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Math Help - Integrate this singularity?

  1. #1
    Junior Member NowIsForever's Avatar
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    Integrate this singularity?

    I found this problem recently on Yahoo Answers! It is gone now because it went unanswered after four days, and it is the policy of YA! to delete such questions.


    Given:

    {I(n)=\int^{\pi/2}_{-\pi/2}\frac{\cos(2nx)}{(1+e^x)\cos(2x)}dx}


    Prove:

    I(n + 2) = -I(n), for all natural numbers n. (Since "natural number" is ambiguous, let's say for non negative—or possibly positive integers.)


    I couldn't get passed the singularities at -\pi/4 and \pi/4 . WolframAlpha couldn't provide an answer either, even for (e.g.), n=2.


    I have my doubts about this being true; however, if it is I would like to see a proof.
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  2. #2
    MHF Contributor
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    Re: Integrate this singularity?

    Hey NowIsForever.

    Use the fact that cos(a+b) = cos(a)cos(b) - sin(a)sin(b) where a = 2nx and b = 2x. This gives cos(2nx)cos(2x) - sin(2nx)sin(2x) for which you get some cancellation and simplification for the cos and sin of 2x.
    Thanks from topsquark
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  3. #3
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    Re: Integrate this singularity?

    The integral has to be understood as a Cauchy integral (i.e. as Principal Value) since it is not convergent except for n=1.
    On this condition, I(n+2) = -I(n) , as shown in attachment and I(1) = pi/2
    Attached Thumbnails Attached Thumbnails Integrate this singularity?-cauchyintegral.jpg  
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  4. #4
    Junior Member NowIsForever's Avatar
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    Re: Integrate this singularity?

    Quote Originally Posted by JJacquelin View Post
    The integral has to be understood as a Cauchy integral...
    Ah, yes, that does it; however, you have (n + 2) in the last four integrals where it should be (n + 1).

    You side stepped the singularity by clever use of trigonometric identities, and the identity \frac{1}{1+e^{-x}}=1-\frac{1}{1+e^x}.

    I thought it would require something like the Cauchy principal value calculated with limits, but you've made that unnecessary, thanks.

    Nice—and elegant.
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  5. #5
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    Re: Integrate this singularity?

    Hi !
    We saw that I(1)=pi/2 which leads to I(n)=((-1)^((n-1)/2))*(Pi/2) in case of odd n.
    In case of even n, we have to compute I(0) which requires to follow the computation rules for the principal value of a Cauchy integral. The result is I(0)=0.
    As a consequence I(n)=0 in case of even n
    .
    Quote Originally Posted by NowIsForever View Post
    you have (n + 2) in the last four integrals where it should be (n + 1).
    Thanks you for bringing the typo to attention. Of course, it doesn't change the result.
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