Register to reply

The integration of e^(x^2)

by Owais Iftikhar
Tags: integration
Share this thread:
yasiru89
#19
Nov7-08, 04:42 AM
P: 107
Quote Quote by Reedeegi View Post
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.

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
#20
Nov7-08, 05:33 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Quote Quote by flouran View Post
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
#21
Dec20-09, 03:32 PM
P: 1
Do a little research on the Chain Rule. This is how you solve these kinds of problems.
arildno
#22
Dec20-09, 03:55 PM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Quote Quote by 13neaultar View Post
Do a little research on the Chain Rule. This is how you solve these kinds of problems.
Wrong.
thepatient
#23
Dec21-09, 01:50 PM
P: 140
How about integration using summation of (x^n)/n! ?
arildno
#24
Dec21-09, 02:23 PM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Quote Quote by thepatient View Post
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):
[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
#25
Dec21-09, 10:20 PM
P: 402
Itīs 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
#26
Dec22-09, 06:59 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Quote Quote by 13neaultar View Post
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
#27
Dec23-09, 05:55 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
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
#28
Jan1-10, 10:28 AM
P: 24
[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
#29
Apr10-10, 10:02 PM
P: 2
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
#30
Apr11-10, 12:54 AM
P: 140
Quote Quote by LTA85 View Post
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.
Redbelly98
#31
Apr11-10, 07:42 AM
Mentor
Redbelly98's Avatar
P: 12,071
Quote Quote by HallsofIvy View Post
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.
My best guess, this number is aleph-naught: infinite, but countable.

I have assumed an infinite lifetime for the forum.
vitruvianman
#32
Apr11-10, 09:08 AM
P: 24
Quote Quote by Redbelly98 View Post
I have assumed an infinite lifetime for the forum.
lol

a nice point of view
LTA85
#33
Apr11-10, 06:40 PM
P: 2
Quote Quote by thepatient View Post
No.


If u = x^2. du = 2xdx, not 2dx.
You're right. What a stupid error I made.

u = x^2, du = 2x
multiply du * e^u = 2x e^x^2.
thepatient
#34
Apr11-10, 11:27 PM
P: 140
Quote Quote by LTA85 View Post
You're right. What a stupid error I made.

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


[tex]\int [/tex]ex2dx 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.


Register to reply

Related Discussions
Integration problems. (Integration by parts) Calculus & Beyond Homework 5
Integration help Calculus 3
Dirac delta functions integration Calculus & Beyond Homework 1
How should I integrate this differential equation? Calculus 9
Question arrising from integration homework (advanced integration i guess?) Calculus 9