1. I worked out a new integration technique for cases where substitutiuon of variables don't work.

i.e. where you can't get the intergand f(x) in terms of ONLY u and du.

It works for any amount of products that may contain f(g(x)) type terms.

Would like someone's rating of the importance of this.

2.

3. huh? So you can integrate e^(x^2)?

4. :-D

Yes I can. It actually makes a nice sequence:

Integral e^(x^2) dx = e^(x^2)*(1/2x + 1/(4x^3) + 3/(8x^5)

+15/(16 x^7) + ... )

where the general term in the sum is:

a_k / (2^(k+1)x^(2k+1))

and a_k is determined successively from:

a_0 = 1, a_1 = 1, a_(k+1) = a_k * (2k-1)

I have the paper with the technique for sale.

5. mmm, I assume you mean a_k / (2^(k+1) x^(2k+1) as I can't see any reason why you can't get a series which converges close to 0 but lets ignore questions of convergence. (though it really doesn't matter as you get a similar contradiction using your original equation)

Your series expansion for the integral e^(x^2) is incorrect. As an exercise in generating functions someone can do the working out (as i'm not going to try convert that to normal text without the help of latex).

If you want to use infinite series though, the easiest solution would be C + (x + 1/3 x<sup>3</sup> + 1/10 x<sup>5</sup> + ... + 1/[(2n + 1) n!] x<sup>2n+1</sup> + ... ) where C is a constant.

6. Did you test my formula by computation?

I will receck the formula, it must sum to close to 1 at x= 0, but my integratio ntechnique is well proven.

7. well to my checks it does not work in this case, assuming your solution you arrive at a contradiction - anyway the bigger question is what does it do so differently to a normal series expansion and integrating term by term?

8. Here's the deal, talanum1. Colloquially, "to integrate" something means to integrate it using "elementary" means--using known antiderivatives, substitution (simple, trigonometric, etc.), integration by parts, high school algebra, and the like, and not using, say, infinite series or complex analysis. If you:

1. start with a function which is built out of a finite number of applications of basic arithmetic operations and the usual transcendental functions (exponentiation, logarithms, trigonometric functions (usual and hyperbolic));

2. limit yourself to the techniques described above; and

3. can successfully integrate by said techniques

then your result must be of the form outlined in point 1. I believe the functions in point 1. are typically called "elementary functions," so if we agree to call the techniques described as "elementary techniques," then this can be more succinctly stated as, "If you can integrate an elementary function by elementary techniques, then the result must be an elementary function."

It's well-known that f(x) = e^{x^2} does not have an elementary derivative, so you cannot, in the colloquial sense, integrate f. However, in a broader sense of integrate, as pointed out by river_rat, you can integrate f. The easiest way is definitely to use the power series expansion of f, as the series expansion is well-known and easy to describe, and integration of power series is as easy as integration of polynomials.

The point of the above exposition? When you say you can integrate a function, you have to be clear what you mean. Your original language suggested that you believed you could integrate any composite function, for example f, in an elementary manner. Your more detailed description of your answer shows that you did not accomplish this impossible feat.

Now you seem to have come up with another method which gives a series representation for an antiderivative of f. We're interested in seeing what your method is. river_rat and I are perfectly capable of checking an expression to see whether it gives the correct derivative, but doing so would not divulge to us the meat of your result, which is the method. Please share!

9. :?

My technique is new and has a 2 step proof.

Set: g(x) = x^2 and

My tecnique produces the result:

f(x) = e^(g(x))/g'(x) - integral e^(g(x))* (1/g'(x))' dx

You can check that:

f ' (x) = e^(g(x))g'(x)/g'(x)

so f(x) is the solution, probably except at g'(x) = 0.

the first term of f(x) is e^(x^2)/2x and the rest of the terms is produced by the integral in f(x).

See if you can find fault with this.

10. well for starters the derivative of

f(x) = e^(g(x))/g'(x) - integral e^(g(x))/ (1/g'(x))' dx
is

e<sup>g(x)</sup> ( 1 + (g'(x))<sup>2</sup>/g''(x) - g''(x)/(g'(x))<sup>2</sup>)

and not

f ' (x) = e^(g(x))g'(x)/g'(x)

11. This looks like integration by parts to me. In fact, set u = 1/g'(x) and v = e^(g(x)) so that du = -(g''(x)/(g'(x))^2)dx and dv = g'(x)e^(g(x))dx. Then
using:

int u dv = u v - int v du

we get your formula. Care to show us your general method? I have a sneaking suspicion that it, too, reduces to integration by parts.

12. :?

If you take the derivative of the first term of f(x) as a product of functions:

e^g(x) * (1/g'(x))

and use the product rule you get:

g'(x) e^g(x) * (1/g'(x)) + e^g(x) * (1/g'(x))'

isn't that right?

Now the second term is simply the cancellation of the integral:

- e^g(x) * (1/g'(x))'

wich cancells the above second term.

I have it that integration by parts produces the same result if the integrand is of the form:

f(x)g(x) or f(x)

but not if it something like f(g(x)) and you don't transform it into f(u).

It seems that for cases like j(h(x)) where h'(x) is of order x^n and both my method (IPR) and integration by parts (IbP) produces an infinity of integrals IbP's series converges faster for small x and IPR's series converges faster for x large.

13. Apologies in advance for my question, which is rather tangential to the topic being discussed:

I understand that the error function (essentially the integral of exp[-x<sup>2</sup>]) cannot be expressed in closed form in terms of standard elementary functions. Something I am not clear about is whether this simply means that no one has managed to do it, or that it has been proven to be impossible. Any enlightenment on this peripheral question would be of interest.

14. It has been proved to be impossible, I believe, but I've not seen the proof.

15. 8)

Old Fool: does an infinite sum count? You can do this by my integration technique.

The first term would be:

- e^(-x^2)/2x

For e^(x^2):

The series does't converge at x=0 but the constant of integration may compensate.

serpicojr: your choice of dv is not standard integration by parts, you just convinced me my technique is not worthless, therefore the article is still for sale.

16. What's not standard about my choice of dv? Just because you haven't seen something like it in your calculus I textbook doesn't mean it's not standard.

Additionally, if you had been paying attention to the conversation and not getting lost in illusions of grandeur, you would have noticed that "elementary function" has already been defined and that its definition does not include arbitrary infinite series. A quick search (try Wikipedia) reveals that the indefinite integral of e^(x^2) is not elementary. Therefore, the correct answer to Old Fool's question is, "You cannot express the error function as an elementary function," and not, "Let me ignore the terminology of your question and allude to some integration technique I came up with which gives me some sort of ugly expression which I'm going to bandy about as the correct answer. You can buy my paper if you want the details ok lol kthxbye! 8) :? "

17. does an infinite sum count?

A result in the form of an infinite series is not in closed form. It would be straightforward to integrate a function such as exp[x<sup>2</sup>] by writing it out as an infinite series in powers of x and then integrating the series term by term. However, such a result would have limited usefulness.

18. :?

serpicojr: You came up with the same formula!!! Ugly as hell.

My way converges fast for large x.

19. The formula I was referring to was your series representation, and I most certainly did not derive that anywhere in this thread.

I'll admit that your series expression isn't that ugly. However, it's not as nice as the standard power series expansion that everyone else (river_rat, Old Fool, and myself) has referred to:

\sum_{n=0}^\infty x^{2n+1}/(2n+1)n!

The derivation of this series is also much nicer than your method, basically because e^x has a nice Taylor series.

And finally, your series does converge fast for large x, but you also need to calculate e^(x^2). I'm assuming you're concerned about convergence for purposes of estimation, in which case you presumably would be estimating e^(x^2) by its Taylor expansion about 0. But then your expression is pretty poor: the Taylor series for e^(x^2) converges slowly for large x, and the series in 1/x in your expression converges slowly for small x. So it converges slowly on both ends. In this case, you'd be better off using the standard power series representation, as it at least converges quickly for small x.

20. Originally Posted by Old Fool
Apologies in advance for my question, which is rather tangential to the topic being discussed:

I understand that the error function (essentially the integral of exp[-x<sup>2</sup>]) cannot be expressed in closed form in terms of standard elementary functions. Something I am not clear about is whether this simply means that no one has managed to do it, or that it has been proven to be impossible. Any enlightenment on this peripheral question would be of interest.
Hi old fool

The area of mathematics that deals with that question is called Differential Galois Theory. I can try and dig up a proof and see how readable it is, but i suspect there is a lot of field extension rubbish required.

21. I can try and dig up a proof and see how readable it is, but i suspect there is a lot of field extension rubbish required.

In that case, please don't bother :-D

22. What's wrong with field extensions? I rather enjoy them! :-D

23. I give the details so that you can check the series yourself:

Method for integrating k(x):

1. Intellegntly guess a function product G(x)H(x) such that:

d/dx (G(x)H(x) = (terms that cancell)k(x) +/- h(x) (1)

2. Test G(x)H(x) by differentiation for fitting this condition,

h(x) is whatever else is produced by the product rule.

Then the solution is:

Integral k(x) dx = G(x)H(x) -/+ integral h(x) dx (2)

Proof:

Differentiate RS of (2):

d/dx (G(x)H(X) -/+ integral h(x) dx) = d/dx G(x)H(x) -/+ h(x) (3)

Substitute RS (1) into RS (3):

d/dx (G(x)H(X) -/+ integral h(x) dx) = k(x) +/- h(x) -/+ h(x) = k(x)

integrate both sides gives the result.

Rules for guessing well:

1. For k(x) = g(x)f(h(x)) a good guess is:

g(x)F(h(x))(1/h'(x))

where F(x) is the integral of f(x) with the constant of integration left out.

This rule generates the series if applied repeatedly on any produced integral as long as the function e^x is chosen to be integrated every time.

24. It looks like e^(x^2) has infinitely many formulae.

25. Well shit, so does everything!

26. I mean not arbitrarily constructed formulae (like: f(x) + x - x + x^2 - x^2) but derived formulae.

27. Seriously. Everything does.

28. :?

Can you give me an example then? (one not amounting to adding/subtracting zero or multiplication/division by one or a function from operation(anti-operation(f(x)))

29. Real-analytic functions have Taylor series centered at any point where they're defined. Period functions have Fourier expansions with respect to infinitely many different bases. I'm a number theorist, I study Dirichlet series. In addition to their series representation, they often have an infinite product representation (Euler product) and can be represented as integral transforms of theta series/modular forms/other functions defined on the upper half-plane.

Satisfied? If not, I can go on.

30. 8)

Does some of them increase in complexity indefinitely? The above mentioned formulae becomes frightfully complex while describing a simple graph.

31. Originally Posted by talanum1
Does some of them increase in complexity indefinitely? The above mentioned formulae becomes frightfully complex while describing a simple graph.
This is a good thing?

32. 8)

No but it is interesting.

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