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Improper integral question

  • Thread starter dsta
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  • #1
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Establish convergence/divergence of the following improper integral:

integral from 0 to infinity of 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

My attempt at a solution was to break it up into 3 intergrals: 0 to 1, 1 to 5, and 5 to infinity...I showed that the first two of these integrals converges by comparing them to 1/x^(1/3) and 1//x-5/^(1/3) respectively. I can't establish convergence/divergence of last intergral from 5 to infinity, as hard as I have tried.

Thanks, any help is much appreciated.
 

Answers and Replies

  • #2
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Establish convergence/divergence of the following improper integral:

integral from 0 to infinity of 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

My attempt at a solution was to break it up into 3 intergrals: 0 to 1, 1 to 5, and 5 to infinity...I showed that the first two of these integrals converges by comparing them to 1/x^(1/3) and 1//x-5/^(1/3) respectively. I can't establish convergence/divergence of last intergral from 5 to infinity, as hard as I have tried.

Thanks, any help is much appreciated.
Is there some reason you wrote /x - 5/ using the slashes?
Why did you choose 1 as a limit of integration in two of your integrals? There is nothing unusual happening at x = 1.
In your integrals, what do you have for the integrands? Please show us the work you did.

BTW, there is no such word as "intergral."
 
  • #3
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By /x-5/, I mean the absolute value of (x-5).

I chose 1 because don't you have to break up the original improper integral into appropriate intervals to see what's going on? There are unusual things happening at x=0 and x=5, and 1 was chosen to 'link up' these separate intervals (for want of a better phrase).

So I have this:
[0 to 1] 1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) ) + [1 to 5]1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) ) + [5 to infinity]1 / ( x^(1/3)*(/x-5/^(1/3))*(1 + sqrt(x))^0.7) )

The first two integrals I have shown to be convergent (by comparison), but I am stuck on the last one. Mabye I am going about this the wrong way altogether, but I'm sort of following a similar example from my lecture notes.

(Sorry about the messiness, I am trying to learn Latex)
 
  • #4
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1+sqrt(x)>sqrt(x), and for large x, x-5>0.5x
 
  • #5
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P.S. You also should break [5,infty) into two intervals.
 
  • #6
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OK, I understand now why you are using 1 as a limit of integration. For absolute values, you can use this character: |.

To help you out with your LaTeX, your last integral is
[tex]\int_5^{\infty}\frac{1}{x^{1/3}|x - 5|^{1/3}(1 + \sqrt{x})^{0.7}}dx[/tex]
You'll need to break this into two integrals, say from a to 6 and from 6 to b, and look at the limit of each as a approaches 5 from above and as b approaches infinity. You can click on the integral and see the LaTeX code, and can copy and paste it to create your own.

Since in both integrals, x >= 5, you can drop the absolute value signs so you have (x^2 - 5x)^(1/3). I doubt that the integral can be found by ordinary means, so you'll probably need to use another comparison.

I'll need to think about this one...
 
  • #7
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4,860
The denominator is roughly comparable with x61/60, with that exponent obtained by adding 2/3 and 7/20. Since this exponent is > 1, the integral of this function, from 6 to infinity, converges. If you can show that your denominator is > x61/60, and thereby that your integrand is < 1/x61/60, you're home free.
 
  • #8
9
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Ahh okay, that makes alot more sense now, thanks alot Mark.

When I finish writing it up I will post my final explanation.
 

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