# The integration of e^(x^2)

by Owais Iftikhar
Tags: integration
P: 107
 Quote by 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.
An approximation with this particular function where the terms increase for real $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. Math Emeritus Sci Advisor Thanks PF Gold P: 39,682  Quote by 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. 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.  P: 1 Do a little research on the Chain Rule. This is how you solve these kinds of problems. Sci Advisor HW Helper PF Gold P: 12,016  Quote by 13neaultar Do a little research on the Chain Rule. This is how you solve these kinds of problems. Wrong.  P: 140 How about integration using summation of (x^n)/n! ? Sci Advisor HW Helper PF Gold P: 12,016  Quote by thepatient 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!  P: 402 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. Math Emeritus Sci Advisor Thanks PF Gold P: 39,682  Quote by 13neaultar 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.  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: $$I=\int_{A}^{B}{e}^{x^{2}}dx=\int_{A}^{B}\frac{2x}{2x}e^{x^{2}}dx=\frac{ 1}{2x}e^{x^{2}}\mid_{A}^{B}+\int_{A}^{B}\frac{1}{2x^{2}}e^{x^{2}}dx=\fr ac{1}{2x}e^{x^{2}}\mid_{A}^{B}+\frac{1}{2C^{2}}I$$ 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.  P: 24 $$erf(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{-t^2}dt$$ when $$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$$  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) P: 140  Quote by 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) No. If u = x^2. du = 2xdx, not 2dx. Mentor P: 12,074  Quote by 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 [itex]e^{x^2}$ 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.
P: 24
 Quote by Redbelly98 I have assumed an infinite lifetime for the forum.
lol

a nice point of view
P: 2
 Quote by thepatient 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.
P: 140
 Quote by LTA85 You're right. What a stupid error I made. u = x^2, du = 2x multiply du * e^u = 2x e^x^2.

$$\int$$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.

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