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I have to integrate the complex function $$ \frac{e^z-1}{z^5} $$ over the curve $\gamma(t)=1+re^{-5it}$ where $t \in [0,2\pi]$. The curve has winding number -5 with respect to a point inside the disc whose boundary is $\gamma$. I tried to apply Cauchy's formula in order to obtain (case $r>1$) $$ -5\cdot \frac{2\pi i}{4!} (e^z-1)^{(4)}|_{z=0} = \frac{-5\pi i}{12} . $$

Question 1: is that reasoning valid?

Question 2: in the case $r<1$, since the function is holomorphic over the curve $\gamma$, can we use Cauchy's theorem to affirm that $$ \int_\gamma \frac{e^z-1}{z^5}\, dz = 0 \quad \text{?} $$ Question 3: how to treat the case $r=1$?

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You are doing fine. For question $(3)$ you have a pole on the path of integration, so you need to consider Cauchy principal value contour integration. –  Mhenni Benghorbal Jan 20 '13 at 16:18
    
@mrf the contour is centered at $z=1$, so there is a difference. –  Antonio Vargas Jan 20 '13 at 16:20
    
@AntonioVargas Oops, sorry about that, I didn't notice the $1+$. –  mrf Jan 20 '13 at 16:24
    
@MhenniBenghorbal in this case the principal value does not exist since the pole is of even order and the Laurent series of the integrand centered at $z=0$ has real coefficients. –  Antonio Vargas Jan 20 '13 at 17:43
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1 Answer 1

up vote 3 down vote accepted

The integral cannot be evaluated in the usual sense when $r=1$ since there is a pole of order $4$ on the contour (at the origin). Remember that the $p$-test says that only poles of order $<1$ are integrable.

The usual method to "assign a value" to divergent integrals like this is the Cauchy principal value method, which in this case would look like

$$ \begin{align*} &\text{PV} \oint_\gamma \frac{e^z-1}{z^5}\,dz \\ &\qquad = -5 \cdot \lim_{\epsilon \to 0^+} \left[ \int_0^{\pi-\epsilon}\frac{\exp\left(1+e^{it}\right)-1}{\left(1+e^{it}\right)^5}\,ie^{it}\,dt + \int_{\pi+\epsilon}^{2\pi} \frac{\exp\left(1+e^{it}\right)-1}{\left(1+e^{it}\right)^5}\,ie^{it}\,dt \right]. \end{align*} $$

Here I've parameterized the circle by $z = 1+e^{it}$ but removed the arc of the circle of length $2\epsilon$ which passes through the pole at $z=0$, as in the picture below.

enter image description here

However, the even order of the pole presents an issue. As $\epsilon \to 0$ we'll be approaching the pole along paths tangent to the imaginary axis, and along this axis the real part of the integrand is even:

$$ \operatorname{Re}\left[\frac{e^{iy}-1}{(iy)^5}\right] = \frac{\sin y}{y^5}. $$

Here's a plot of this:

enter image description here

So, since

$$ \lim_{y \to 0} \frac{\sin y}{y^5} = \infty, $$

no cancellation will occur when calculating the principal value and the result as $\epsilon \to 0$ will be $\infty$.

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