Why Does Integration of Certain Functions Result in Infinity?

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Discussion Overview

The discussion revolves around the integration of certain functions, specifically the integral of \(\frac{1}{\sqrt{p(1-p)}}\) from 0 to 1, and the implications of the function approaching infinity at the endpoints. Participants explore the nature of the integral and the behavior of the function near its singularities.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant notes that integrating \(\frac{1}{\sqrt{p(1-p)}}\) results in \(\pi\), despite the function approaching infinity at \(p = 0\) and \(p = 1\).
  • Another participant questions the reasoning behind the expectation that the area under the curve should be infinite, suggesting that the integral could still yield a finite result.
  • A further response emphasizes that the denominator \(\sqrt{p(1-p)}\) leads to infinity at the endpoints, implying that the area under the curve might also be infinite.
  • One participant encourages others to consider why the area might be infinite and suggests working through simpler integrals to clarify understanding.

Areas of Agreement / Disagreement

Participants express differing views on whether the area under the curve is infinite or finite, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants have not reached a consensus on the nature of the integral, and there are unresolved assumptions regarding the behavior of the function at its singularities.

touqra
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Using Maple 8, I integrate this:

[tex]\int_0^1\frac{1}{\sqrt(p(1-p))} dp[/tex]

and I get [tex]\pi[/tex]
but, this function goes to infinity at p = 0 and 1.
How can this be possible ?
 
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How can this be possible ?
Why wouldn't it be?
 
Hurkyl said:
Why wouldn't it be?

The denominator has sqrt[p*(1-p)], which will give infinity on both p = 0 and 1. And so, the area under this curve is infinity too, right ?
 
Can you think of a reason why that should be true, though? If you're convinced the area should be infinite, then try to work out a proof of it -- such exercises are often really good at clearing up misunderstandings.

Incidentally, it might help to play with simpler functions, e.g.

[tex]\int_0^1 \frac{1}{\sqrt{x}} \, dx[/tex]

or

[tex]\int_1^\infty \frac{1}{y^2} \, dy[/tex]
 

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