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Help me solving this complex integral

  1. Jul 9, 2011 #1

    could you please help me solving this integral:

    [itex]\oint \frac{e^{-(a+b)+az+\frac{b}{z}}}{z(z-1)}dz[/itex]

    over the unit circle, where [itex] a, b[/itex] are two positive constants (it is not a homework)
    thanks a lot in advance
  2. jcsd
  3. Jul 9, 2011 #2


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    There are two poles, one in the centre (no problem) and on on the boundary (not that much of a problem, you just have to deform your contour slightly. I think you main problem is going to be exp(b/z) term.
  4. Jul 10, 2011 #3
    I think the problem is exactly related to the pole which is placed at [itex] z=0 [/itex]. it is of order infinity. Am I right?
  5. Jul 10, 2011 #4


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    So the integral basically becomes:
    According to the sources I have read, you have to compute the Laurent series for [itex]e^{z}[/itex] and te Laurent series of [itex]e^{\frac{1}{z}}[/itex] along with all the other functions involved and just pick out the coefficient of the [itex]1/z[/itex] term. Sorry, but it is going to take a lot of algebra on this one.
  6. Jul 10, 2011 #5
    The answer is as follows:

    Residue at pole z=1 is [itex] 2\pi \frac{1}{2} [/itex]
    and residue at pole z=0 is [itex] -2\pi e^{-(a+b)}\sum _{n=0}^{\infty} \frac{a^n}{n!} \sum_{m=0}^{n}\frac{b^m}{m!} [/itex]

    So, we conclude the result as:

    [itex] \pi - 2\pi e^{-(a+b)}\sum _{n=0}^{\infty} \frac{a^n}{n!} \sum_{m=0}^{n}\frac{b^m}{m!} [/itex]

    Is everything Ok?
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