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Checking my values for a complex integral

  1. Aug 10, 2008 #1
    I am going to provide my answer to a complex integral and i was just seeking a few pointers as to weather i was on the right track or was there something i completely forgot...happens quite a bit...lol

    [tex]\oint exp(z+(1/z))[/tex] around the path [tex]\left |z|\right=1[/tex]

    now i converted that to a Laurent series....to get

    [tex]\sum ^{inf} _{0} (1/n!) (z+(1/z))^n [/tex]

    then using the residue theorem i can have that the integral is equal to 2*pi*i given that b1=1 for taking the series around z=0

    am i right???
  2. jcsd
  3. Aug 10, 2008 #2
    No, you must expand each of the powers and then extract the coefficient of 1/z and add them up. I think the sum of the series can be expressed in terms of the Bessel function of order 1 at -2 i (but I have to check to verify this)

    Now, you can find that result directly simply by putting
    z = exp(i theta), and integrating from theta = 0 to 2 pi. You then get an integral that is the same up to some factor to the Bessel function of order 1 of imaginary argument.
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