Computing the asymptotics of a principal value integral I have been looking at the following principal value integral (with $x>0$ and $0< \sigma < 1$)
$$
\mathrm{P.V.} \int_0^\infty \frac{v^{-\sigma} e^{-x v}}{1 - v} dv,
$$
and I would like to know how to expand this integral for small values of $x$.
One option is to go through special functions and by inspection, it looks like incomplete Gamma functions can be used. This is confirmed by Mathematica, which can evaluate it to be
$$
\mathrm{P.V.} \int_0^\infty \frac{v^{-\sigma} e^{-x v}}{1 - v} dv = e^{-x}(-i\pi + e^{-i \pi \sigma} \sigma \Gamma(-\sigma) \Gamma(\sigma, -x)),
$$
and it has the expansion
$$
\mathrm{P.V.} \int_0^\infty \frac{v^{-\sigma} e^{-x v}}{1 - v} dv = -\pi \cot(\pi\sigma)(1 - x + O(x^2))  - \Gamma(-\sigma)x^{\sigma} \left(1 - \tfrac1{1+\sigma} x + O(x^2)\right),
$$
However, my question is specifically: How do I derive either of the above two equalities?
 A: We start with $(8.6.4)$:
$$\tag{1}
\Gamma (\sigma ,z) = \frac{{z^\sigma  e^{ - z} }}{{\Gamma (1 - \sigma )}}\int_0^{ + \infty } {\frac{{t^{ - \sigma } e^{ - t} }}{{z + t}}dt} 
$$
valid for $|\arg z|<\pi$ and $\Re\sigma<1$. Using $(8.2.10)$, we find
$$
\Gamma (\sigma ,z) =  - \frac{{2\pi i}}{{\Gamma (1 - \sigma )}}e^{\pi i\sigma }  + e^{2\pi i\sigma } \Gamma (\sigma ,ze^{ - 2\pi i} ).
$$
Hence
$$\tag{2}
\Gamma (\sigma ,z) =  - \frac{{2\pi i}}{{\Gamma (1 - \sigma )}}e^{\pi i\sigma }  + \frac{{z^\sigma  e^{ - z} }}{{\Gamma (1 - \sigma )}}\int_0^{ + \infty } {\frac{{t^{ - \sigma } e^{ - t} }}{{z + t}}dt} 
$$
provided $\pi  < \arg z < 3\pi$ and $\Re\sigma<1$. Approaching $z = xe^{\pi i}$, $x>0$, in $(1)$ and $(2)$ and taking the average of the two expressions, we find
$$
\Gamma (\sigma , - x) =  - \frac{{\pi i}}{{\Gamma (1 - \sigma )}}e^{\pi i\sigma }  + e^{\pi i\sigma } \frac{{x^\sigma  e^x }}{{\Gamma (1 - \sigma )}}{\rm P}{\rm .V}{\rm .}\int_0^{ + \infty } {\frac{{t^{ - \sigma } e^{ - t} }}{{ - x + t}}dt} ,
$$
or with $t=vx$,
$$
\Gamma (\sigma , - x) =  - \frac{{\pi i}}{{\Gamma (1 - \sigma )}}e^{\pi i\sigma }  - e^{\pi i\sigma } \frac{{e^x }}{{\Gamma (1 - \sigma )}}{\rm P}{\rm .V}{\rm .}\int_0^{ + \infty } {\frac{{v^{ - \sigma } e^{ - xv} }}{{1 - v}}dv} .
$$
Rearranging yields
\begin{align*}
{\rm P}{\rm .V}{\rm .}\int_0^{ + \infty } {\frac{{v^{ - \sigma } e^{ - xv} }}{{1 - v}}dv} & = e^{ - x} ( - \pi i - e^{ - \pi i\sigma } \Gamma (1 - \sigma )\Gamma (\sigma , - x)) \\ & = e^{ - x} ( - \pi i + e^{ - \pi i\sigma } \sigma \Gamma (\sigma )\Gamma (\sigma , - x)).
\end{align*}
Now just apply $(8.7.3)$ and the Maclaurin series of $e^{-x}$ to get the desired power series about $0$.
