# Explanation of coefficient when evaluating contour around a branch for fractional version of Cauchy's Integral Formula

I am working on fractional derivatives which are defined by taking the Cauchy Integral formula and letting the order of the derivative be non-integer. Specifically, $$\begin{equation} f^{(\alpha)}(z)=\frac{\Gamma(\alpha+1)}{2\pi i}\oint f(t)(t-z)^{-(\alpha+1)}\text{d}t \end{equation}$$ Where our contour in the $$t$$ plane begins at the origin, encloses $$z$$ once in the positive sense and returns to the origin. Now, for simplicity let's assume that $$f(t)$$ is analytic in a region containing both the origin and $$z$$. Then, our contour integral has a branch on the line segment from the origin to $$z$$. So, we can deform our contour to go from the origin, up the line segment, loop around $$z$$, down the line segment and back to the origin. The paper I am following expresses this contour as $$f^{(\alpha)}(z)=\frac{\Gamma(\alpha+1)}{2\pi i}\int_{0}^{z^{+}} f(t)(t-z)^{-(\alpha+1)}\text{d}t$$ They then immediately claim that $$f^{(\alpha)}(z)=\frac{\Gamma(\alpha+1)}{2\pi i}\left[1-\exp{(-2\pi i(\alpha+1))}\right]\int_{0}^{z} f(t)(t-z)^{-(\alpha+1)}\text{d}t$$ Where the integral is now your standard Riemann integral, that can be evaluated via the Fundamental Theorem of Calculus. My question is: why is this true? There is something about contour integrals that I am not aware of that they are using to make the jump from Eq. 2 to Eq. 3.

Also, if it helps, I believe I am working with a proto-Riemann-Liouville derivative, as Eq.3 becomes the Riemann-Liouville derivative when substituting out appropriate coefficients via the the Euler-Reflection formula for the Gamma function.

It comes directly from tracking the branch of $$(t-z)^{-(\alpha+1)}$$, that in the returning line from $$z$$ to $$0$$ the $$(t-z)^{-(\alpha+1)}$$ is multiplied by $$\exp(-2\pi i(\alpha+1))\neq 1$$.