
#19
Nov708, 04:42 AM

P: 107

As has been said, there is no elementary antiderivative (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. 



#20
Nov708, 05:33 AM

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#21
Dec2009, 03:32 PM

P: 1

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




#22
Dec2009, 03:55 PM

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#23
Dec2109, 01:50 PM

P: 140

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




#24
Dec2109, 02:23 PM

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We get then, as an antiderivative 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! 



#25
Dec2109, 10:20 PM

P: 403

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.




#26
Dec2209, 06:59 AM

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#27
Dec2309, 05:55 AM

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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. 



#28
Jan110, 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] 



#29
Apr1010, 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) 



#30
Apr1110, 12:54 AM

P: 140

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



#31
Apr1110, 07:42 AM

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P: 11,985

I have assumed an infinite lifetime for the forum. 



#32
Apr1110, 09:08 AM

P: 24

a nice point of view 



#33
Apr1110, 06:40 PM

P: 2

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



#34
Apr1110, 11:27 PM

P: 140

[tex]\int [/tex]e^{x2}dx If: u = x^{2} 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 antiderivative of apples can't be oranges. In no nice way is this function integrable. 


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