# §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 $\displaystyle F\left(1,1;2;z\right)$ $\displaystyle=-z^{-1}\ln\left(1-z\right),$ 15.4.2 $\displaystyle F\left(\tfrac{1}{2},1;\tfrac{3}{2};z^{2}\right)$ $\displaystyle=\frac{1}{2z}\ln\left(\frac{1+z}{1-z}\right),$ 15.4.3 $\displaystyle F\left(\tfrac{1}{2},1;\tfrac{3}{2};-z^{2}\right)$ $\displaystyle=z^{-1}\operatorname{arctan}z,$ 15.4.4 $\displaystyle F\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2};z^{2}\right)$ $\displaystyle=z^{-1}\operatorname{arcsin}z,$ 15.4.5 $\displaystyle F\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2};-z^{2}\right)$ $\displaystyle=z^{-1}\ln\left(z+\sqrt{1+z^{2}}\right).$
 15.4.6 $\displaystyle F\left(a,b;a;z\right)$ $\displaystyle=(1-z)^{-b},$ $\displaystyle F\left(a,b;b;z\right)$ $\displaystyle=(1-z)^{-a},$ ⓘ Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $z$: complex variable, $a$: real or complex parameter and $b$: real or complex parameter Notes: See §15.2(ii) for treatment of the case when the third parameter is a nonpositive integer. Referenced by: §10.22(iv), §15.2(ii), §15.4(i), Erratum (V1.0.16) for Subsections 15.4(i), 15.4(ii) Permalink: http://dlmf.nist.gov/15.4.E6 Encodings: TeX, TeX, pMML, pMML, png, png Addition (effective with 1.0.16): This equation was expanded by adding the formula $F\left(a,b;a;z\right)=(1-z)^{-b}$. See also: Annotations for §15.4(i), §15.4 and Ch.15

where the limit interpretation (15.2.6), rather than (15.2.5), has to be taken when the third parameter is a nonpositive integer. See the final paragraph in §15.2(ii).

 15.4.7 $F\left(a,\tfrac{1}{2}+a;\tfrac{1}{2};z^{2}\right)=\tfrac{1}{2}\left((1+z)^{-2a% }+(1-z)^{-2a}\right),$
 15.4.8 $F\left(a,\tfrac{1}{2}+a;\tfrac{1}{2};-{\tan}^{2}z\right)=(\cos z)^{2a}\cos% \left(2az\right).$
 15.4.9 $F\left(a,\tfrac{1}{2}+a;\tfrac{3}{2};z^{2}\right)=\frac{1}{(2-4a)z}\left((1+z)% ^{1-2a}-(1-z)^{1-2a}\right),$
 15.4.10 $F\left(a,\tfrac{1}{2}+a;\tfrac{3}{2};-{\tan}^{2}z\right)=(\cos z)^{2a}\frac{% \sin\left((1-2a)z\right)}{(1-2a)\sin z}.$
 15.4.11 $F\left(-a,a;\tfrac{1}{2};-z^{2}\right)=\tfrac{1}{2}\left(\left(\sqrt{1+z^{2}}+% z\right)^{2a}+\left(\sqrt{1+z^{2}}-z\right)^{2a}\right),$
 15.4.12 $F\left(-a,a;\tfrac{1}{2};{\sin}^{2}z\right)=\cos\left(2az\right).$
 15.4.13 $F\left(a,1-a;\tfrac{1}{2};-z^{2}\right)=\frac{1}{2\sqrt{1+z^{2}}}\left(\left(% \sqrt{1+z^{2}}+z\right)^{2a-1}+\left(\sqrt{1+z^{2}}-z\right)^{2a-1}\right),$
 15.4.14 $F\left(a,1-a;\tfrac{1}{2};{\sin}^{2}z\right)=\frac{\cos\left((2a-1)z\right)}{% \cos z}.$
 15.4.15 $F\left(a,1-a;\tfrac{3}{2};-z^{2}\right)=\frac{1}{(2-4a)z}\left(\left(\sqrt{1+z% ^{2}}+z\right)^{1-2a}-\left(\sqrt{1+z^{2}}-z\right)^{1-2a}\right),$
 15.4.16 $F\left(a,1-a;\tfrac{3}{2};{\sin}^{2}z\right)=\frac{\sin\left((2a-1)z\right)}{(% 2a-1)\sin z}.$
 15.4.17 $F\left(a,\tfrac{1}{2}+a;1+2a;z\right)=\left(\tfrac{1}{2}+\tfrac{1}{2}\sqrt{1-z% }\right)^{-2a},$ ⓘ Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $z$: complex variable and $a$: real or complex parameter Notes: See §15.2(ii) for treatment of the case when the third parameter is a nonpositive integer. Referenced by: §15.4(i), §15.4(i) Permalink: http://dlmf.nist.gov/15.4.E17 Encodings: TeX, pMML, png See also: Annotations for §15.4(i), §15.4 and Ch.15
 15.4.18 $F\left(a,\tfrac{1}{2}+a;2a;z\right)=\frac{1}{\sqrt{1-z}}\left(\tfrac{1}{2}+% \tfrac{1}{2}\sqrt{1-z}\right)^{1-2a},$ ⓘ Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $z$: complex variable and $a$: real or complex parameter Notes: See §15.2(ii) for treatment of the case when the third parameter is a nonpositive integer. Referenced by: §15.4(i), §15.4(i) Permalink: http://dlmf.nist.gov/15.4.E18 Encodings: TeX, pMML, png See also: Annotations for §15.4(i), §15.4 and Ch.15
 15.4.19 $F\left(a+1,b;a;z\right)=\left(1-(1-(\ifrac{b}{a}))z\right)(1-z)^{-1-b}.$ ⓘ Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $z$: complex variable, $a$: real or complex parameter and $b$: real or complex parameter Notes: See §15.2(ii) for treatment of the case when the third parameter is a nonpositive integer. Referenced by: §15.4(i), §15.4(i) Permalink: http://dlmf.nist.gov/15.4.E19 Encodings: TeX, pMML, png See also: Annotations for §15.4(i), §15.4 and Ch.15

In (15.4.17), (15.4.18) and (15.4.19) when the third entry is a nonpositive integer one has to use the limit interpretation (15.2.6), rather than (15.2.5). Compare the final paragraph in §15.2(ii).

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

## §15.4(ii) Argument Unity

If $\Re\left(c-a-b\right)>0$, then

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

If $c=a+b$, then

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

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

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

If $\Re\left(c-a-b\right)<0$, then

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

### Chu–Vandermonde Identity

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

### Dougall’s Bilateral Sum

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

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

## §15.4(iii) Other Arguments

 15.4.26 $F\left(a,b;a-b+1;-1\right)=\frac{\Gamma\left(a-b+1\right)\Gamma\left(\tfrac{1}% {2}a+1\right)}{\Gamma\left(a+1\right)\Gamma\left(\tfrac{1}{2}a-b+1\right)}.$
 15.4.27 $F\left(1,a;a+1;-1\right)=2^{-a}F\left(a,a;a+1;\tfrac{1}{2}\right)=\tfrac{1}{2}% a\left(\psi\left(\tfrac{1}{2}a+\tfrac{1}{2}\right)-\psi\left(\tfrac{1}{2}a% \right)\right).$ ⓘ Symbols: $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $\psi\left(\NVar{z}\right)$: psi (or digamma) function and $a$: real or complex parameter Source: Intermediate equality follows from (15.8.1). Referenced by: §15.4(iii), Erratum (V1.1.2) for Additions Permalink: http://dlmf.nist.gov/15.4.E27 Encodings: TeX, pMML, png Addition (effective with 1.1.2): Intermediate equality was added. See also: Annotations for §15.4(iii), §15.4 and Ch.15
 15.4.28 $F\left(a,b;\tfrac{1}{2}a+\tfrac{1}{2}b+\tfrac{1}{2};\tfrac{1}{2}\right)=\sqrt{% \pi}\frac{\Gamma\left(\tfrac{1}{2}a+\tfrac{1}{2}b+\tfrac{1}{2}\right)}{\Gamma% \left(\tfrac{1}{2}a+\tfrac{1}{2}\right)\Gamma\left(\tfrac{1}{2}b+\tfrac{1}{2}% \right)}.$
 15.4.29 $F\left(a,b;\tfrac{1}{2}a+\tfrac{1}{2}b+1;\tfrac{1}{2}\right)=\frac{2\sqrt{\pi}% }{a-b}\Gamma\left(\tfrac{1}{2}a+\tfrac{1}{2}b+1\right)\*\left(\frac{1}{\Gamma% \left(\tfrac{1}{2}a\right)\Gamma\left(\tfrac{1}{2}b+\tfrac{1}{2}\right)}-\frac% {1}{\Gamma\left(\tfrac{1}{2}a+\tfrac{1}{2}\right)\Gamma\left(\tfrac{1}{2}b% \right)}\right).$
 15.4.30 $F\left(a,1-a;b;\tfrac{1}{2}\right)=\frac{2^{1-b}\sqrt{\pi}\Gamma\left(b\right)% }{\Gamma\left(\tfrac{1}{2}a+\tfrac{1}{2}b\right)\Gamma\left(\tfrac{1}{2}b-% \tfrac{1}{2}a+\tfrac{1}{2}\right)}.$
 15.4.31 $F\left(a,\tfrac{1}{2}+a;\tfrac{3}{2}-2a;-\tfrac{1}{3}\right)=\left(\frac{8}{9}% \right)^{-2a}\frac{\Gamma\left(\tfrac{4}{3}\right)\Gamma\left(\tfrac{3}{2}-2a% \right)}{\Gamma\left(\tfrac{3}{2}\right)\Gamma\left(\tfrac{4}{3}-2a\right)}.$
 15.4.32 $F\left(a,\tfrac{1}{2}+a;\tfrac{5}{6}+\tfrac{2}{3}a;\tfrac{1}{9}\right)=\sqrt{% \pi}\left(\frac{3}{4}\right)^{a}\frac{\Gamma\left(\tfrac{5}{6}+\tfrac{2}{3}a% \right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{3}a\right)\Gamma\left(\tfrac{5}{6}% +\tfrac{1}{3}a\right)}.$
 15.4.33 $F\left(3a,\tfrac{1}{3}+a;\tfrac{2}{3}+2a;{\mathrm{e}}^{\ifrac{\mathrm{i}\pi}{3% }}\right)=\sqrt{\pi}{\mathrm{e}}^{\ifrac{\mathrm{i}\pi a}{2}}\left(\frac{16}{2% 7}\right)^{(3a+1)/6}\frac{\Gamma\left(\frac{5}{6}+a\right)}{\Gamma\left(\frac{% 2}{3}+a\right)\Gamma\left(\frac{2}{3}\right)},$ ⓘ Symbols: $\Gamma\left(\NVar{z}\right)$: gamma function, $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit and $a$: real or complex parameter Notes: See §15.2(ii) for treatment of the case when the third parameter is a nonpositive integer. Referenced by: §15.4(iii), §15.4(iii) Permalink: http://dlmf.nist.gov/15.4.E33 Encodings: TeX, pMML, png See also: Annotations for §15.4(iii), §15.4 and Ch.15
 15.4.34 $F\left(3a,a;2a;{\mathrm{e}}^{\ifrac{\mathrm{i}\pi}{3}}\right)=\sqrt{\pi}{% \mathrm{e}}^{\ifrac{\mathrm{i}\pi a}{2}}\frac{2^{2a}\Gamma\left(\frac{1}{2}+a% \right)}{3^{(3a+1)/2}}\left(\frac{1}{\Gamma\left(\frac{1}{3}+a\right)\Gamma% \left(\frac{2}{3}\right)}+\frac{1}{\Gamma\left(\frac{2}{3}+a\right)\Gamma\left% (\frac{1}{3}\right)}\right),$ ⓘ Symbols: $\Gamma\left(\NVar{z}\right)$: gamma function, $F\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ or $F\left({\NVar{a},\NVar{b}\atop\NVar{c}};\NVar{z}\right)$: $={{}_{2}F_{1}}\left(\NVar{a},\NVar{b};\NVar{c};\NVar{z}\right)$ Gauss’ hypergeometric function, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit and $a$: real or complex parameter Source: From §15.5(ii) one obtains that $w(a)={\mathrm{e}}^{-\ifrac{\mathrm{i}\pi a}{2}}2^{-2a}3^{(3a+1)/2}F\left(3a,a;% 2a;{\mathrm{e}}^{\ifrac{\mathrm{i}\pi}{3}}\right)\Gamma\left(\frac{1}{3}+a% \right)/\Gamma\left(\frac{1}{2}+a\right)$ is a solution of the difference equation $(3a+5)w(a+2)-6(a+1)w(a+1)+(3a+1)w(a)=0$. Two linearly independent solutions of this difference equation are $w_{1}(a)=1$ and $w_{2}(a)=\Gamma\left(\frac{1}{3}+a\right)/\Gamma\left(\frac{2}{3}+a\right)$. Combining (15.6.1) with Laplace’s method (see §2.4(iii)) we obtain that $w(a)\sim\sqrt{\pi}\left(\frac{1}{\Gamma\left(\frac{2}{3}\right)}+\frac{a^{-1/3% }}{\Gamma\left(\frac{1}{3}\right)}\right)$, as $a\to\infty$. Hence, $w(a)=\sqrt{\pi}\left(\frac{1}{\Gamma\left(\frac{2}{3}\right)}+\frac{w_{2}(a)}{% \Gamma\left(\frac{1}{3}\right)}\right)$. Notes: See §15.2(ii) for treatment of the case when the third parameter is a nonpositive integer. Referenced by: §15.4(iii), Erratum (V1.1.2) for Additions Permalink: http://dlmf.nist.gov/15.4.E34 Encodings: TeX, pMML, png Addition (effective with 1.1.2): This equation was added. See also: Annotations for §15.4(iii), §15.4 and Ch.15

where the limit interpretation (15.2.6), rather than (15.2.5), has to be taken when in (15.4.33) $a=-\frac{1}{3},-\frac{4}{3},-\frac{7}{3},\dots$, and in (15.4.34) $a=0,-1,-2,\dots$. Compare the final paragraph in §15.2(ii).