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Integral test of convergence

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  1. Jan 6, 2017 #1

    Rectifier

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    The problem
    I am trying to show that the following integral is convergent
    $$ \int^{\infty}_{2} \frac{1}{\sqrt{x^3-1}} \ dx $$


    The attempt
    ## x^3 - 1 \approx x^3 ## for ##x \rightarrow \infty##.

    Since ## x^3 -1 < x^3 ## there is this relation:
    ##\frac{1}{\sqrt{x^3-1}} > \frac{1}{\sqrt{x^3}}## (the denominator is smaller for function with ## x^3-1 ## and hence bigger quotient).

    I know since earlier that the integral ## \int^{\infty}_{2} \frac{1}{\sqrt{x^3}} \ dx ## is convergent. But the problem is that function ##\frac{1}{\sqrt{x^3}}## is smaller than the other one and therefore always "below" that other function. The other functions integral can diverge and ## \int^{\infty}_{2} \frac{1}{\sqrt{x^3}} \ dx ## will still be convergent like nothin' happened.

    Please help me to solve this mystery.
     
  2. jcsd
  3. Jan 6, 2017 #2

    fresh_42

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    Why don't you find an integral that is always "above" that other function?
     
  4. Jan 6, 2017 #3

    Rectifier

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    Will ##\int^{\infty}_{2} \frac{1}{\sqrt{x^3}} + 1## do the job? Smells fishy though.
     
  5. Jan 6, 2017 #4

    Ray Vickson

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    For ##x > 1## we have ##\sqrt{x^3-1} = x^{3/2} \sqrt{1-x^{-3}}##. Try to find a nice constant lower bound on ##\sqrt{1-x^{-3}}## that is good for ##x \geq 2## (or, at least, for large positive ##x##).
     
  6. Jan 6, 2017 #5

    fresh_42

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    I don't know. I hope the ##1## isn't meant to be part of the integrand. How about ##x^{-\frac{5}{4}}##?
     
  7. Jan 6, 2017 #6

    FactChecker

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    Use a change of variable x = a+1 and (after a little manipulation) compare the integral to the integral of 1/(a3/2)
     
  8. Jan 6, 2017 #7

    Rectifier

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    On which integral?
     
  9. Jan 6, 2017 #8

    Rectifier

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    Why do I need to find a constant lower bound? Isn't the bound x=2 already?
     
  10. Jan 6, 2017 #9

    Rectifier

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    That one is convergent too but I have to show that it is always above the original function.

    In other words:

    I must show that
    ## \frac{1}{\sqrt{x^3-1}}< \frac{1}{x^{\frac{5}{4}}} ## for all x:es

    Any suggestions on how I can do that?
     
  11. Jan 6, 2017 #10

    fresh_42

    Staff: Mentor

    Do some algebra. Remove the quotients, the square roots and so on.
     
  12. Jan 6, 2017 #11

    Rectifier

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    I solved it another way by showing that the quotient of the first function and the second is not infinite for ## x \rightarrow \infty ##. If you have any alternative solution strategies please share them here.
     
  13. Jan 6, 2017 #12

    fresh_42

    Staff: Mentor

    I did (##x>0## ) : ##\frac{1}{\sqrt{x^3-1}}<\frac{1}{x^\frac{5}{4}} \Longleftrightarrow x^\frac{5}{4}<\sqrt{x^3-1} \Longleftrightarrow x^\frac{5}{2}<x^\frac{6}{2}-1 \Longleftrightarrow x^6 > x^5 + 2x^3 - 1## which is true for ##x\geq 2## .
     
  14. Jan 6, 2017 #13

    Ray Vickson

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    There are other ways of doing your problem, but finding a positive constant ##c## such that ##\sqrt{1-x^{-3}} > c## for large ##x>0## is certainly one way to do it. Your previously-expressed worries about ##1/\sqrt{x^3}## not being an upper bound on ##1/\sqrt{x^3-1}## would then go away.
     
    Last edited: Jan 6, 2017
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