrm d}t =
-i\sqrt{\frac{\pi}2}\operatorname{erf}(ix)[/tex]
JasonRox
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#5
Dec12-05, 08:32 AM
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The best way to know if you are right is to take the derivative of the anti-derivative.

The derivative of e^(x^2)*2*x, with respect to x, does not equal e^(x^2).
omagdon7
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#6
Dec12-05, 01:59 PM
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Use polar coordinates and you'll make progress
inFinie
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#7
Dec12-05, 02:02 PM
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If you want a series representation you can expand e^x to taylor series near 0, substitute x with x^2, integrate.
mepcotterell
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#8
Dec17-05, 10:47 AM
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-isqrt(pi)erf(ix)/2
kevin_timmins
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#9
Oct21-08, 06:42 AM
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what on erf are you talking about?
Tac-Tics
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#10
Oct21-08, 08:31 AM
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It's called the gaussian function.

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

It's integral is non-elementary, meaning you can't find a polynomial, trigonomic, or exponential function (or combination of those functions) which describe it. The integral is sometimes called "erf" (though it might be scaled differently), also called the "error function".

The proof of why the integral is the square root of pi is really interesting, and is mentioned on the wiki article.
D H
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#11
Oct21-08, 08:57 AM
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Quote Quote by kevin_timmins View Post
what on erf are you talking about?
Why on earth are you necromancing a three year old thread just to make a bad pun?
ice109
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#12
Oct21-08, 09:33 AM
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hahahahaha what a good reason to bring back and old thread
thharrimw
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#13
Oct21-08, 12:15 PM
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what dose erf stand for?
flouran
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#14
Nov3-08, 07:32 PM
P: 64
I was wondering if it would in fact be possible to express the integral of e^(x^2) in terms of elementary functions. Couldn't I just use a bilateral Laplace transform on e^(x^2) and convert it into the s domain. Then, I would integrate with respect to s (since switching functions from the time domain to the s domain makes it linear, thus making it easy to integrate), and then convert that function back to the time domain using the Bromwich Integral?
Mute
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#15
Nov3-08, 07:50 PM
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Quote Quote by flouran View Post
I was wondering if it would in fact be possible to express the integral of e^(x^2) in terms of elementary functions. Couldn't I just use a bilateral Laplace transform on e^(x^2) and convert it into the s domain. Then, I would integrate with respect to s (since switching functions from the time domain to the s domain makes it linear, thus making it easy to integrate), and then convert that function back to the time domain using the Bromwich Integral?
What confidence do you have that finding the antiderivative of the laplace transform of f(t) with respect to s then inverse laplace transforming it would give you the antiderivative of f(t)? Even if it did, I would wager that you wouldn't be able to do the resulting inverse laplace transform in terms of elementary functions!

For instance, if we calculate the bilateral laplace transform for [itex]\exp(-t^2/2) [/itex]:

[tex]F(s) = \int_{-\infty}^{\infty}dt~e^{-st}e^{-t^2/2}[/tex]

Complete the square in the exponent:

[tex]\int_{-\infty}^{\infty}dt~e^{-\frac{1}{2}(t^2 + 2st + s^2 - s^2)} = \int_{-\infty}^{\infty}dt~e^{-\frac{1}{2}(t+s)^2}e^{s^2/2}[/tex]

which gives
[tex]\sqrt{2\pi}e^{s^2/2}.[/tex]

So, it looks like in order to integrate the bilateral laplace transform, you still need to integrate a Gaussian! (Well, Gaussian-like, due to the sign of the exponent).

(Edit: I'm pretty sure the integrating thing wouldn't work anyways. If you take the regular laplace transform for t from 0 to infinity, then consider L[t] = 1/s^2. Integrating with respect to s gives -1/s. Taking the inverse lapace transform of that gives the step function -H(t), which is certainly not the derivative of t.)
flouran
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#16
Nov3-08, 08:24 PM
P: 64
I know it sounds like a stupid question but, instead of setting f(t)=e^(-t^2/2), why can't we set it to be e^(t^2), and see if we can integrate its Laplace transform (since e^(-t^2/2) is completely different from e^(t^2))? I especially pose this problem to Mute.
flouran
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#17
Nov3-08, 10:51 PM
P: 64
i guess you are right, Mute, e^(x^2) doesn't have a true integral in elementary terms. Thus, I agree with the others, that the only way to solve Iftikhar's problem is by expressing the integral of the function in terms of erf.

THAT'S THE ONLY WAY, for now at least.
Reedeegi
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#18
Nov7-08, 12:45 AM
P: 99
Correct me if I am wrong, but can't you integrate the series expansion of e^x^2 term by term? Using numerical methods, the area under the curve of e^x^2 is very close to about 6 iterations of the term-by-term integration of the series.


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