1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integral Evaluation with Partial Fractions

  1. Feb 27, 2014 #1
    1. The problem statement, all variables and given/known data

    I(alpha) = ∫1/((x+alpha^2)(x+1)) dx between the limits of 0 and infinity

    Evaluate the integral above depending on the parameter alpha using partial fractions.

    3. The attempt at a solution

    1/((x+alpha^2)(x+1)) = A/(x+alpha^2) + B/(x+1)

    1 = A(x+1) + B(x+alpha^2)

    let x = -1 then B = 1/(-1+alpha^2)

    let x = (-alpha^2) then A = 1/(-alpha^2+1)

    therefore the integral becomes two integrals that can be added together:

    1/(-alpha^2+1)∫1/(x+alpha^2)dx + 1/(-1+alpha^2)∫1/(x+1)dx

    both integrals are definite integrals between 0 and infinity

    When I integrate 1/(x+1) between zero and infinity I obtain infinity as an answer. This is the same as when I integrate the other partial fraction 1/(x+alpha^2) between zero and infinity. Therefore when I multiply this by the constants that I have already taken outside of the integral my answer to the evaluation of the integral will be infinity.

    This seems to be correct as when the x value increases, the graph will not reach a zero point so as this continues forever, the area under the graph will be equal to infinity.

    What action do I need to take from the point of having the partial fractions in order to evaluate this integral properly?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 27, 2014 #2
    When you get to the part with all the natural logs, instead of taking the limit at this point, try combining all the natural logs into one and using l'Hospital's Rule.
  4. Feb 28, 2014 #3
    Thank you,

    I have combined the logs to get:

    1/(1-alpha^2) (ln((x+alpha^2)/(x+1))) with the limits 0 and infinity around the ln section, the first part is the constant as before.

    When the infinite limit is applied we get ln((infinity +alpha^2)/(infinity +1)) so I used L'hopital's rule:

    If lim(x → c) f(x)/g(x) = L then, lim(x → c) f'(x)/g'(x) = L

    For my question...

    f(x) = x+alpha^2 f'(x) = alpha^2
    g(x) = x+1 g'(x) = 1

    so my integral becomes:

    1/(1-alpha^2) [ln(alpha^2/1] with the x limits around []

    = 1/(1-alpha^2) [lnalpha^2-lnalpha^2]
    = 1/(1-alpha^2) [ln1]
    = 0

    Is this correct? I am sure I have made a mistake as I now need to use my result to find I'(alpha)?
  5. Feb 28, 2014 #4
    I won't be near a computer until later tonight, but after integrating you should have 4 terms. When you plug in 0 to one of the terms, it turns to zero and goes away. Now you are left with three terms. For one of the terms you need to factor out a -1 to make the coefficients all line up. Then you can coming the natural logs and then take the limit. Perhaps someone else can write this out more clearly, if not I will reply later tonight.
  6. Feb 28, 2014 #5
    Okay, so you have ##\displaystyle\lim_{a\to\infty} \left[\frac{1}{1-\alpha^2}\int_0^a \frac{1}{x+\alpha^2} dx + \frac{1}{\alpha^2-1} \int_0^a \frac{1}{x+1} dx\right]## and then you correctly obtained natural logarithms from the integrations:

    ##\displaystyle\lim_{a\to\infty} \left[\frac{1}{1-\alpha^2} ln(x+\alpha^2) + \frac{1}{\alpha^2-1} ln(x+1)\right]_0^a##

    Before you take the limit as a goes to infinity, plug in the a's and 0's and combine the natural logs into one natural log. You'll have a common factor of ##\frac{1}{1-\alpha^2}## and you can thus combine the logs into one. When you do, you can finally take the limit with no complications.
  7. Feb 28, 2014 #6
    Don't forget to treat the special case of alpha=1.
  8. Feb 28, 2014 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'll assume you did the integration correctly. You don't need to use the Hospital rule here. Instead, just evaluate the limit the normal way:
    $$\lim_{x \to \infty} \frac{x+\alpha^2}{x+1} = \lim_{x \to \infty} \frac{1+\frac{\alpha^2}{x}}{1+\frac1x} = ?$$

    If you insist on using the Hospital rule, make sure you differentiate correctly. Check f'.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted