The integration of e^(x^2)

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

An approximation with this particular function where the terms increase for real [itex]x[/tex], makes any such process largely meaningless.

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.

HallsofIvy

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

Wrong.

thepatient

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

arildno

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):
[tex]F(x)=C+\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)n!}[/tex]
This is most definitely not a nice function!

JSuarez

It´s not possible to express the primitive of [tex]e^{x^{2}}[/tex] 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

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:

[tex]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=\fr ac{1}{2x}e^{x^{2}}\mid_{A}^{B}+\frac{1}{2C^{2}}I[/tex]
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:
[tex]I=\frac{\frac{1}{2x}e^{x^{2}}\mid_{A}^{B}}{1-\frac{1}{2C^{2}}}[/tex]

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

vitruvianman

[tex] erf(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt [/tex]

when

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

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

No.


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

Redbelly98

My best guess, this number is aleph-naught: infinite, but countable.

I have assumed an infinite lifetime for the forum.

vitruvianman

lol

a nice point of view

LTA85

You're right. What a stupid error I made.

u = x^2, du = 2x
multiply du * e^u = 2x e^x^2.

thepatient

[tex]\int [/tex]e^x2dx If:

u = x²
du = 2xdx

You can't substitute dx by du, since dx = du/(2x), not just du. You would have dx in terms of two variables, you can't integrate x in terms of u. The anti-derivative of apples can't be oranges.


In no nice way is this function integrable.