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Math Help - Integrate - getting stuck

  1. #1
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    Integrate - getting stuck

    I want to solve a problem.

    \displaystyle \int_{|z|=8} \frac{dz}{(1-e^z)}

    I thought I could get something decent by doing a Taylor expantion. But no!

    How should i solve it?
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  2. #2
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    Quote Originally Posted by liquidFuzz View Post
    I want to solve a problem.

    \displaystyle \int_{|z|=8} \frac{dz}{(1-e^z)}

    I thought I could get something decent by doing a Taylor expantion. But no!

    How should i solve it?

    Calculate the residue of \frac{1}{1-e^z} at z=0 ...

    Tonio
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    Yes, I know that. But how do I do that?
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  4. #4
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    To help you a bit in finding the residue at z=0, since our aim is to compute the residue at the singularity z=0,

    consider  1/(1-e^{z}) = 1/(1-(1+z+z^{2}/2+...)) (write the Taylor's expansion for f(z)=e^{z})

     = 1/(-z-z^{2}/2-...) = (-1/z)(1/(1+z/2+...))

    Now reduce this into a Laurent series expansion about z=0 and hence find the residue as coefficient of 1/z
    Last edited by bandedkrait; October 30th 2010 at 07:40 AM. Reason: more elaborate
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    I'm sorry if I was vague when I described my problem. I was on a train tapping the message on my phone.


    bandedkrait
    As I stated in my first post, I actually did the expansion, but I'm not sure of how to interpret the result. When I played around with expansions I used the similarity to geometric series to get the next step from where you stopped and got this.
    \displaystyle - \frac{1}{z} +\frac{1}{2} - \frac{z}{12} + O(z^3)

    This is about it. I don't know what the next step is...
    Last edited by liquidFuzz; October 30th 2010 at 09:58 AM. Reason: corrected an thing, 1/6 to 1/12
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    HUH!!! I googled residue lauernt series and got an interesting result.

    Quote Originally Posted by wikipedia
    The coefficient a−1 of the Laurent expansion of such a function is called the residue of ƒ(z)
    So my residue is -1, or?
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  7. #7
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    Yep, in this case it is. The residue is the coefficient of the term (z-z0)^{-1} in the Laurent Series expansion of a function about a point z0 In this case, it happens to be -1 as ur simplifications shows. here z0=0
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    Quote Originally Posted by liquidFuzz View Post
    HUH!!! I googled residue lauernt series and got an interesting result.



    So my residue is -1, or?

    Er...yes. You didn't know that before attempting to solve such an integral?!?

    Tonio
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    Quote Originally Posted by tonio View Post
    Er...yes. You didn't know that before attempting to solve such an integral?!?

    Tonio
    Actually, there might be just a tiny bit more mathematics I don't know...
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  10. #10
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    Quote Originally Posted by tonio View Post
    Calculate the residue of \frac{1}{1-e^z} at z=0 ...

    Tonio
    The residue at z = 2 \pi i will also need to be calculated (since the closed contour |z| = 8 encloses both the isolated singular points z = 0 and z = 2 \pi i of \displaystyle \frac{1}{1 - e^z}).
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