Consider the integral
$$
g(z)=\int_{-i \infty}^{i \infty} G(t) \Gamma(-t)(-z)^t d t
$$
where $G(t)$ is regular in the lalf plane $\operatorname{Re} t>0,\left[G(t) z^t / \Gamma(t+1)\right]$ approaches zero for large values of $\operatorname{Re} t$. The contour is along the imaginary axis except for a small semicircle going clockwise about the origin. Show by closing the contour with a semicircle of large radius in the half plane $\operatorname{Re} t>0$ that
$$
g(z)=2 \pi i \sum_{n=0}^{\infty} G(n)\left(\frac{z^n}{n!}\right)
$$
Hence show that
$$
\begin{aligned}
F(a, b|c| z) & =\frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n)}{\Gamma(c+n)}\left(\frac{z^n}{n!}\right) \\
& =\frac{\Gamma(c)}{2 \pi i \Gamma(a) \Gamma(b)} \int_{-i \infty}^{i \omega} \frac{\Gamma(a+t) \Gamma(b+t)}{\Gamma(c+t)} \Gamma(-t)(-z)^t d t
\end{aligned}
$$
for $\operatorname{Re} a$ and $\operatorname{Re} b$ less than zero, $a$ and $b$ not integers. Show that the contour may also be closed by a circle in the negative real plane. Deduce that
$$
\begin{aligned}
\frac{\Gamma(a) \Gamma(b)}{\Gamma(c)} F^{\prime}(a, b|c| z) & =\frac{\Gamma(a) \Gamma(a-b)}{\Gamma(a-a l}(-z)^{-a} F\left(a, 1-c+a|1-b+a| \frac{1}{z}\right) \\
& +\frac{\Gamma(b) \Gamma(b-a)}{\Gamma(b-c)}(-z)^{-b} F\left(b, 1-c+b!1 \sim a+b \left\lvert\, \frac{1}{z}\right.\right) \\
& \text { ssee Eq. (5.2.+9)]. }
\end{aligned}
$$