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 $r=1$?
Best Answer
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.
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:
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 $\infty$.