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Complex line integrallll

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

    for [itex]n=1,2,3,...[/itex] , evaluate the integral,

    [itex]I= \int_C\frac{e^{iz}}{z^n}dz[/itex]
    where [itex]C [/itex] is a curve like [itex]z(t)=e^{it}[/itex] and [itex]0 \leq t \leq 2\Pi[/itex]

    2. Relevant equations

    3. The attempt at a solution
    I tried to use Cauchy integral formula; that

    [itex]f^{(n)}(z)=\frac{n!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}d\zeta[/itex]

    then we can obtain,

    [itex]f^{(n-1)}(z)=\frac{(n-1)!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n}}d\zeta[/itex]

    [itex]f^{(n-1)}(z)=\frac{(n-1)!}{2 \Pi i}\int_C\frac{f(\zeta)}{(\zeta-z)^{n}}d\zeta[/itex]

    [itex](e^{iz})^{(n-1)}(z) \Big\vert_{z=0}=\frac{(n-1)!}{2 \Pi i}\int_C\frac{e^{iz}}{z^{n}}d\zeta[/itex]

    [itex]i^{n-1}=\frac{(n-1)!}{2 \Pi i}\int_C\frac{e^{iz}}{z^{n}}d\zeta [/itex]

    [itex]\int_C\frac{e^{iz}}{z^{n}}d\zeta=\frac{i^{n-1} 2 \Pi i} {(n-1)!} [/itex]

    [itex]~~~~~~~~=\frac{i^n 2 \Pi} {(n-1)!} [/itex] .

    can you check, is it right??????
  2. jcsd
  3. Sep 14, 2011 #2


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    Yes, that's correct.

    You could also expand the integrand as a Laurent series and use the residue theorem.
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