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Improper integral, divergence/convergence

  • Thread starter usn7564
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  • #1
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Homework Statement


Evaluate if the integral diverges or converges using the comparison theorem.

[tex]\int^ \infty_2 \frac{dx}{\sqrt{x^3+1}}[/tex]

Having trouble with this question, the exercises I have managed I generally guessed if it was convergent/divergent, and then found a smaller of bigger function at the interval and went by that. Here I assumed it was convergent and used [tex]\frac{1}{\sqrt{x^2+1}}[/tex], though that was divergent which says absolutely nothing. Ran into the same issue with other attempts.

Is my reasoning the issue or have I just not found the correct manipulation? Not really approaching with anything more involved than "Find something bigger and hope it converges".
Any help would be appreciated, thanks.
 

Answers and Replies

  • #2
HallsofIvy
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It seems to me that a simpler comparison is [itex]x^3< x^3+ 1[/itex] so that [itex]\frac{1}{\sqrt{x^3+ 1}}< \frac{1}{\sqrt{x^3}}= x^{-3/2}[/itex].
 
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  • #3
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I feel slightly moronic now, was too focused on manipulating the x term for whatever reason.

Thanks, glad I didn't get it all backwards at least.
 

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