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Finding convergence/divergence of improper integral

  • #1
217
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Determine the convergence or divergence of this following improper integral:

[tex]\int_2^∞ \frac {1}{(x^3+7)^{\frac{1}{3}}}[/tex]

So I'm trying to find something easy to compare this to, any help?
 

Answers and Replies

  • #2
828
2
Well, as I'm sure you can tell, it is a lot like 1/x. So, I would try to find something that is close to 1/x to which you can compare this.
 
  • #3
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Why can't I just compare it to 1/x?
 
  • #4
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I don't know if you have gone over this, but you can compare it to

[tex]\int\limits_1^\infty {\frac{1}{{{x^n}}}} [/tex]

If n<2, it is divergent. If n≥2, then it is convergent.
 
Last edited:
  • #5
828
2
Well,
[tex]\frac{1}{(x^3+7)^{\frac{1}{3}}} < \frac{1}{x} [/tex]

isn't it? Since [
[tex]\int_2^∞ \frac {1}{x}[/tex]
diverges, this isn't any help.

So, I'd try showing that
[tex](x^3+7)^{1/3} < x + b[/tex]

where b is a number. (In particular, b is a number that appears in your problem).

Then realize that

[tex]\int_2^∞ \frac {1}{x+b}[/tex]

diverges, and you're done.
 
  • #6
217
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How can I show that [tex]\int_2^∞ \frac {1}{x+7}[/tex] is divergent?
 
  • #7
828
2
How can I show that [tex]\int_2^∞ \frac {1}{x+7}[/tex] is divergent?
First of all, I was kind of wrong when I said that "b" was in your problem. Use [tex]7^{1/3}[/tex] in stead of [tex]7[/tex]. As for showing that

[tex]\int_2^∞ \frac {1}{x+7^{1/3}}[/tex]

is divergent, is there some theorem in your book that would help? I don't know if there is or not. But, you could say that

[tex]lim \frac{1}{x+7^{1/3}} = \frac{1}{x}[/tex]

and argue from there. Or, you could just integrate


[tex]\int_2^∞ \frac {1}{x+7^{1/3}}[/tex]

You have done so to show that


[tex]\int_2^∞ \frac {1}{x}[/tex]

is divergent, right? Just use a change of variables.
 

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