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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.

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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 $\exp(-2\pi i(\alpha+1))\neq 1$.

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