# §15.4 Special Cases

## §15.4(i) Elementary Functions

The following results hold for principal branches when $|z|<1$, and by analytic continuation elsewhere. Exceptions are (15.4.8) and (15.4.10), that hold for $|z|<\ifrac{\pi}{4}$, and (15.4.12), (15.4.14), and (15.4.16), that hold for $|z|<\ifrac{\pi}{2}$.

 15.4.1 $\displaystyle\mathop{F\/}\nolimits\!\left(1,1;2;z\right)$ $\displaystyle=-z^{-1}\mathop{\ln\/}\nolimits\!\left(1-z\right),$ 15.4.2 $\displaystyle\mathop{F\/}\nolimits\!\left(\tfrac{1}{2},1;\tfrac{3}{2};z^{2}\right)$ $\displaystyle=\frac{1}{2z}\mathop{\ln\/}\nolimits\!\left(\frac{1+z}{1-z}\right),$ 15.4.3 $\displaystyle\mathop{F\/}\nolimits\!\left(\tfrac{1}{2},1;\tfrac{3}{2};-z^{2}\right)$ $\displaystyle=z^{-1}\mathop{\mathrm{arctan}\/}\nolimits z,$ 15.4.4 $\displaystyle\mathop{F\/}\nolimits\!\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{% 2};z^{2}\right)$ $\displaystyle=z^{-1}\mathop{\mathrm{arcsin}\/}\nolimits z,$ 15.4.5 $\displaystyle\mathop{F\/}\nolimits\!\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{% 2};-z^{2}\right)$ $\displaystyle=z^{-1}\mathop{\ln\/}\nolimits\!\left(z+\sqrt{1+z^{2}}\right).$ 15.4.6 $\displaystyle\mathop{F\/}\nolimits\!\left(a,b;b;z\right)$ $\displaystyle=(1-z)^{-a};$

compare §15.2(ii).

 15.4.7 $\mathop{F\/}\nolimits\!\left(a,\tfrac{1}{2}+a;\tfrac{1}{2};z^{2}\right)=\tfrac% {1}{2}\left((1+z)^{-2a}+(1-z)^{-2a}\right),$
 15.4.8 $\mathop{F\/}\nolimits\!\left(a,\tfrac{1}{2}+a;\tfrac{1}{2};-{\mathop{\tan\/}% \nolimits^{2}}z\right)=(\mathop{\cos\/}\nolimits z)^{2a}\mathop{\cos\/}% \nolimits\!\left(2az\right).$
 15.4.9 $\mathop{F\/}\nolimits\!\left(a,\tfrac{1}{2}+a;\tfrac{3}{2};z^{2}\right)=\frac{% 1}{(2-4a)z}\left((1+z)^{1-2a}-(1-z)^{1-2a}\right),$
 15.4.10 $\mathop{F\/}\nolimits\!\left(a,\tfrac{1}{2}+a;\tfrac{3}{2};-{\mathop{\tan\/}% \nolimits^{2}}z\right)=(\mathop{\cos\/}\nolimits z)^{2a}\frac{\mathop{\sin\/}% \nolimits\!\left((1-2a)z\right)}{(1-2a)\mathop{\sin\/}\nolimits z}.$
 15.4.11 $\mathop{F\/}\nolimits\!\left(-a,a;\tfrac{1}{2};-z^{2}\right)=\tfrac{1}{2}\left% (\left(\sqrt{1+z^{2}}+z\right)^{2a}+\left(\sqrt{1+z^{2}}-z\right)^{2a}\right),$
 15.4.12 $\mathop{F\/}\nolimits\!\left(-a,a;\tfrac{1}{2};{\mathop{\sin\/}\nolimits^{2}}z% \right)=\mathop{\cos\/}\nolimits\!\left(2az\right).$
 15.4.13 $\mathop{F\/}\nolimits\!\left(a,1-a;\tfrac{1}{2};-z^{2}\right)=\frac{1}{2\sqrt{% 1+z^{2}}}\left(\left(\sqrt{1+z^{2}}+z\right)^{2a-1}+\left(\sqrt{1+z^{2}}-z% \right)^{2a-1}\right),$
 15.4.14 $\mathop{F\/}\nolimits\!\left(a,1-a;\tfrac{1}{2};{\mathop{\sin\/}\nolimits^{2}}% z\right)=\frac{\mathop{\cos\/}\nolimits\!\left((2a-1)z\right)}{\mathop{\cos\/}% \nolimits z}.$
 15.4.15 $\mathop{F\/}\nolimits\!\left(a,1-a;\tfrac{3}{2};-z^{2}\right)=\frac{1}{(2-4a)z% }\left(\left(\sqrt{1+z^{2}}+z\right)^{1-2a}-\left(\sqrt{1+z^{2}}-z\right)^{1-2% a}\right),$
 15.4.16 $\mathop{F\/}\nolimits\!\left(a,1-a;\tfrac{3}{2};{\mathop{\sin\/}\nolimits^{2}}% z\right)=\frac{\mathop{\sin\/}\nolimits\!\left((2a-1)z\right)}{(2a-1)\mathop{% \sin\/}\nolimits z}.$
 15.4.17 $\mathop{F\/}\nolimits\!\left(a,\tfrac{1}{2}+a;1+2a;z\right)=\left(\tfrac{1}{2}% +\tfrac{1}{2}\sqrt{1-z}\right)^{-2a},$
 15.4.18 $\mathop{F\/}\nolimits\!\left(a,\tfrac{1}{2}+a;2a;z\right)=\frac{1}{\sqrt{1-z}}% \left(\tfrac{1}{2}+\tfrac{1}{2}\sqrt{1-z}\right)^{1-2a}.$
 15.4.19 $\mathop{F\/}\nolimits\!\left(a+1,b;a;z\right)=\left(1-(1-(\ifrac{b}{a}))z% \right)(1-z)^{-1-b}.$

For an extensive list of elementary representations see Prudnikov et al. (1990, pp. 468–488).

## §15.4(ii) Argument Unity

If $\Re{(c-a-b)}>0$, then

 15.4.20 $\mathop{F\/}\nolimits\!\left(a,b;c;1\right)=\frac{\mathop{\Gamma\/}\nolimits\!% \left(c\right)\mathop{\Gamma\/}\nolimits\!\left(c-a-b\right)}{\mathop{\Gamma\/% }\nolimits\!\left(c-a\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)}.$

If $c=a+b$, then

 15.4.21 $\lim_{z\to 1-}\frac{\mathop{F\/}\nolimits\!\left(a,b;a+b;z\right)}{-\mathop{% \ln\/}\nolimits\!\left(1-z\right)}=\frac{\mathop{\Gamma\/}\nolimits\!\left(a+b% \right)}{\mathop{\Gamma\/}\nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits% \!\left(b\right)}.$

If $\Re{(c-a-b)}=0$ and $c\neq a+b$, then

 15.4.22 $\lim_{z\to 1-}(1-z)^{a+b-c}\left(\mathop{F\/}\nolimits\!\left(a,b;c;z\right)-% \frac{\mathop{\Gamma\/}\nolimits\!\left(c\right)\mathop{\Gamma\/}\nolimits\!% \left(c-a-b\right)}{\mathop{\Gamma\/}\nolimits\!\left(c-a\right)\mathop{\Gamma% \/}\nolimits\!\left(c-b\right)}\right)=\frac{\mathop{\Gamma\/}\nolimits\!\left% (c\right)\mathop{\Gamma\/}\nolimits\!\left(a+b-c\right)}{\mathop{\Gamma\/}% \nolimits\!\left(a\right)\mathop{\Gamma\/}\nolimits\!\left(b\right)}.$

If $\Re{(c-a-b)}<0$, then

 15.4.23 $\lim_{z\to 1-}\frac{\mathop{F\/}\nolimits\!\left(a,b;c;z\right)}{(1-z)^{c-a-b}% }=\frac{\mathop{\Gamma\/}\nolimits\!\left(c\right)\mathop{\Gamma\/}\nolimits\!% \left(a+b-c\right)}{\mathop{\Gamma\/}\nolimits\!\left(a\right)\mathop{\Gamma\/% }\nolimits\!\left(b\right)}.$

### Chu–Vandermonde Identity

 15.4.24 $\mathop{F\/}\nolimits\!\left(-n,b;c;1\right)=\frac{{\left(c-b\right)_{n}}}{{% \left(c\right)_{n}}},$ $n=0,1,2,\dots$.

### Dougall’s Bilateral Sum

This is a generalization of (15.4.20). If $a,b$ are not integers and $\Re{(c+d-a-b)}>1$, then

 15.4.25 $\sum_{n=-\infty}^{\infty}\frac{\mathop{\Gamma\/}\nolimits\!\left(a+n\right)% \mathop{\Gamma\/}\nolimits\!\left(b+n\right)}{\mathop{\Gamma\/}\nolimits\!% \left(c+n\right)\mathop{\Gamma\/}\nolimits\!\left(d+n\right)}=\frac{\pi^{2}}{% \mathop{\sin\/}\nolimits\!\left(\pi a\right)\mathop{\sin\/}\nolimits\!\left(% \pi b\right)}\*\frac{\mathop{\Gamma\/}\nolimits\!\left(c+d-a-b-1\right)}{% \mathop{\Gamma\/}\nolimits\!\left(c-a\right)\mathop{\Gamma\/}\nolimits\!\left(% d-a\right)\mathop{\Gamma\/}\nolimits\!\left(c-b\right)\mathop{\Gamma\/}% \nolimits\!\left(d-b\right)}.$

## §15.4(iii) Other Arguments

 15.4.26 $\mathop{F\/}\nolimits\!\left(a,b;a-b+1;-1\right)=\frac{\mathop{\Gamma\/}% \nolimits\!\left(a-b+1\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}a+1% \right)}{\mathop{\Gamma\/}\nolimits\!\left(a+1\right)\mathop{\Gamma\/}% \nolimits\!\left(\tfrac{1}{2}a-b+1\right)}.$
 15.4.27 $\mathop{F\/}\nolimits\!\left(1,a;a+1;-1\right)=\tfrac{1}{2}a\left(\mathop{\psi% \/}\nolimits\!\left(\tfrac{1}{2}a+\tfrac{1}{2}\right)-\mathop{\psi\/}\nolimits% \!\left(\tfrac{1}{2}a\right)\right).$
 15.4.28 $\mathop{F\/}\nolimits\!\left(a,b;\tfrac{1}{2}a+\tfrac{1}{2}b+\tfrac{1}{2};% \tfrac{1}{2}\right)=\sqrt{\pi}\frac{\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1% }{2}a+\tfrac{1}{2}b+\tfrac{1}{2}\right)}{\mathop{\Gamma\/}\nolimits\!\left(% \tfrac{1}{2}a+\tfrac{1}{2}\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2% }b+\tfrac{1}{2}\right)}.$
 15.4.29 $\mathop{F\/}\nolimits\!\left(a,b;\tfrac{1}{2}a+\tfrac{1}{2}b+1;\tfrac{1}{2}% \right)=\frac{2\sqrt{\pi}}{a-b}\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}a% +\tfrac{1}{2}b+1\right)\*\left(\frac{1}{\mathop{\Gamma\/}\nolimits\!\left(% \tfrac{1}{2}a\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}b+\tfrac{1}{% 2}\right)}-\frac{1}{\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}a+\tfrac{1}{% 2}\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{1}{2}b\right)}\right).$
 15.4.30 $\mathop{F\/}\nolimits\!\left(a,1-a;b;\tfrac{1}{2}\right)=\frac{2^{1-b}\sqrt{% \pi}\mathop{\Gamma\/}\nolimits\!\left(b\right)}{\mathop{\Gamma\/}\nolimits\!% \left(\tfrac{1}{2}a+\tfrac{1}{2}b\right)\mathop{\Gamma\/}\nolimits\!\left(% \tfrac{1}{2}b-\tfrac{1}{2}a+\tfrac{1}{2}\right)}.$
 15.4.31 $\mathop{F\/}\nolimits\!\left(a,\tfrac{1}{2}+a;\tfrac{3}{2}-2a;-\tfrac{1}{3}% \right)=\left(\frac{8}{9}\right)^{-2a}\frac{\mathop{\Gamma\/}\nolimits\!\left(% \tfrac{4}{3}\right)\mathop{\Gamma\/}\nolimits\!\left(\tfrac{3}{2}-2a\right)}{% \mathop{\Gamma\/}\nolimits\!\left(\tfrac{3}{2}\right)\mathop{\Gamma\/}% \nolimits\!\left(\tfrac{4}{3}-2a\right)}.$
 15.4.32 $\mathop{F\/}\nolimits\!\left(a,\tfrac{1}{2}+a;\tfrac{5}{6}+\tfrac{2}{3}a;% \tfrac{1}{9}\right)=\sqrt{\pi}\left(\frac{3}{4}\right)^{a}\frac{\mathop{\Gamma% \/}\nolimits\!\left(\tfrac{5}{6}+\tfrac{2}{3}a\right)}{\mathop{\Gamma\/}% \nolimits\!\left(\tfrac{1}{2}+\tfrac{1}{3}a\right)\mathop{\Gamma\/}\nolimits\!% \left(\tfrac{5}{6}+\tfrac{1}{3}a\right)}.$
 15.4.33 $\mathop{F\/}\nolimits\!\left(3a,\tfrac{1}{3}+a;\tfrac{2}{3}+2a;e^{\ifrac{% \mathrm{i}\pi}{3}}\right)=\sqrt{\pi}e^{\ifrac{\mathrm{i}\pi a}{2}}\left(\frac{% 16}{27}\right)^{(3a+1)/6}\frac{\mathop{\Gamma\/}\nolimits\!\left(\frac{5}{6}+a% \right)}{\mathop{\Gamma\/}\nolimits\!\left(\frac{2}{3}+a\right)\mathop{\Gamma% \/}\nolimits\!\left(\frac{2}{3}\right)}.$