Help with C is Euler's constant and $\Gamma(0)=\infty$ in paper I am referring to a paper by S. Nadarajah & S. Kotz.
The notation is simple enough to understand, however i having trouble with $C$ is Euler’s constant and $\Gamma(0)=\infty$ 
by equation (2.3) I have
$$F(z)= \displaystyle\lambda\int_{0}^{\infty}y^{-1-1}exp\left(-\frac{\lambda}{y}\right)erfc\left(-\frac{z}{\sqrt{2}\sigma} y\right)~dy - 1$$
by lemma 1(defined $\displaystyle p>0 , \alpha < 0 , \arrowvert \text{arg}(c) \arrowvert < \frac{\pi}{4}$) , then 
\begin{align}
\displaystyle F(z) &= \lambda\left[\frac{1}{\lambda} + \frac{2z}{\sqrt{2\pi}\sigma}\Gamma(0)G\left(\frac{1}{2};\frac{3}{2},1,\frac{1}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) + \frac{z}{\sqrt{2\pi}\sigma}\Gamma(0)G\left(\frac{1}{2};\frac{3}{2},\frac{1}{2},1;-\frac{\lambda^2z^2}{8\sigma^2}\right) \right. \\
& \hspace{10mm} \left. + \frac{z^2 \lambda}{4\sqrt{\pi}\sigma^2}\Gamma \left(-\frac{1}{2}\right) G\left(1;2,\frac{3}{2},\frac{3}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) \right]-1
\end{align}
note that $\Gamma(-\frac{1}{2})=-2\sqrt{\pi}$ and $\Gamma(0)=\infty$
so $~~~~~~~~\displaystyle F(z) = \frac{\lambda z}{\sqrt{2}\sigma}\left[\frac{3\Gamma(0)}{\sqrt{\pi}}G\left(\frac{1}{2};\frac{3}{2},1,\frac{1}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) - \frac{\lambda z}{\sqrt{2}\sigma}G\left(1;2,
\frac{3}{2},\frac{3}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right)\right] $
but the equation is not equal to equation(2.1)
I would be really happy if someone could help me.help please
 A: Notes: 


*

*The paper referenced is unclear about the parameters $p$ and how $z$ is defined. It would seem that somehow $p \sim \Gamma(0)$. This would work if another parameter is defined, say $\alpha$, such that $p = \alpha \, \Gamma(0) \to \beta$ where $\beta$ is a finite quantity. 

*It is of interest to note that 
\begin{align}
I &= \int_{0}^{\infty} e^{-\frac{p}{x}} \, erfc(c x) \, \frac{dx}{x^{2}} \\
&= \frac{1}{p} - \frac{3 \, C \, \Gamma(0)}{\sqrt{\pi}} \, G\left(\frac{1}{2}; \frac{3}{2}, 1, \frac{1}{2}; - \frac{c^{2} p^{2}}{4} \right) - p \, C^{2} \, G\left(1; 2, \frac{3}{2}, \frac{3}{2}; - \frac{c^{2} p^{2}}{4} \right) 
\end{align}
and upon making the change of variable $u = p/x$ then performing integration by parts, then making the change of variable $t = (cp)/u$ the integral becomes
\begin{align}
I = \frac{2 \, c^{2} \, p^{2}}{\sqrt{\pi}} \, \int_{0}^{\infty} e^{- t^{2} - \frac{c p}{t}} \, \frac{dt}{t}
\end{align}



Let
\begin{align}
I &= \int_{0}^{\infty} e^{-\frac{p}{x}} \, erfc(c x) \, \frac{dx}{x^{2}} \\
\end{align}
and make the change $u = \frac{p}{x}$ to obtain
$$I = \int_{0}^{\infty} e^{-u} \, erfc\left(\frac{c p}{u}\right) \, du.$$
Now, by integration by parts, where 
$$\partial_{u} \, ercf\left(\frac{c p}{u}\right) = \frac{2 c p}{\sqrt{\pi} \, u} \, e^{- \frac{c^{2} p^{2}}{u^{2}}},$$
the integral becomes
$$I = \left[ - e^{-u} \, erfc\left(\frac{c p}{u}\right) \right]_{0}^{\infty} + \frac{2 c p}{\sqrt{\pi}} \, \int_{0}^{\infty} e^{- \frac{c^{2}p^{2}}{u^{2}} - u} \, \frac{du}{u}.$$
Since $erfc(\infty) = 0$ then the non-integral term vanishes and upon making the change $x = \frac{c p}{u}$ the resulting integral becomes
$$I = \frac{2 c^{2} p^{2}}{\sqrt{\pi}} \, \int_{0}^{\infty} e^{- x^{2} - \frac{c p}{x}} \, \frac{dx}{x}.$$
A: Look at this part of the formula given in Lemma 1: 
$$
\begin{multline}
-\frac{2cp^{\alpha+1}}{\sqrt{\pi}}\Gamma(-\alpha-1)G\left(\frac{1}{2},\frac{3}{2},\frac{3+\alpha}{2},1+\frac{\alpha}{2}; -\frac{c^2p^2}{4}\right)  \\ 
+ \frac{1}{c^\alpha\sqrt{\pi}\alpha}\Gamma\left(\frac{\alpha+1}{2}\right) G\left(-\frac{\alpha}{2},1-\frac{\alpha}{2},\frac{1}{2},\frac{1-\alpha}{2}; -\frac{c^2p^2}{4}\right)
\end{multline}
$$
We cannot substitute $\alpha$ for $-1$ in this expression because it would yield two occurences of $\Gamma(0)$. But this gives $-\infty  +\infty$ and there possibly is a limit here.
Below is an heuristic derivation of the limit at $\alpha=-1$. Substitute $\alpha$ for $-1$ except in the $\Gamma(\cdot)$:
$$
\begin{multline}
-\frac{2c}{\sqrt{\pi}}\Gamma(-\alpha-1)G\left(\frac{1}{2},\frac{3}{2},1,\frac{1}{2}; -\frac{c^2p^2}{4}\right) \\ -
\frac{c}{\sqrt{\pi}}\Gamma\left(\frac{\alpha+1}{2}\right) G\left(\frac{1}{2},\frac{3}{2},\frac{1}{2},1; -\frac{c^2p^2}{4}\right) \\
= -\frac{c}{\sqrt{\pi}}\left(2\Gamma(-\alpha-1)+\Gamma\left(\frac{\alpha+1}{2}\right) \right)G\left(\frac{1}{2},\frac{3}{2},1,\frac{1}{2}; -\frac{c^2p^2}{4}\right)
\end{multline}
$$
Now, I have not tried to know why, but it seems that $$2\Gamma(-\alpha-1)+\Gamma\left(\frac{\alpha+1}{2}\right) \to -3C \quad \text{when $\alpha \to -1$:}$$ 
> f <- function(x) 2*gamma(-x-1) + gamma((1+x)/2)
> f(-0.999999999)/3
[1] -0.5772157
> digamma(1) # this is the Euler constant
[1] -0.5772157

