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Improper Integrals

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex] [tex]\frac{dx}{x\sqrt{x^2-4}}[/tex] from 2 to infinity

    2. Relevant equations

    Trigonometric substitution, improper integrals

    3. The attempt at a solution

    [tex]\int[/tex] [tex]\frac{dx}{x\sqrt{x^2-4}}[/tex] from 2 to infinity

    = [tex]\underbrace{lim}_{t->inf}[/tex] [tex]\int[/tex] [tex]\frac{dx}{x\sqrt{x^2-4}}[/tex] from 2 to t

    x = 2 sec [tex]\theta[/tex]
    dx = 2 sec [tex]\theta[/tex] tan [tex]\theta[/tex]
    (x^2-4)^(1/2) = 2 tan [tex]\theta[/tex]

    = [tex]\underbrace{lim}_{t->inf}[/tex] [tex]\int[/tex] [tex]\frac{2sec\theta tan\theta d\theta}{2sec\theta2tan\theta}[/tex] from 2 to t

    = [tex]\underbrace{lim}_{t->inf}[/tex] [tex]\int[/tex] [tex]\frac{d\theta}{2}[/tex] from 2 to t

    = [tex]\underbrace{lim}_{t->inf} \theta/2[/tex]

    = [tex]\underbrace{lim}_{t->inf}[/tex] [tex](1/2)(arc tan(\frac{2}{\sqrt{x^(2)-4)}} [/tex] and plugging in the limits of the integrand ...

    = [tex]\underbrace{lim}_{t->inf}[/tex] [tex](1/2)(arc tan(\frac{2}{\sqrt{x^(2)-4)}} - ((1/2)(arc tan(\frac{2}{\sqrt{(2)^(2)-4)}})[/tex]

    What I'm wondering is why the answer to the integral I got isn't negative (I checked on an online integral calculator and it has -.5arctan ...) because I just don't see where that would come into play! Also, what do I do now that I plugged that 2 in and got an undefined number? Should I have split this up into the product of two integrals? How would I do that and was all this work done in vain? (I hope not!!)
  2. jcsd
  3. Sep 27, 2008 #2

    Vanadium 50

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    Somehow your TeX was mangled, making your message hard to figure out. However...

    Your answer has to be positive, since 1/x is positive from 2 to infinity, and sqrt(anything) is defined to be positive. So your integrand is always positive, and adding up a lot of positive things will be positive.
  4. Sep 27, 2008 #3


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    If x=2*sec(theta), then x=2 corresponds to theta=0 and x->infinity to theta->pi/2. Just use the theta limits to evaluate the integral instead of going all the way back to x and having to use a limiting process.
  5. Sep 27, 2008 #4
    If I do it that way, I get the answer as pi/4 and that is the end of it. I checked the answer online and it said the answer to the larger limit was 0, not pi/4. Am I neglecting something by just plugging in pi/2 for t?
  6. Sep 27, 2008 #5


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    You didn't solve for theta correctly. theta=arctan(sqrt(x^2-4)/2). You have the argument of the arctan upside down. That's where the extra minus is coming from.
  7. Sep 27, 2008 #6
    I completely understand now, thank you! Just one last question ... how does t -> infinity correspond to theta -> pi/2? I don't see how I could just plug infinity into sqrt(x^2-4) or something
  8. Sep 27, 2008 #7


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    You have to take the limit, not 'plug in'. As x->infinity, sqrt(x^2-4)/2->infinity. If you look at the graph of arctan, you'll see it approaches pi/2 as the argument goes to infinity.
  9. Sep 27, 2008 #8
    Oh, ok, I misinterpreted your original statement. Thank you for your help!
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