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

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

