Trying to solve the integral $\int_{0}^{\infty} \frac{e^{-x}}{x^{\alpha}(x-(a+i b))} dx$, $a,b \in \mathbb{R}$, $\alpha<1$ I am trying to solve the integral
$$I_{\alpha}:=\int_{0}^{\infty} \frac{e^{-x}}{x^{\alpha}(x-(a+i b))} dx,$$ $a,b \in \mathbb{R}$, $\alpha <1$.
Mathematica gives
$$I_{\alpha}=e^{-z} \left(-\frac{1}{z}\right)^{\alpha} \Gamma (1-\alpha) \Gamma (\alpha,-z)$$
provided that $\Im(z)\neq 0\lor \Re(z)<0$. Here $z=a+ i b$.
Assuming that $a,b>0$, and trying to use the residue theory on the contour
( ($\arg{z}=0,\, 0<|z|<R$), ($0 \leq \arg{z}\leq \pi/2,\, |z|=R$), ($\arg{z}=\pi/2,\, 0<|z|<R$) ),  one gets  an integral on the segment that lies on the imaginary axis that is  equally difficult to (but different from) the integral one gets on the contour segment that lies on the real axis. This is easy to see just by looking at the factor $x-(a+i b)$ in the denominator of the integrand.
If we use real-variable methods, we must solve
$$J_{\alpha}:=\int_{0}^{\infty} \frac{e^{-x}(x-a)}{x^{\alpha}(
(x-a)^2+b^2)} dx,$$
and
$$H_{\alpha}:=\int_{0}^{\infty} \frac{e^{-x}b}{x^{\alpha}(
(x-a)^2+b^2)} dx.$$
Then $I_{\alpha}=J_{\alpha}+i H_{\alpha}$.
Any suggestions ?
 A: We start by introducing a new function.
$$I(a,t,z)=\int_0^\infty\frac{x^ae^{-t(x+z)}}{x+z}dx$$
provided that $t≥0$.
We easily find $I(a,\infty,z)=0$
When we take the derivative with respect to $t$, our integral simplifies using the gamma function after subbing $u=tx$
$$\partial_t\space I(a,t,z)=-\int_0^\infty x^ae^{-t(x+z)}dx=-t^{-a-1}e^{-zt}\Gamma(a+1).$$
Using $I(a,\infty,z)=0$, we get
$$J(a,z)=\int_0^\infty\frac{x^ae^{-x}}{x+z}dx=e^z\left(I(a,1,z)-I(a,\infty,z)\right)=e^z\int_\infty^1\partial_t\space I(a,t,z)dt$$
$$=e^z\int_\infty^1\left(-t^{-a-1}e^{-zt}\Gamma(a+1)\right)=e^z\Gamma(a+1)\int_1^\infty t^{-a-1}e^{-zt}dt.$$
Substitute $u=zt$
$$J(a,z)=\int_0^\infty\frac{x^ae^{-x}}{x+z}dx=z^ae^z\Gamma(a+1)\int_z^\infty u^{-a-1}e^{-u}du=z^ae^z\Gamma(a+1)\Gamma(-a,z)$$
Finally:
$$\int_0^\infty\frac{e^{-x}}{x^a(x-z)}dx=J(-a,-z)=(-z)^ae^{-z}\Gamma(1-a)\Gamma(a,-z)$$
A: Assume that $\Re z<0$ and $\alpha<1$. Then
\begin{align*}
\int_0^\infty  {\frac{{e^{ - x} x^{ - \alpha } }}{{x - z}}dx} & = \int_0^\infty  {e^{ - x} x^{ - \alpha } \left( {\int_0^\infty  {e^{ - (x - z)t} dt} } \right)dx} \\ & = \int_0^{ \infty } {e^{zt} \left( {\int_0^\infty  {e^{ - (1 + t)x} x^{ - \alpha } dx} } \right)dt} 
\\ & = \int_0^\infty  {e^{zt} (1 + t)^{\alpha  - 1} \left( {\int_0^\infty  {e^{ - s} s^{ - \alpha } ds} } \right)dt} \\ & = \Gamma (1 - \alpha )\int_0^\infty  {e^{zt} (1 + t)^{\alpha  - 1} dt} 
\\ & = e^{ - z} \Gamma (1 - \alpha )\int_1^\infty  {e^{zu} u^{\alpha  - 1} du} \\ & = e^{ - z} \Gamma (1 - \alpha )( - z)^{ - \alpha } \int_{ - z}^\infty  {e^{ - v} v^{\alpha  - 1} dv} \\ & = e^{ - z} \Gamma (1 - \alpha )( - z)^{ - \alpha } \Gamma (\alpha , - z).
\end{align*}
