Help with C is Euler's constant and $\Gamma(0)=\infty$ in paper

Question

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

Answer 1

Notes:

1. 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.
2. 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}

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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 $$I = \frac{2 c^{2} p^{2}}{\sqrt{\pi}} \, \int_{0}^{\infty} e^{- x^{2} - \frac{c p}{x}} \, \frac{dx}{x}.$$

Answer 2

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