The context of my question is as follows. Consider a rational function $K\left(z,w\right)$ admitting a “partial fractions” decomposition of the form: $$K\left(z,w\right)=P\left(z,w\right)\sum_{n=1}^{N}\frac{c_{n}}{w-z^{q_{n}}}$$ for some polynomial $P\in\mathbb{C}\left[z,w\right]$, some integer $N\geq1$, some non-zero complex constants $c_{1},\ldots,c_{N}$, and some positive rational numbers $q_{1},\ldots,q_{N}$, with the $q_{n}$s being pair-wise distinct with respect to $n$. By integrating against this “kernel” $K$, we get a linear operator $T$ on the space $\mathcal{A}\left(\mathbb{D}\right)$ of all holomorphic functions $f:\mathbb{D}\rightarrow\mathbb{C}$: $$T\left\{ f\right\} \left(z\right)=\frac{1}{2\pi i}\oint_{r\partial\mathbb{D}}K\left(z,w\right)f\left(w\right)dw,\textrm{ }\forall r\in\left(0,1\right),\textrm{ }\forall\left|z\right|<r^{\prime}$$
The contour is taken over the circle centered at $0$ of radius $r$ (where $r$ is fixed) because not all $f\in\mathcal{A}\left(\mathbb{D}\right)$ will be integrable along the unit circle. The constant $r^{\prime}$ is a real number in $\left(0,1\right)$ which is larger than $r$ and, furthermore, depends solely on $r$ and $K$. Finally, $K$ must also have the property that, for any fixed $r$ and any fixed $z$ with $\left|z\right|<r^{\prime}$ , all of the $w$-poles of $K\left(z,w\right)$ must lie within the open disk $r\mathbb{D}$ (centered at $0$ of radius $r$).
Now, let $B\left(z\right)$ be a rational function such that for any pole $p$ of $B\left(z\right)$, $p$ is of degree $1$, and $\left|p\right|>r$. Fixing $z$ with $\left|z\right|<r^{\prime}$, let $P_{z}\left(K\right)$ denote the set of $w$-poles of $K\left(z,w\right)$, and let $P\left(B\right)$ denote the set of poles of $B$. Then, the Inside-Outside Theorem allows us to write: $$T\left\{ B\right\} \left(z\right)=\sum_{p\in P_{z}\left(K\right)}\textrm{Res}_{w=p}\left[K\left(z,w\right)B\left(w\right)\right]=-\sum_{p\in P\left(B\right)}\textrm{Res}_{w=p}\left[K\left(z,w\right)B\left(w\right)\right]$$ Since all the $w$-poles of $K\left(z,w\right)$ have magnitude less than $r$, and since all the poles of $B$ are of degree $1$, we have:$$\sum_{p\in P\left(B\right)}\textrm{Res}_{w=p}\left[K\left(z,w\right)B\left(w\right)\right]=\sum_{p\in P\left(B\right)}K\left(z,p\right)\textrm{Res}_{p}\left[B\right]$$ and so: $$T\left\{ B\right\} \left(z\right)=-\sum_{p\in P\left(B\right)}K\left(z,p\right)\textrm{Res}_{p}\left[B\right]$$ Generalizing from here, when the poles of $B$ are allowed to have degrees greater than $1$, we end up modifying this linear combination formula with partial derviatives of $K$ with respect to w.
However: what happens when we allow poles of fractional order?
Case in point, I would like to know if there is an analogue of the above computation when: $$B\left(z\right)=\frac{1}{\left(z-\xi\right)^{\alpha}}$$ where $\xi$ is a point on the unit circle and $\alpha$ positive non-integer real number.
Also, how does the choice of branch cut play into this? The natural choice, to me, would be to take a cut along the portion of the ray emanating from the origin and passing through $\xi$ which lies outside of the unit disk, but I don't know how to deal with the integration around the slit.