Convergence/Divergence of Integrals

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Discussion Overview

The discussion centers on determining the convergence or divergence of the integral \(\int_{0}^{1-}\frac{dx}{\sqrt{1-x^4}}\). Participants explore various approaches and reasoning related to the behavior of the integrand near the problematic point and the implications for convergence without evaluating the integral itself.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that the function behaves like \(1/x^{2}\) near the problematic point, which leads to a belief that it should be integrable in that domain.
  • Others argue that the critical point for convergence is at \(x=1\), where \(\sqrt{1 - x^4}\) behaves like \(2\sqrt{1-x}\), indicating potential divergence.
  • One participant expresses confidence that the integral converges but does not provide a clear rationale for this assertion.
  • There is a discussion about the evaluation of the integral in terms of the Gamma function, with references to elliptic integrals and the results obtained from computational tools like Mathematica and Maple.
  • Some participants clarify misunderstandings about the integrability of the function and the implications of their earlier statements.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the integral converges or diverges, with multiple competing views and reasoning presented throughout the discussion.

Contextual Notes

Participants express uncertainty regarding the behavior of the integrand near the limits of integration and the implications for convergence. There are also unresolved mathematical steps related to the evaluation of the integral and its connection to the Gamma function.

amcavoy
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Determine whether the following integral converges or diverges without evaluating it.

[tex]\int_{0}^{1-}\frac{dx}{\sqrt{1-x^4}}[/tex]

Thanks for your help.
 
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That function behaves ~1/x^{2},so it should be integrable in that domain.U have to find a function which is greater than your function,however,it's integral on [0,1] needs to be finite...

[tex]\int_{0}^{1} \frac{dx}{\sqrt{1-x^{4}}} =\frac{1}{4}\left(\sqrt{\pi}\right)^{3}\frac{\sqrt{2}}{\Gamma^{2}\left(\frac{3}{4}\right)}[/tex]

Daniel.
 
The problem point is x=1, not x=0. √(1 - x4) behaves like 2√(1-x) there.

P.S. 1/x2 isn't integrable over [0, 1]. :-p
 
I didn't say "integrable".And i didn't say =1/x^{2}...:wink:

Daniel.
 
i believe it certainly converges. now let me think why i say that.

if y^2 = 1-x^4 then 2ydy = -4x^3 dx, so dx/sqrt(1-x^4) = dx/y = -2dy/4x^3. which makes perfectly good sense at x = 1.


huh?

formulas have always seemed like magic to me.
 
Last edited:
Yeah,but it's still infinite,because your limits of integration will still involve 0...

Daniel.
 
dextercioby said:
That function behaves ~1/x^{2},so it should be integrable in that domain.U have to find a function which is greater than your function,however,it's integral on [0,1] needs to be finite...

[tex]\int_{0}^{1} \frac{dx}{\sqrt{1-x^{4}}} =\frac{1}{4}\left(\sqrt{\pi}\right)^{3}\frac{\sqrt{2}}{\Gamma^{2}\left(\frac{3}{4}\right)}[/tex]

Daniel.

Can you explain how this integral is evaluated in terms of the Gamma function?
 
I dunno.Basically,it returned my a value for the Legendre elliptic integral and then expressed this one in terms of Gamma factor...

Daniel.
 
If u make a substitution,u can transform the integral to

[tex]\int_{1}^{+\infty} \frac{dx}{\sqrt{x^{4}-1}}[/tex]

whose antiderivative,Mathematica finds

Daniel.

P.S.My Maple gave me the results...
 

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