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Homework Help: (Improper) Integral -2x/(1+x^2) dx from -INF to INF

  1. Oct 25, 2013 #1


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    1. The problem statement, all variables and given/known data
    Consider the integral

    INT -2x/(1+x^2) dx FROM -INF TO INF (The attached TheIntegral.jpg file shows this in a more aesthetically-pleasing manner.)

    If the integral is divergent, state that it is so. Otherwise, evaluate the integral.

    2. Relevant equations
    U substitution.

    3. The attempt at a solution
    I turned the integral the problem gives into:

    lim t-> inf INT -2x/(1+x^2) dx FROM -t TO t

    Letting u = 1+x^2, du/dx = 2x, dx = du/2x

    Lower limit: u = 1 + (-t)^2 = 1 + t^2, Upper limit: u = 1 + (t)^2 = 1 + t^2
    lim t-> inf INT -1/u du FROM (1 + t^2) TO (1 + t^2)
    lim t-> inf [-ln|u|] FROM (1 + t^2) TO (1 + t^2) = 0

    Using Wolfram Alpha, I know the integral is divergent but, when I use Maxima, I get Principal Value: 0 (and, I don't know what means, but it's what I get).

    In addition to me agreeing with the value that Maxima gives, I also used another piece of software called Kmplot which is how I got the image I attached in this thread, TheGraph.jpg, which shows a symmetric graph with areas that, to my intuition at least, seem to cancel out.

    Could someone please explain to me why this integral is a divergent integral rather than one that converges to a value of 0?

    Attached Files:

  2. jcsd
  3. Oct 25, 2013 #2
    Improper integral ##\int_{-\infty}^\infty f(x)\,dx## is converged if both integrals ##\int_{-\infty}^af(x)\,dx## and ##\int_a^\infty f(x)\,dx## are convergent. Cauchy principal value is a limit ##\lim\limits_{a\to\infty}\int_{-a}^af(x)\,dx## and it may exist even if ##\int_{-\infty}^\infty f(x)\,dx## is divergent.
  4. Oct 25, 2013 #3


    Staff: Mentor

    I posted a reply, but I deleted it and am rethinking what I said...
  5. Oct 25, 2013 #4


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    Here is the Wolfram Alpha link.:

    Looking at the above link again, it does say that the integral equals to zero below "Cauchy Principal Value:", which is beyond what I'm currently learning, but it also says that the integral does not converge.

    According to the answer from the problem I am doing, the integral is in fact diverging (but, I still don't understand why).
  6. Oct 25, 2013 #5


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    Szynkasz, that was the information I needed!

    Thank you both! :)
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