How do infinite keyhole contours converge? I am trying to use what I know about contours to evaluate this integral but for fixed complex values $(c, Re \{c \} \geq 0)$ instead of $n$ 
$$\int_{-\infty}^{y}(y-t)^{c}dt.$$
The only place to really take a branch cut here is along the negative axis, but in doing so, I can no longer integrate along that axis, which means I now must extend into the complex plane from negative infinity, to an arc centered at $y=t,$ then back to negative infinity, forming a bullet or keyhole shape (or an equivalent deformation) such as the one mentioned here
Inverse Gamma function for integers (Hankel)
The integral should be then $$\oint_{\Gamma}(y-z)^{c}dz$$
Now, I don't completely know if this is correct, so partly I want to check if I am interpreting this correctly, but as far as I know, if I take this extension into the complex plane by some offset $\epsilon >0$ parallel to the real axis, this contour can be broken down into
$$\oint_{\Gamma}(y-z)^{c} = \int_{-\infty}^{y}(y-t+\epsilon)^{c}dt+\int_{-\pi}^{\pi}(y-|r|e^{i \theta})^{c}d\theta + \int_{y}^{-\infty}(y-t-\epsilon)^{c}dt, \ \ \ \ \ |r| = \epsilon,$$
Firstly, is this setup correct?
And secondly, to show this is well defined, I assume I take the limit as $\epsilon \rightarrow 0$ with absolute value brackets around the expression. What kind of cancellation am I looking for exactly to ensure this converges? I don't think I need the residue theorem since there's no singularities for the given condition, but if there is and it is a problem, take $Re \{c \} \geq 1$ instead.
 A: Suppose you have an integral $\int_{\rm Hankel} e^{t}t^{-c}\,dt$ (see Fig. left). It converges for any value of $c$. Simply because the convergence guaranteed by the behavior of the exponetnial function. Now, to actually compute the integral you need to simplify the contour. So you deform it into a combination of two straight lines and an infinitesimal circle (see Fig. right). 
The point is: you compute this integral at any suitable value of your parameter. Say, at $c<1$. Then the integral round the circle is estimated as $\int_{-\pi}^\pi|\varepsilon|e^{i\varphi(1+c)}id\varphi |\varepsilon|^{-c}\sim |\varepsilon|^{1-c}\Big|_{\varepsilon\rightarrow 0}\rightarrow 0$. But your answer will be an analytical function of $c$ and will be correct for any value of $c$ since the original integral is an analytical function of $c$ (this is called an analytical continuation.) So, parameterizing the complex variable $t$ on the lower bank of the branch cut $t=\rho e^{-i\pi}$ and on the upper bank of the branch cut as $t=\rho e^{i\pi} $the integral along the Hankel contour will be smply reduced to:
\begin{gather}
  \int_{\rm Hankel} =\left[\int_{+\infty}^{-\varepsilon} e^{-\rho}t^{-c} e^{i\pi c}(-d\rho)+\int_{-\varepsilon}^\infty e^{-\rho}t^{-c} e^{-i\pi c}(-d\rho)\right]_{\varepsilon\rightarrow 0 }=2i\sin\pi c \Gamma(1-c)
\end{gather}
