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2nd order pole while computing residue in a complex integral

  1. Apr 16, 2015 #1

    I am trying to understand how to get the residue as given by wolfram :


    The issue I am facing is - since it is a second order pole, when I try to different e^{\sqrt{x^+1}} I get a \sqrt{x^+1} in the denominator. And hence I cannot substitute x = i to get the residue. How do I deal with a pole which comes about on differentiating?

  2. jcsd
  3. Apr 16, 2015 #2


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    The standard way to deal with a second-order pole is to expand the nominator using Taylor's theorem around the pole. Dividing by the denominator will give you the value of the residue. Take the pole at z=i: [itex]e^{\sqrt{z^{2}+1}}=e^{0}+f_{1}(z)(z-i)+f_{2}(z)(z-i)^{2} [/itex]. Dividing by [itex] (z-i)^{2}[/itex], we get [itex]\frac{e^{\sqrt{z^{2}+1}}}{(z-i)^{2}}=\frac{e^{0}}{(z-i)^{2}}+\frac{f_{1}(z)}{(z-i)}+f_{2}(z) [/itex]. So, f1(i) is the residue at z=i.
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