I would like to get the Asymptotic Value of the following Meijer G-function when z tends to zero.
I found that I can calculate the residue at the closest pole to the contour from the right since $z \rightarrow o$, which in my case is a 2nd order pole $s= -\alpha$, where $\alpha$ is a positive real number
\begin{align} f(z)=& ~G^{3,1}_{2,3}\left(z\bigg|{-\alpha,1-\alpha\atop 0,-\alpha,-\alpha}\right)% = \frac{1}{2 \pi i} \int_C \frac{\Gamma^2(-\alpha-s)\Gamma(-s)}{1}\frac{\Gamma(1+\alpha+s) }{\Gamma(1-\alpha-s)} z^s ds \end{align}
This leads to having two poles to the right of the contour, simple pole at s=0 and 2nd order pole at $s= -\alpha$, then the closest pole is $s= -\alpha$. So, to calculate the residue at the 2nd order pole, I tried the following: \begin{align} \bar{f}(z)=& \operatorname*{Res}_{s=-\alpha}\left(\frac{\Gamma(s)\Gamma(s+1-\alpha)\Gamma(1-s)}{s}z^{-s}\right)\\% =& \frac{1}{(2-1)!} \lim_{s\rightarrow -\alpha} \frac{d}{ds} \big[(s+\alpha)^2~\frac{\Gamma^2(-\alpha-s)\Gamma(-s)}{1}\frac{\Gamma(1+\alpha+s) }{\Gamma(1-\alpha-s)} z^s\big] \end{align}
I do not know how to proceed from here. Thank you in advance
The integrand is $\Gamma^2(-\alpha - s) g(s)$, where $g$ is regular at $-\alpha \not \in \mathbb Z$. Then $$-\operatorname*{Res}_{s = -\alpha} \Gamma^2(-\alpha - s) g(s) = -\operatorname*{Res}_{s = -\alpha} \frac {\pi^2 g(s)} {\Gamma^2(1 + \alpha + s)} \, \csc^2 \pi (\alpha + s) = \\ -\frac d {ds} \frac {g(s)} {\Gamma^2(1 + \alpha + s)} \bigg\rvert_{s = -\alpha} = -\Gamma(\alpha) z^{-\alpha} (\ln z - \psi(\alpha)).$$