Testing for integral convergence

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Homework Statement



Test the following integral for convergence:
[tex] \int^{∞}_{-∞}\frac{dx}{\sqrt{x^{4}+1}} [/tex]

Homework Equations





The Attempt at a Solution


I was able to use the ratio test to show that the integral converges if and only if [itex] \int^{∞}_{-∞}\frac{dx}{x^{2}} [/itex] converges, but I haven't been able to show whether this particular integral converges or diverges.
This integral is not continuous, but it is almost continuous since the discontinuity is a countable set.

NOTE: The integral on the bottom actually diverges, and yet the original integral converges!!! But HOW?!!!


BiP
 
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Answers and Replies

  • #2
STEMucator
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Homework Statement



Test the following integral for convergence:
[tex] \int^{∞}_{-∞}\frac{dx}{\sqrt{x^{4}+1}} [/tex]

Homework Equations





The Attempt at a Solution


I was able to use the ratio test to show that the integral converges if and only if [itex] \int^{∞}_{-∞}\frac{dx}{x^{2}} [/itex] converges, but I haven't been able to show whether this particular integral converges or diverges.
This integral is not continuous, but it is almost continuous since the discontinuity is a countable set.

BiP

Do you remember the p comparison test for :

[tex]\int_{-∞}^{∞} \frac{1}{x^p}[/tex]
 
  • #3
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Do you remember the p comparison test for :

[tex]\int_{-∞}^{∞} \frac{1}{x^p}[/tex]

Yup, but note that the function is discontinuous in the interval.

BiP
 
  • #4
STEMucator
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Yup, but note that the function is discontinuous in the interval.

BiP

Break the integrand up and re-arrange your limits of integration.
 
  • #5
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Break the integrand up and re-arrange your limits of integration.

Ok, I edited my original post. Please re-read it.

BiP
 
  • #6
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Ok, I edited my original post. Please re-read it.

BiP

Are you sure 1/x^2 converges? I just ran through this really quickly and I'm getting that this definitely diverges?
 
  • #7
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Are you sure 1/x^2 converges? I just ran through this really quickly and I'm getting that this definitely diverges?

Yes, which is why I edited on the NOTE to say that it does indeed diverge.
The original problem however happens to converge, even though the ratio test implied they would behave the same way.

BiP
 
  • #8
Dick
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1/x^2 converges on the intervals [1,infinity) and (-infinity,-1]. You'll have to use a different function on the interval [-1,1] for a comparison test. Then add the three together.
 
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  • #9
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I've been searching for a while but can't find a function to compare with for convergence. What do you guys suggest?

BiP
 
  • #10
Dick
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I've been searching for a while but can't find a function to compare with for convergence. What do you guys suggest?

BiP

I hope you've broken it up into intervals. If so, then for [-1,1] how about just using 1? That's bigger than your integrand.
 
  • #11
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I hope you've broken it up into intervals. If so, then for [-1,1] how about just using 1? That's bigger than your integrand.

It converges! Thank you!

BiP
 
  • #12
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I've been searching for a while but can't find a function to compare with for convergence. What do you guys suggest?

BiP

Assume that your integral dx/(x^4+1) is denoted by f(x). Assume the function you're comparing it to 1/x^2 is g(x).

Now I'm not sure you're allowed to do this, but find :

[tex]lim_{x} \rightarrow ∞ \frac{f(x)}{g(x)}[/tex]

This will tell you whether both functions converge or both functions diverge. If your limit is positive and finite, but less than infinity, then your original series will converge.
 
  • #13
Dick
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Assume that your integral dx/(x^4+1) is denoted by f(x). Assume the function you're comparing it to 1/x^2 is g(x).

Now I'm not sure you're allowed to do this, but find :

[tex]lim_{x} \rightarrow ∞ \frac{f(x)}{g(x)}[/tex]

This will tell you whether both functions converge or both functions diverge. If your limit is positive and finite, but less than infinity, then your original series will converge.

There are premises that need to be satisfied to use that test. For one thing 1/x^2 needs to be continuous on the domain of integration. It's not.
 

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