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January 30

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Generalization of integral 2

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It is known .

Consider the following generalization

where satisfies and for all x.

What is the result then? --AnalysisAlgebra (talk) 11:04, 30 January 2013 (UTC)[reply]

You might find Cauchy–Schwarz inequality useful, otherwise I don't think there is much one can say. Dmcq (talk) 23:26, 30 January 2013 (UTC)[reply]
OK. I want to find this limit: --AnalysisAlgebra (talk) 02:54, 31 January 2013 (UTC)[reply]
Hint: what is the maximum value of x(1-x) in that interval, and what value of x does it occur at? Looie496 (talk) 03:38, 31 January 2013 (UTC)[reply]
1/4 and 1/2, respectively. Are you trying to bound the expression or something? But that tells you nothing about the maximum value of with the f in there.--AnalysisAlgebra (talk) 05:28, 31 January 2013 (UTC)[reply]
As n tends to infinity with f remaining constant, becomes a more significant factor in determining . -- Meni Rosenfeld (talk) 05:46, 31 January 2013 (UTC)[reply]
Is that also true for any in --AnalysisAlgebra (talk) 05:57, 31 January 2013 (UTC)[reply]
Then the dominant factor is . -- Meni Rosenfeld (talk) 14:33, 31 January 2013 (UTC)[reply]
Are you saying the limit is independent of ? The limit is if but that's not the case if for example ?--AnalysisAlgebra (talk) 09:33, 31 January 2013 (UTC)[reply]
It depends on f, but only on its behavior in a specific area. -- Meni Rosenfeld (talk) 14:33, 31 January 2013 (UTC)[reply]
Or the integral can be done exactly using the above identity by just substituting f for its taylor series evaluated at either x=0 or x=1. As a note, naively treating the distribution as a nascent delta could lead you to erroneously conclude that if f(1/2)=0 the integral is zero. — Preceding unsigned comment added by 123.136.64.14 (talk) 06:13, 31 January 2013 (UTC)[reply]
Only if you know , and also only if is analytic. --AnalysisAlgebra (talk) 09:33, 31 January 2013 (UTC)[reply]

@User:AnalysisAlgebra: Please do not remove other's comments at Help Desk/Math as you did here: [1] This is considered bad form. See here for the guideline. El duderino (abides) 11:36, 31 January 2013 (UTC)[reply]

If the function is of the form then the integral is a beta function. So write your function as a linear combination of such functions, . Bo Jacoby (talk) 12:41, 31 January 2013 (UTC).[reply]

What if can't be written as a linear combo of those functions? Even if you allow an infinite sum, I'm pretty sure there are functions satisfying the conditions above which are not linear combinations of such functions.--AnalysisAlgebra (talk) 12:47, 31 January 2013 (UTC)[reply]
I doubt you'll find any reasonable closed form for the general case. The limit however should be quite easy. -- Meni Rosenfeld (talk) 14:33, 31 January 2013 (UTC)[reply]
If the function f is continuous, then the Stone–Weierstrass theorem tells you that it can be approximated by polynomials. So don't be pretty sure that it can't be done. Bo Jacoby (talk) 08:27, 1 February 2013 (UTC).[reply]
So, I was right, then. --AnalysisAlgebra (talk) 09:27, 1 February 2013 (UTC)[reply]
No, you were wrong. Bo Jacoby (talk) 13:01, 1 February 2013 (UTC).[reply]
If f is sufficiently differentiable at 1/2, the asymptotics of should only depend on the even derivatives of f at 1/2, and probably only the first nonzero (even) derivative. — Arthur Rubin (talk) 17:32, 3 February 2013 (UTC)[reply]
When put that way, the result is obvious. If f is differentiable at 1/2, and integrable over (0,1), then:
Arthur Rubin (talk) 17:48, 3 February 2013 (UTC)[reply]