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Complex integration

  1. Dec 23, 2011 #1
    1. The problem statement, all variables and given/known data

    Integrate $\Sigma \frac{1}{n!}\int z^{n}e^{1/z}dz$

    2. Relevant equations



    3. The attempt at a solution

    Wrote out the first couple of terms, with $\frac{1}{z}=w$, making the integral $\Sigma \frac{1}{n!} (-w^{2-n}e^{w}+(2-n)(w^{1-n}e^{w}+(1-n)(w^{1-n}e^w)-(1-n)^2(w^(-n)e^w)...)$

    But wasn't sure how to go from here.
     
  2. jcsd
  3. Dec 23, 2011 #2
    Need to use tex delimiters and not $ as in:

    [tex]
    \int
    [/tex]

    Do a quote of my post to see that. Also, what happen to the integral signs. Would be easy if those were just integrations around closed contours about the origin. Otherwise, the antiderivatives are not elementary.
     
  4. Dec 23, 2011 #3
    [tex]
    \Sigma \frac{1}{n!}\int z^{n}e^{1/z}dz
    [/tex]


    [tex]
    \Sigma \frac{1}{n!} (-w^{2-n}e^{w}+(2-n)(w^{1-n}e^{w}+(1-n)(w^{1-n}e^w)-(1-n)^2(w^{-n}e^w)...)
    [/tex]

    Sorry I was used to a different forum. Depending on the forum, a different style is used, I guess. Thanks for pointing this out.

    Yes, the contour integration is done on a unit circle, |z|=1, but it's multiply connected because of the isolated singularity at the origin.
     
  5. Dec 23, 2011 #4
    So you want to find:

    [tex]\sum_{n=1}^{\infty}\frac{1}{n!}\mathop\oint\limits_{|z|=1} z^n e^{1/z}dz[/tex]

    How about using the Residue theorem? You can do that easily (I think, haven't gone over all of it yet) by taking the residue at infinity.
     
  6. Dec 23, 2011 #5
    Right, the residue is
    [tex]
    \frac{1}{(n+1)!}
    [/tex]

    So then the expression becomes
    [tex] \Sigma \frac{1}{n!(n+1)!} [/tex]
     
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