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Comprison test for Integrals

  1. Apr 12, 2008 #1
    Hey guys. I'm a little new on the comparison test, if you could just check that i'm on the right track, it would be great.

    1. The problem statement, all variables and given/known data
    Using comparison test to decide whether or not [tex]\frac{\sqrt{x}}{\ln{x} + x^2}[/tex] converges.

    2. Relevant equations
    (Check 3.)

    3. The attempt at a solution
    Okay, so first off, I guess that this converges. So find a function that is larger than the one that is given and test if that converges, yes?

    Would: [tex]\frac{\sqrt{x}}{\ln{x} + x^2} < \frac{\sqrt{x}}{x^2}[/tex] be a good choice?

    So would I be right in saying that as [tex]x^{-3/2}[/tex] converges, that also the one given converges?

    I'm not sure if i've gotten this backwards or not. Could someone please clear it up for me?

  2. jcsd
  3. Apr 12, 2008 #2


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    First of all, it doesn't make any sense to ask "whether [tex]\frac{\sqrt{x}}{\ln{x} + x^2}[/tex] converges". Do you mean to determine whether or not an integral of that converges? If so then you still need to specify whether you are asking about the improper integral [tex]\int_a^\infty \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex] with a> 0 or [tex]\int_0^a \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex], again with a> 0.
  4. Apr 12, 2008 #3
    ... Whoops. So sorry, forgot to include the integral.

    But yes, check if the Integral converges. Sorry about that.

    [tex]\int_1^\infty \frac{\sqrt{x}}{\ln{x} + x^2} dx[/itex]

    Not sure why, but my upper bound won't stay as infinity. (Today's the first time i've used LaTeX.)

    Bounds are 1 to Infinity.

    Oh, and whenever I said in the first post whether or not things converge, it's when taking the limit as x approaches infinity. Sorry about not making that clear in the first post.
    Last edited: Apr 12, 2008
  5. Apr 12, 2008 #4


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    Your choice of x^(-3/2) to compare with is a good one. And yes, it works. You might want to pound that point in by saying why the inequality is true.
  6. Apr 12, 2008 #5
    Alright, thanks. :)

    I'll add in an extra step saying that x^2 < ln(x) + x^2 and explaining it. Thanks.
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