The integration of e^(x^2)

[Owais Iftikhar]

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?

 

[arildno]

You won't be able to find a nice anti-derivative to the function $e^{x^{2}}$

 

[HallsofIvy]

I don't know how many times we have to say this- it's given in any good calculus book. The anti-derivative of $e^{x^2}$ is not an "elementary" function. Because $e^{-x^2}$ is used so much in statistics, the "error" function, erf(x), is defined as an anti-derivative of $e^{-x^2}$ but I don't believe anyone has ever bothered to give a name to the anti-derivative of $e^{x^2}$.

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

 

[D H]

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

$$\int_0^x{\mathrm e}^{t^2}{\mathrm d}t =<br /> -i\sqrt{\frac{\pi}2}\operatorname{erf}(ix)$$

 

[JasonRox]

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]

Polar coordinates

Use polar coordinates and you'll make progress

 

[inFinie]

If you want a series representation you can expand e^x to taylor series near 0, substitute x with x^2, integrate.

 

[mepcotterell]

-isqrt(pi)erf(ix)/2

 

[kevin_timmins]

what on erf are you talking about?

 

[Tac-Tics]

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]

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]

hahahahaha what a good reason to bring back and old thread

 

[thharrimw]

what dose erf stand for?

 

[flouran]

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]

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 $\exp(-t^2/2)$:

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

Complete the square in the exponent:

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

which gives
$$\sqrt{2\pi}e^{s^2/2}.$$

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]

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]

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]

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.

 

[yasiru89]

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.

An approximation with this particular function where the terms increase for real [itex]x[/tex], makes any such process largely meaningless.<br /> <br /> As has been said, there is no elementary anti-derivative (barring the infinite series version), for e^(-x^2) from 0 to infinity this is handled by converting to a polar integral, but for this one, that will obviously be infinite.[/itex]

 

[HallsofIvy]

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.

Why would you think that changing from t²/2 to t² would make a difference in integrating? A simple subtitution can obviously change either one into the other. If one cannot be integrated in terms of elementary functions, the other clearly cannot either.

 

[13neaultar]

Do a little research on the Chain Rule. This is how you solve these kinds of problems.

 

[arildno]

Do a little research on the Chain Rule. This is how you solve these kinds of problems.

Wrong.

 

[thepatient]

How about integration using summation of (x^n)/n! ?

 

[arildno]

How about integration using summation of (x^n)/n! ?

That is definitely an option; however, we do not regard an infinite power series as a nice anti-derivative, nor as a finite combination of elementary functions.

We get then, as an anti-derivative F(x):
$$F(x)=C+\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)n!}$$
This is most definitely not a nice function!

 

[JSuarez]

It´s not possible to express the primitive of $$e^{x^{2}}$$ in terms of algebraic combinations of "elementary" functions (polynomials, trignometric, exponentials and their inverses). The reason is far from trivial and lies in a field called Differential Galois Theory, that is an analogue for differential equations of the algebraic Galois Theory for polynomials. The inexpressability of this primitive (and many others), is akin to the one regarding roots of polynomials with degree greater than four.

 

[HallsofIvy]

Do a little research on the Chain Rule. This is how you solve these kinds of problems.

No, that is how you differentiate functions. This question was about integrating.

 

[arildno]

To Thepatient:

You ought to see that the above power series is only useful for tiny x's, i.e, when just the few first terms are strictly dominant in magnitude.

For large x's (say A and B), the following scheme might be used:

$$I=\int_{A}^{B}{e}^{x^{2}}dx=\int_{A}^{B}\frac{2x}{2x}e^{x^{2}}dx=\frac{1}{2x}e^{x^{2}}\mid_{A}^{B}+\int_{A}^{B}\frac{1}{2x^{2}}e^{x^{2}}dx=\frac{1}{2x}e^{x^{2}}\mid_{A}^{B}+\frac{1}{2C^{2}}I$$
where C is some number between A and B.
(the existence of such a C is guaranteed by the intermediate value theorem)

Thus, we get:
$$I=\frac{\frac{1}{2x}e^{x^{2}}\mid_{A}^{B}}{1-\frac{1}{2C^{2}}}$$

when C can be regarded as a big number, the numerator by its own is a fairly accurate estimate.

 

[vitruvianman]

$$erf(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$

when

$$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$$

 

[LTA85]

You can use integration by substitution.

Problem: e^x^2

Solution: Set u = x^2, then find the derivative of u, du = 2 dx
solve for dx. dx = du/2
and multiply it by the problem. dx*e^x^2 = 1/2*e^x^2.

Answer: 1/2 e^x^2 + C (c is a constant)

 

[thepatient]

You can use integration by substitution.

Problem: e^x^2

Solution: Set u = x^2, then find the derivative of u, du = 2 dx
solve for dx. dx = du/2
and multiply it by the problem. dx*e^x^2 = 1/2*e^x^2.

Answer: 1/2 e^x^2 + C (c is a constant)

No.


If u = x^2. du = 2xdx, not 2dx.