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Help for this improper integral

  1. Mar 14, 2014 #1
    Let $$λ \in R$$

    $$I=\int_{0}^{\infty} \left(\frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1}\right)dx $$

    I need to find λ for which this would return a number (not infinity) . I tried writing Numerators as derivatives but not sure about the correctness and results.


    eg $$\fracλ2\int\frac{d(2x+1)}{2x+1}$$

    Any idea how to solve this ?

    I dont know how to find the antiderivative of the first fraction.
     
  2. jcsd
  3. Mar 14, 2014 #2

    jbunniii

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    When ##x## is large,
    $$\frac{x+1}{3x^2 + \lambda} \approx \frac{1}{3x}$$
    and
    $$\frac{\lambda}{2x+1} \approx \frac{\lambda}{2x}$$
    so
    $$\frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1} \approx \frac{1/3 - \lambda/2}{x}$$
    Can you use these approximations to determine what ##\lambda## must be, given the fact that the "tail" of ##1/x## is too "heavy" to be integrated?
     
  4. Mar 14, 2014 #3
    A hint was to use Laplace for this . I know your solution makes sense but I dont know how it should be written or if there is another way more formal
     
  5. Mar 14, 2014 #4

    jbunniii

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    Do you mean Laplace transforms? I'm not sure how they would be useful here, but probably there's some equivalent formulation that uses them.

    If you solve for the value of ##\lambda## using the approach I outlined, you can simply replace it back into the original integrand and simplify it. You should be able to easily verify that the result is integrable.
     
  6. Mar 14, 2014 #5
    Hm but is this approach correct ? (how can I proof that this is correct ?And put this simpler version within integral ?)
    Here is how the book does it
    http://i.imgur.com/59IHuKQ.jpg
     
  7. Mar 14, 2014 #6
    Do you know how to perform trigonometric substitution?
     
  8. Mar 14, 2014 #7

    jbunniii

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    Well, I can't read the Greek, and I'm not sure how they are going to solve for ##\lambda## after the last line in the image. Even if it works, the method seems very complicated compared with the straightforward one I suggested.

    As I said, to see that my method is correct, first solve for the value of ##\lambda## that will eliminate the problematic ##1/x##: obviously you want ##1/3 - \lambda / 2 = 0##, so ##\lambda = 2/3##.

    Then to see that this value of ##\lambda## works, just plug it into the original integrand and simplify it:

    $$
    \frac{x+1}{3x^2 + \lambda} - \frac{\lambda}{2x+1} =
    \frac{x+1}{3x^2 + 2/3} - \frac{2/3}{2x+1}
    $$
    I'll let you do the algebra, but the key is that the ##x## term in the numerator will vanish, leaving you with an integrand which is asymptotically a multiple of ##1/x^2## as ##x \rightarrow \infty## (and bounded as ##x \rightarrow 0##), hence integrable. No other value of ##\lambda## will do this for you.
     
  9. Mar 15, 2014 #8
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