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Limit integral evaluation

  • Thread starter FNMwacki
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  • #1
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I've evaluated the integral of my problem to be (ln(x^2 + 1)/2*pi), and need to evaluate this at infinity and negative infinity...not sure where to proceed from here to evaluate these limits.

Actually i need to prove that it doesnt exist (I am proving the first moment of the Cauchy Distribution does not exist)

Thanks for any help in advance.
 

Answers and Replies

  • #2
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Is it sufficient to state that both these limits (at positive and Neg infinity) = infinity and that (infinity - infinity) is not defined??
 
  • #3
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I've evaluated the integral of my problem to be (ln(x^2 + 1)/2*pi), and need to evaluate this at infinity and negative infinity...not sure where to proceed from here to evaluate these limits.
Since your limits of integration appear to be -infinity and +infinity, you are dealing with an improper integral, so what you have done is incorrect.

I believe your original integral looks something like this.

[tex]\int_{-\infty}^{\infty} f(x) dx[/tex]

The way to approach it is to split it into two improper integrals, like so.
[tex]\int_{-\infty}^0 f(x) dx + \int_0^{\infty} f(x) dx [/tex]

To evaluate these improper integrals, use limits, like this.
[tex]\lim_{a \to -\infty} \int_a^0 f(x) dx + \lim_{b \to \infty} \int_0^b f(x) dx [/tex]


Actually i need to prove that it doesnt exist (I am proving the first moment of the Cauchy Distribution does not exist)

Thanks for any help in advance.
 
  • #4
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Thanks for the response, I see your point, but even evaluating as such I would get (∞ -0) + (0 - ∞) = (∞ - ∞) which is still undefined...
 
  • #5
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PS - Thanks for the "Helpful symbols" ;)
 
  • #6
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If the first limit is infinite, then the first integral diverges. Same for the second integral.
 
  • #7
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Thanks again for the response, but I'm not sure I follow the point you are trying to make here...

If the integrals diverge, and knowing Cauchy Distribution is symmetric, we can use the reflection property and also say it is 2*(infinity - 0)...
 
  • #8
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Based on your result in the first post, it seems to me that your integral was roughly this:
[tex]\int_{-\infty}^{\infty} \frac{2x}{x^2 + 1} dx[/tex]

I have omitted the constants, so some adjustment would be necessary to get exactly what you got for the antiderivative, which was (1/2pi) * ln(x2 + 1).

The problem is that the integrand above is NOT an even function, so the two halves of the graph aren't symmetric about the y-axis. This makes me think you have made a mistake in your integration.

Per Wikepedia, the standard Cauchy distribution has a pdf of f(x; 0, 1) = 1/(pi * (1 + x2)). The antiderivative here is not a log.
 
  • #9
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You are correct, f(x; 0, 1) = 1/(pi * (1 + x^2)) is my PDF, but i am trying to prove the Expectation of this pdf, so ∫ x*f(x)dx from -∞ to ∞

I pull out the 1/pi, use U substitution for (1+x^2), so i have (1/pi)*∫ x/(2xu) du, where du=2xdx => dx=du/2x

then i call pull out the 2 from the denominator to have (1/2*pi)*∫ 1/(u) du >>the x's cancel

the integral of 1/u = log(u); substitute back in for u to get (log(x^2 + 1)/(2*pi))

or am i missing something here??
 
  • #10
33,503
5,191
Thanks again for the response, but I'm not sure I follow the point you are trying to make here...

If the integrals diverge, and knowing Cauchy Distribution is symmetric, we can use the reflection property and also say it is 2*(infinity - 0)...
Your distribution is symmetric (it's an even function), but your integrand is not symmetric, at least not in the same way. The integrand is an odd function, meaning that the graph is symmetric across the origin.

You are correct, f(x; 0, 1) = 1/(pi * (1 + x^2)) is my PDF, but i am trying to prove the Expectation of this pdf, so ∫ x*f(x)dx from -∞ to ∞

I pull out the 1/pi, use U substitution for (1+x^2), so i have (1/pi)*∫ x/(2xu) du, where du=2xdx => dx=du/2x

then i call pull out the 2 from the denominator to have (1/2*pi)*∫ 1/(u) du >>the x's cancel

the integral of 1/u = log(u); substitute back in for u to get (log(x^2 + 1)/(2*pi))

or am i missing something here??
I didn't put it together that you were trying to find E(X), so the work above looks fine. I think it would be sufficient to say that each of the two integrals diverges. (One diverges to infinity and the other to neg. infinity.)
 
  • #11
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Ok, thanks again for the help. Looking back now I probably could have been more clear what I was doing in the original post...
 
  • #12
SammyS
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On the other hand, you might look at this as:

[tex]\lim_{a\to\infty}\int_{-a}^{a} \frac{2x}{x^2 + 1}\, dx \,,[/tex]

which would give you zero.
 
  • #13
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5,191
On the other hand, you might look at this as:

[tex]\lim_{a\to\infty}\int_{-a}^{a} \frac{2x}{x^2 + 1}\, dx \,,[/tex]

which would give you zero.
For the reason this won't work, see this wiki article - http://en.wikipedia.org/wiki/Cauchy_distribution - especially the section titled Mean.
 
  • #14
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On the other hand, you might look at this as:

[tex]\lim_{a\to\infty}\int_{-a}^{a} \frac{2x}{x^2 + 1}\, dx \,,[/tex]

which would give you zero.
Ya, thanks for the input, but i already know the expectation should not exist...

Or you could see Wiki as Mark pointed out...
 

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