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Two improper integrals

  1. Jun 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine the divergence or convergence of the improper integral. Evaluate the integral if it converges.

    1. [tex]\int^{4}_{2}[/tex][tex]\frac{1}{\sqrt{x^{2}-4}}[/tex] dx

    2. [tex]\int^{2}_{0}[/tex][tex]\frac{1}{\sqrt[3]{x-1}}[/tex] dx

    2. Relevant equations



    3. The attempt at a solution

    1.

    t = x - 2
    dt = dx

    [tex]\int^{4}_{2}[/tex][tex]\frac{1}{\sqrt{x^{2}-4}}[/tex] dx
    =[tex]\int^{2}_{0}[/tex][tex]\frac{1}{\sqrt{(t+2)^{2}-4}}[/tex] dx
    =[tex]\int^{2}_{0}[/tex][tex]\frac{1}{\sqrt{t^{2}+4t}}[/tex] dx

    [tex]\stackrel{lim}{x\rightarrow\infty}[/tex] [tex]\int^{2}_{N}[/tex][tex]\frac{1}{\sqrt{t^{2}+4t}}[/tex] dt

    I'm not sure how to integrate [tex]\int^{2}_{N}[/tex][tex]\frac{1}{\sqrt{t^{2}+4t}}[/tex] dt from here.


    2.

    [tex]\int^{2}_{0}[/tex][tex]\frac{1}{\sqrt[3]{x-1}}[/tex] dx = [tex]\int^{1}_{0}[/tex][tex]\frac{1}{\sqrt[3]{x-1}}[/tex] + [tex]\int^{2}_{1}[/tex][tex]\frac{1}{\sqrt[3]{x-1}}[/tex]

    t = 1 - x
    dt = -dx

    [tex]\int^{1}_{0}[/tex][tex]\frac{1}{\sqrt[3]{x-1}}[/tex] = [tex]\int^{0}_{1}[/tex][tex]\frac{-1}{\sqrt[3]{-t}}[/tex] dt = [tex]\int^{1}_{0}[/tex][tex]\frac{1}{\sqrt[3]{-t}}[/tex] dt


    [tex]\stackrel{lim}{x\rightarrow\infty}[/tex] [tex]\int^{1}_{N}[/tex][tex]-t^{-1/3}[/tex] dt
    = [tex]\stackrel{lim}{x\rightarrow\infty}[/tex] -[tex]\frac{3}{4}(-t)^{4/3}[/tex][tex]|^{1}_{N}[/tex] = [tex]\stackrel{lim}{x\rightarrow\infty}[/tex] [tex]\left[\frac{3}{4} + \frac{3}{4}(-N)^{4/3}\right][/tex] = [tex]\infty[/tex]

    So I got that this integral diverges, however, the book answers are:


    Book Answers:

    1. [tex] ln (2 + \sqrt{3}) [/tex]
    2. 0
     
    Last edited: Jun 28, 2009
  2. jcsd
  3. Jun 28, 2009 #2
  4. Jun 28, 2009 #3
    Hmm, I see. I would like to know how this formula was derived though. Also do you see something wrong with #2?
     
  5. Jun 28, 2009 #4

    Dick

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    For the first one you do a trig substitution, like x=2*sec(t). For the second one the "improper" part of your integrand is it's divergence at x=1. You should be letting x->1 in both halves of the integral, not x->infinity.
     
    Last edited: Jun 28, 2009
  6. Jun 29, 2009 #5
    In the first one, [tex]\lim_{N \to 0}[/tex] not [tex]\infty[/tex]
     
  7. Jun 29, 2009 #6
    But I set t = x - 1 specifially to make the discontinuity at 0; with t the problem occurs at 0, then don't we take the limit to infinity?
     
  8. Jun 29, 2009 #7

    Dick

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    Sure, in terms of t the integrand is (-t)^(-1/3) so the discontinuity is at 0. So for example integrate it from epsilon to 1 and let epsilon approach 0. Why would you send anything to infinity? BTW the antiderivative of (-t)^(-1/3) isn't (-3/4)*(-t)^(4/3). What's the correct power?
     
  9. Jun 29, 2009 #8
    ah I just flipped through my textbook and you are right. I took the limit to 0 and got the right answer. the first I still am not sure how to integrate, there doesn't seem to be a trig substitution that works
     
  10. Jun 29, 2009 #9
    the power you asked for is 2/3
     
  11. Jun 29, 2009 #10

    Dick

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    Your variable changes aren't really helping you. You are just moving the discontinuity around. To do the first integral stick with the form 1/sqrt(x^2-4) and substitute x=2*sec(t).
     
  12. Jun 29, 2009 #11

    Dick

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    That's better.
     
  13. Jun 29, 2009 #12
    I don't understand how you came up with 2 sec t
     
  14. Jun 29, 2009 #13

    Dick

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    If x=2*sec(t) then x^2-4 becomes 4*sec(t)^2-4=4(sec(t)^2-1). sec(t)^2-1=tan(t)^2. Now the quantity inside the square root is a perfect square.
     
  15. Jun 30, 2009 #14
    I tried doing the substitution but then got stuck again. This is what I did:

    [tex]x = 2 sec (t)[/tex]
    [tex]dt = 2 sec(t) tan (t) dx [/tex]
    [tex]t = sec^{-1} (x/2)[/tex]

    [tex]\int^{\pi/3}_{0}[/tex][tex]\frac{1}{\sqrt{(2 sec(t)^{2} - 4} - 4}(2 sec(t) tan (t)}[/tex][tex]dt[/tex]

    = [tex]\int^{\pi/3}_{0}[/tex][tex]\frac{1}{\sqrt{4(sec^{2}t -1)}(2 sec(t) tan (t)}[/tex][tex]dt[/tex]

    = [tex]\int^{\pi/3}_{0}[/tex][tex]\frac{1}{\sqrt{4 tan^{2})t)}(2 sec(t) tan (t)}[/tex][tex]dt[/tex]


    = [tex]\int^{\pi/3}_{0}[/tex][tex]\frac{1}{2 tan(t)(2 sec(t) tan (t)}[/tex][tex]dt[/tex]

    = [tex]\int^{\pi/3}_{0}[/tex][tex]\frac{1}{2 tan^{2}(t)(2 sec(t))}[/tex][tex]dt[/tex]


    However, I still am not sure how to integrate this expression.
     
  16. Jun 30, 2009 #15

    Dick

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    If x=2*sec(t) then dx=2*sec(t)*tan(t)*dt. NOT dt=2*sec(t)*tan(t)*dx. The tan's will cancel and you should be left with just the integral of sec(t).
     
  17. Jun 30, 2009 #16
    Obviously you did something wrong.
    [tex]x=2sec(t)[/tex]

    [tex]dx=2tan(t)sec(t)dt[/tex]

    Now,
    [tex]\int^{\pi/3}_{0}\frac{1}{\sqrt{\frac{4}{cos^2(t)}-4}}*2tan(t)sec(t)dt=[/tex]
    [tex]=\int^{\pi/3}_{0}\frac{cos(t)}{\sqrt{4(1-cos^2(t))}}*2tan(t)sec(t)dt=[/tex]
    [tex]=\int^{\pi/3}_{0}\frac{cos(t)}{2sin(t)}*2tan(t)sec(t)dt=[/tex]

    Do you know how to continue from now on?

    You got here cotan(t)*tan(t)=1 and its pretty easy to solve it.

    Regards.
     
  18. Jun 30, 2009 #17
    Finally solved it! Thank you! I would have never thought to use a trig substitution like that. Thanks.
     
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