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Math Help - Integral with residue, question about poles.

  1. #1
    Newbie dttah's Avatar
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    [Solved]Integral with residue, question about poles.

    Hello everyone, I am having a little problem with the following integral:

    \int \frac{sinz}{z^3(z^2+1)}

    Where D is the Domain D = {z in the complex field : |z|<2}

    Now, my understanding is I have to look where my function goes to zero.
    On the denominator I have :

    z = j (simple pole)
    z = -j (simple pole)
    and z = 0 (pole of order 3)

    Now, on the numerator, I have
    z = 0,pi, 2pi

    Since zero appears both at the numerator and the denominator I understand it becomes a pole of 2order, and not a 3 order one.
    So my integral becomes:

    2\pi j(Res(j) + Res(-j) + Res(0))

    Now what happens is, if I calculate the residue in ZERO it's ZERO! So it means that That's a removable singularity! Now, I double checked with wolfram and it doesn't even bother to calculate the residue in zero.
    Could anyone tell me why? Please?

    sinz/(z^3(z^2+1)) residue - Wolfram|Alpha

    Thanks a lot!!
    Last edited by dttah; April 15th 2013 at 03:06 PM.
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  2. #2
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    Re: Integral with residue, question about poles.

    Vanishing residue says the coefficient of the -1 term in the series expansion is 0( since it is 1/z^2), which doesn't imply the singularity is removable.
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  3. #3
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    Re: Integral with residue, question about poles.

    sin(z)/z would be a removable singularity. sin(z)/z^3 is, as you say, a pole of order 2. Yes, the residue at 0 is 0. That does NOT mean the the singularity is removable.

    sin(z) is analytic and has McLaurin series z- \frac{1}{3!}z^3+ \frac{1}{5!}z^5- \frac{1}{7!}z^7+ \cdot\cdot\cdot

    So the Laurent series for \frac{sin(z)}{z^3} is \frac{sin(z)}{z^3}= z^{-2}- \frac{1}{3!}+ \frac{1}{5!}z^2- \frac{1}{7!}z^4+ \cdot\cdot\cdot
    z= 0 is not a "removable singularity" because that series has negative powers. It is a pole of order 2 because the largest negative power is -2. The residue at z= 0 is the coefficient of the -1 power which is 0.
    Last edited by HallsofIvy; April 15th 2013 at 03:06 PM.
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  4. #4
    Newbie dttah's Avatar
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    Re: Integral with residue, question about poles.

    Alright, I thought that when the residue was 0 it was a removable singularity that's why I had so many doubts! Thanks for clearing that up! .
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