- #1
laleler1
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Homework Statement
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]
Homework Equations
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?