# Complex line integrallll

1. Sep 14, 2011

### laleler1

1. The problem statement, all variables and given/known data

for $n=1,2,3,...$ , evaluate the integral,

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

2. Relevant equations

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

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

then we can obtain,

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

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

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

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

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

$~~~~~~~~=\frac{i^n 2 \Pi} {(n-1)!}$ .

can you check, is it right??????

2. Sep 14, 2011

### vela

Staff Emeritus
Yes, that's correct.

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

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