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Improper Integrals

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the value of the constant C for which the following integral converges. Evaluate the integral for this value of C:

    [tex]\int[/tex] [tex]\frac{x}{x^2+1}[/tex] - [tex]\frac{C}{3x+1}[/tex]dx from 0 to infinity

    2. Relevant equations

    3. The attempt at a solution

    [tex]\stackrel{lim}{t->inf.}[/tex] [tex]\int[/tex] [tex]\frac{x}{x^2+1}[/tex] dx - [tex]\stackrel{lim}{t->inf.}[/tex] [tex]\int[/tex] [tex]\frac{C}{3x+1}[/tex] dx
    for (x^2/(x^2+1):
    u = x^2 + 1
    du = 2xdx
    [tex]\stackrel{lim}{t->inf.}[/tex] (1/2)ln(u) dx
    [tex]\stackrel{lim}{t->inf.}[/tex] (1/2)ln(x^2+1) ] [tex]\stackrel{t}{0}[/tex]
    [tex]\stackrel{lim}{t->inf.}[/tex] (1/2)ln(t^2+1)

    Now I am unsure of what to do. How do I know the limit of this first half? How can I use it to help me find what value of C will make it convergent? Your time and effort is greatly appreciated in helping me understand this :smile:\

    ***Please note (I don't know how to format limits haha) that I mean the limit as t approaches infinity! Thanks!***
  2. jcsd
  3. Sep 27, 2008 #2


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    Don't integrate them separately, they both diverge (they are ~1/x for large x). Combine them algebraically first and see if you can find a value of C that eliminates that divergence.
  4. Sep 27, 2008 #3
    Ok, I will try that, but I am still unsure of my objective in integrating. Should I be looking to eliminate factors from the top and bottom to simplify the integrand before I integrate and take the limit? And I am just looking to make it so that the limit exists?
  5. Sep 27, 2008 #4


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    You are trying to get C to cancel a term in the numerator that's causing the divergence. For example, ((C-1)*x+1)/(x^2+1) diverges if C=2. Since it behaves like 1/x for large x. But if C=1 it converges, since it's just an arctan. You cancelled the x part that's causing problems. That kind of cancellation.
    Last edited: Sep 27, 2008
  6. Sep 27, 2008 #5
    Ahh, ok, I see what you mean by cancellation now. So I am basically trying to cancel out the effect of a smaller power of x over a larger power of x. But I'm not seeing a way to get rid of the x^3. Unless I am supposed to make the whole numerator 1? (I'm sorry I'm being so dense today, calculus tends to do that to me :rofl:)
  7. Sep 27, 2008 #6


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    I think canceling out the x^2 in the numerator would be a good idea. What did you get for the combined integrand anyway?
  8. Sep 28, 2008 #7
    I got:
    [(3-c)x^2+x-c] / [3x^3+x^2+3x+1]

    And then if I plug in for 3=c, I get

    lim as t approaches infinity of [tex]\int[/tex][tex]^{t}_{0}[/tex] [tex]\frac{x-3}{3x^3+x^2+3x+1}[/tex]

    And now, as my luck would have it, I am stuck integrating. I cannot factor the denominator so I think integration by parts is out of the question and I don't think factoring by grouping would yield any sort of cancellation with the numerator. Any tips? :uhh:
  9. Sep 28, 2008 #8


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    x^2*(3x+1)+(3x+1) is another way of writing the denominator. Sure you can't factor it? I'd say try partial fractions.
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