Problem with General Contour integral I am trying to calculate a contour integral $$\oint_{\Gamma}\frac{z^{\alpha}e^z}{z-b}\,dz$$ where $\Gamma$ is the counterclockwise path from $\sigma-i\infty$ to $\sigma+i\infty$ that loops back around and closes on a semicircle toward the left. In this case, $\sigma$ is an arbitrary complex number such that Re$(\sigma$)>b, for real number b. $\alpha$ is a real number such that $0<\alpha<1$.
I wanted to apply the Cauchy integral formula, but I don't think it can be justified because I don't think $s^{\alpha}e^z$ will be holomorphic in the region. Thank you for your help.
 A: Set up the semicircular contour but add a key hole from $z=-R$ to the origin. The integral around the origin
$$\left|\int_\pi^{-\pi} \frac{i\epsilon^{1+\alpha}e^{i\alpha t}e^{\epsilon e^{it}}dt}{\epsilon e^{it} - b}\right| \leq \epsilon^{1+\alpha} \frac{2\pi e^{\epsilon}}{b-\epsilon}$$
goes to $0$ in the limit $\epsilon\to 0^+$. All that's left is the two paths inbetween. In the limit, the upper path takes $-1 = e^{i\pi}$ and the lower path takes $-1 = e^{-i\pi}$
$$\int_0^\infty \frac{x^\alpha e^{-i\alpha \pi}e^{-x}}{x+b}\:dx - \int_0^\infty \frac{x^\alpha e^{i\alpha \pi}e^{-x}}{x+b}\:dx = -2i\sin(\alpha\pi)\int_0^\infty \frac{x^\alpha e^{-x}}{x+b}\:dx$$
To solve the integral we can set up a differential equation:
$$I(s) = \int_0^\infty \frac{x^\alpha e^{-sx}}{x+b}\:dx \implies I'-bI = -\int_0^\infty x^\alpha e^{-sx}\:dx = -\frac{\Gamma(1+\alpha)}{s^{1+\alpha}}$$
This has the homogeneous solution $e^{bs}$ and by variation of parameters one obtains the particular solution
$$I = e^{bs}b^\alpha \Gamma(1+\alpha) \Gamma(-\alpha,bs)$$
where $\Gamma(a,z)$ is the incomplete Gamma function. Taking the limit gives us $\lim_{s\to\infty}I(s) = 0$, so the homogenous solution doesn't contribute at all. This means the value of the integrals that straddle the branch cut is
$$-2ie^bb^{\alpha}\sin(\alpha \pi)\Gamma(1+\alpha)\Gamma(-\alpha,b)$$
Lastly, the residue theorem tells us that
$$\int_{\sigma-i\infty}^{\sigma+i\infty} \frac{z^\alpha e^z}{z-b}\:dz -2ie^bb^{\alpha}\sin(\alpha \pi)\Gamma(1+\alpha)\Gamma(-\alpha,b)  = 2\pi i b^\alpha e^b$$
since the integral on the arc can go to $0$ in a principal value way. Which leaves us with our final result
$$\int_{\sigma-i\infty}^{\sigma+i\infty} \frac{z^\alpha e^z}{z-b}\:dz = 2i e^bb^\alpha\left[\sin(\alpha \pi)\Gamma(1+\alpha)\Gamma(-\alpha,b)+\pi\right] $$
