Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (4.11)

$$\begin{multline*} _{2}{{F}_{1}}\left(\tfrac{1}{2},\tfrac{\nu +1}{2};\tfrac{3}{2};\tfrac{-{{x}^{2}}}{\nu } \right)=\frac{1}{2\sqrt{\tfrac{{{x}^{2}}}{\nu }}}\left\{ {{\left( \frac{{{x}^{2}}}{v} \right)}^{\tfrac{1}{2}}}{\exp(\tfrac{-{{x}^{2}}}{\nu })}h_{\tfrac{-1}{2}}^{\tfrac{-{{x}^{2}}}{\nu }}+ \right. \\ \left( \frac{v-1}{2} \right)\left[ {{\left( \frac{{{x}^{2}}}{v} \right)}^{\tfrac{1}{2}}}{\exp(\tfrac{-{{x}^{2}}}{\nu })}h_{\tfrac{-1}{2}}^{\tfrac{-{{x}^{2}}}{\nu }}-\frac{{{\left( \frac{{{x}^{2}}}{v} \right)}^{\tfrac{1}{2}}}}{0!\frac{1}{2}} \right]+\\ \sum\limits_{i=1}^{\infty } \left\{ {{{\left( \frac{v-1}{2}+i \right)}^{2}}}\cdot \prod\limits_{j=1}^{i-1}{\left( \frac{v-1}{2}+j \right)} \right. \\ \left. \left[ {{\left( \frac{{{x}^{2}}}{v} \right)}^{\tfrac{1}{2}}}{\exp(\tfrac{-{{x}^{2}}}{\nu})}h_{\tfrac{-1}{2}}^{\tfrac{-{{x}^{2}}}{\nu }}-\sum\limits_{k=0}^{i}{{{\left( -1 \right)}^{k}}\frac{{{\left( \frac{{{x}^{2}}}{v} \right)}^{\tfrac{1}{2}+k}}}{k!\left( \frac{1}{2}+k \right)}} \right] \right\}. \end{multline*}$$ Proof:
$$\begin{align*} & _{2}{{F}_{1}}\left( \tfrac{1}{2},\tfrac{\nu +1}{2};\tfrac{3}{2};\tfrac{-{{x}^{2}}}{\nu } \right) \\ &=1-\left( \tfrac{\nu +1}{2} \right)\frac{\tfrac{-{{x}^{2}}}{2\nu }}{1!\tfrac{3}{2}}+\left( \tfrac{\nu +1}{2} \right)\left( \tfrac{\nu +3}{2} \right)\frac{\tfrac{-{{x}^{4}}}{2{{\nu }^{2}}}}{2!\tfrac{5}{2}}-\left( \tfrac{\nu +1}{2} \right)\left( \tfrac{\nu +3}{2} \right)\left( \tfrac{\nu +5}{2} \right)\frac{\tfrac{-{{x}^{4}}}{2{{\nu }^{3}}}}{3!\tfrac{7}{2}}+\cdots \\ & =\frac{1}{2\sqrt{t}}\left( \frac{{{t}^{\tfrac{1}{2}}}}{0!\tfrac{1}{2}}-b\frac{{{t}^{\tfrac{3}{2}}}}{1!\tfrac{3}{2}}+b\left( b+1 \right)\frac{{{t}^{\tfrac{5}{2}}}}{2!\tfrac{5}{2}}-b\left( b+1 \right)\left( b+2 \right)\frac{{{t}^{\tfrac{7}{2}}}}{3!\tfrac{7}{2}}+\cdots \right) \\ & =\frac{1}{2\sqrt{t}}\left\{ \left( \frac{{{t}^{\tfrac{1}{2}}}}{0!\tfrac{1}{2}}-\frac{{{t}^{\tfrac{3}{2}}}}{1!\tfrac{3}{2}}+\frac{{{t}^{\tfrac{5}{2}}}}{2!\tfrac{5}{2}} -\frac{{{t}^{\tfrac{7}{2}}}}{3!\tfrac{7}{2}}+\cdots \right) \right. \\ & \left. +\left[ -\frac{{{t}^{\tfrac{3}{2}}}}{1!\tfrac{3}{2}}\left( b-1 \right)+\frac{{{t}^{\tfrac{5}{2}}}}{2!\tfrac{5}{2}}\left[ b\left( b-1 \right)-1 \right]--\frac{{{t}^{\tfrac{7}{2}}}}{3!\tfrac{7}{2}}\left[ b\left( b+1 \right)\left( b+2 \right)-1 \right] \right] \right\} \\ & =\frac{1}{2\sqrt{t}}\left\{ {{t}^{\tfrac{1}{2}}}\exp \left( -t \right)h_{\tfrac{-1}{2}}^{-t}+\left( b-1 \right)\left[ {{t}^{\tfrac{1}{2}}}\exp \left( -t \right)h_{\tfrac{-1}{2}}^{-t}-\frac{{{t}^{\tfrac{1}{2}}}}{0!\tfrac{1}{2}} \right] \right. \\ & +{{b}^{2}}\left[ {{t}^{\tfrac{1}{2}}}\exp \left( -t \right)h_{\tfrac{-1}{2}}^{-t}-\frac{{{t}^{\tfrac{1}{2}}}}{0!\tfrac{1}{2}}+\frac{{{t}^{\tfrac{3}{2}}}}{1!\tfrac{3}{2}} \right] \\ & \left. +b{{\left( b+1 \right)}^{2}}\left[ {{t}^{\tfrac{1}{2}}}\exp \left( -t \right)h_{\tfrac{-1}{2}}^{-t}-\frac{{{t}^{\tfrac{1}{2}}}}{0!\tfrac{1}{2}}+\frac{{{t}^{\tfrac{3}{2}}}}{1!\tfrac{3}{2}} -\frac{{{t}^{\tfrac{5}{2}}}}{2!\tfrac{5}{2}} \right]+\cdots \right\}, \end{align*}$$
where $t=\frac{{{x}^{2}}}{\nu }$ and $b=\frac{\nu +1}{2}$. Changing $t$ and $b$ back to $x$ and $\nu$, we can conclude the proof.
$\square$

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