Suppose as the simplest example that I want to integrate: $$ \int_{C_{r}(0)} \frac{1}{z+2} \, dz$$, where $C_{r}(0)$ is the circle centred at the origin with radius $r \neq 2$. We have Cauchy's integral formula:
Theorem: Let $f$ be a holomorphic function inside and on a positively oriented contour $\gamma$. If $\alpha$ is inside $\gamma$: $$f(\alpha) = \frac{1}{2\pi i} \int_{\gamma} \frac{f(\xi)}{\xi-\alpha}\, d\xi$$
Thus it seems obvious that we pick $f$ to be the constant function $f(z)=1$ (which is clearly entire). So: $$ 2\pi i = \int_{C_{r>2}(0)} \frac{1}{\xi+2} d\xi $$ [In the case $r<2$, then by Cauchy's other theorem, $\int_{C_{r<2}(0)} \frac{1}{z+2} = 0 $]
However, suppose I choose instead $g(z)= z+3$ (also entire). Then:
$$ 2\pi i = \int_{C_{r>2}(0)} \frac{\xi+3}{\xi+2} d\xi \implies 2\pi i = \int_{C_{r>2}(0)} \frac{3}{\xi+2} \, d\xi + \int_{C_{r>2}(0)} \frac{\xi}{\xi+2} \, d\xi $$ However the integral on the far RHS is equal to zero - recall $\xi(t) = rexp(2\pi i t) $. So we conclude: $$ \int_{C_{r>2}(0)} \frac{1}{\xi +2} \, d\xi = \frac{2\pi i}{3} $$ And so on and so forth for any $n$. Clearly something is a contradiction here? It seems a really stupid question, but I am very puzzled.
The integral on the far RHS is equal to (again by Cauchy's integral formula): $$ 2\pi i f(-2)=2\pi i(-2)=-4 \pi i$$