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Homework Help: Improper Integrals

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data

    do the following integrals converge?

    i) [tex]\int_0^{1}\frac{dx}{x^{3/2}e^{x}}[/tex]

    ii) [tex]\int_0^{1}\frac{x}{\sqrt{1-x^{2}}}dx[/tex]


    3. The attempt at a solution

    looking at them i can guess that they both diverge - proving this is the hard part - this is what i have got but it doesn't prove anything...

    i) [tex]\frac{1}{x^{3/2}}[/tex][tex]\geq1[/tex]

    [tex]\frac{1}{x^{3/2}e^{x}}[/tex][tex]\geq{e^{-x}}[/tex]

    [tex]e^{-x}[/tex] converges



    [tex]\frac{1}{e^{x}}[/tex][tex]\leq1[/tex]

    [tex]\frac{1}{x^{3/2}e^{x}}[/tex][tex]\leq{x^{-3/2}[/tex]

    [tex]x^{-3/2}[/tex] divereges



    ii) [tex]x\leq1[/tex]

    [tex]\frac{x}{\sqrt{1-x^{2}}}[/tex][tex]\leq{\frac{1}{\sqrt{1-x^{2}}}[/tex]

    [tex]{\frac{1}{\sqrt{1-x^{2}}}[/tex] diverges (cannot prove it though)



    [tex]{\frac{1}{\sqrt{1-x^{2}}}\geq1[/tex]

    [tex]{\frac{x}{\sqrt{1-x^{2}}}[/tex][tex]\geq x[/tex]

    x converges

    any help would be much appreciated.
     
    Last edited: Mar 3, 2008
  2. jcsd
  3. Mar 3, 2008 #2
    I take your last statement above to mean that when you integrate both sides of the inequality with respect to x from 0 to 1, the right-hand side converges. That is true. Now what does that tell you about the left-hand side?

    Again, what does that tell you about the left-hand side? Ditto for ii).
     
  4. Mar 3, 2008 #3
    maybe i'm missing something but not much because the left hand side is greater than the right, the right converges, this does not mean the same for the left - it might diverge.

    The same for all the others - the comparisions don't tell us much.

    If all the inequalities were reversed then i wouldn’t have a problem.

    i think...?
     
    Last edited: Mar 3, 2008
  5. Mar 3, 2008 #4
    Yes. Exactly. You need to find an expression so that the inequalities are reversed.
     
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