Für ${\displaystyle {\text{Re}}(b)>0\,}$ und ${\displaystyle {\text{Re}}(c-a-b)>0\,}$ ist

${\displaystyle I:=\int _{0}^{1}x^{b-1}\,(1-x)^{c-b-1}\,(1-x)^{-a}\,dx=B(b,c-a-b)={\frac {\Gamma (b)\,\Gamma (c-a-b)}{\Gamma (c-a)}}}$.

Unter Verwendung der Binomialreihe ${\displaystyle (1-x)^{-a}=\sum _{k=0}^{\infty }{a-1+k \choose k}\,x^{k}}$ ist

${\displaystyle I=\sum _{k=0}^{\infty }{a-1+k \choose k}\int _{0}^{1}x^{b+k-1}\,(1-x)^{c-b-1}\,dx}$

${\displaystyle =\sum _{k=0}^{\infty }{\frac {(a-1+k)!}{(a-1)!\,k!}}\,B(b+k,c-b)=\sum _{k=0}^{\infty }{\frac {\Gamma (a+k)}{\Gamma (a)\,k!}}\,{\frac {\Gamma (b+k)\,\Gamma (c-b)}{\Gamma (c+k)}}}$.

Also ist ${\displaystyle \sum _{k=0}^{\infty }{\frac {\Gamma (a+k)\,\Gamma (b+k)}{k!\,\Gamma (c+k)}}={\frac {\Gamma (a)\,\Gamma (b)\,\Gamma (c-a-b)}{\Gamma (c-a)\,\Gamma (c-b)}}}$.

Wegen ${\displaystyle {\frac {\Gamma (a+k)\,\Gamma (b+k)}{k!\,\Gamma (c+k)}}\sim {\frac {k^{a}\,k^{b}}{k\,\,k^{c}}}=k^{-1-(c-a-b)}}$

konvergiert die Reihe ${\displaystyle \sum _{k=0}^{\infty }{\frac {\Gamma (a+k)\,\Gamma (b+k)}{k!\,\Gamma (c+k)}}}$ auch unter der bloßen Bedingung ${\displaystyle {\text{Re}}(c-a-b)>0\,}$.

Und wird sich auch ohne die Bedingung ${\displaystyle {\text{Re}}(b)>0\,}$ durch ${\displaystyle {\frac {\Gamma (a)\,\Gamma (b)\,\Gamma (c-a-b)}{\Gamma (c-a)\,\Gamma (c-b)}}}$ analytisch fortsetzen lassen.

Diese Formel ist äquivalent zur Formel ${\displaystyle \sum _{k=0}^{\infty }{\frac {\Gamma (a+k)\,\Gamma (b+k)}{k!\,\Gamma (c+k)}}={\frac {\Gamma (a)\,\Gamma (b)\,\Gamma (c-a-b)}{\Gamma (c-a)\,\Gamma (c-b)}}}$ für ${\displaystyle {\text{Re}}(c-a-b)>0\,}$.

Ersetze hierzu ${\displaystyle \Gamma (a+k)\,}$ durch ${\displaystyle {\frac {(-1)^{k}\,\Gamma (a)\,\Gamma (1-a)}{\Gamma (1-a-k)}}}$ und analog ${\displaystyle \Gamma (b+k)\,}$ durch ${\displaystyle {\frac {(-1)^{k}\,\Gamma (b)\,\Gamma (1-b)}{\Gamma (1-b-k)}}}$.

${\displaystyle \sum _{k=0}^{\infty }{\frac {\Gamma (a)\,\Gamma (1-a)\,\Gamma (b)\,\Gamma (1-b)}{k!\,\Gamma (c+k)\,\Gamma (1-a-k)\,\Gamma (1-b-k)}}={\frac {\Gamma (a)\,\Gamma (b)\,\Gamma (c-a-b)}{\Gamma (c-a)\,\Gamma (c-b)}}}$

gleichbedeutend mit ${\displaystyle \sum _{k=0}^{\infty }{\frac {(-a)!\,(-b)!}{k!\,(c-1+k)!\,(-a-k)!\,(-b-k)!}}={\frac {(c-1-a-b)!}{(c-1-a)!\,(c-1-b)!}}}$

Führe nun die Umbennenung ${\displaystyle (a,b,c)=(-\beta ,-\gamma ,\alpha +1)\,}$ durch.

${\displaystyle \sum _{k=0}^{\infty }{\frac {\beta !\,\gamma !}{k!\,(\alpha +k)!\,(\beta -k)!\,(\gamma -k)!}}={\frac {(\alpha +\beta +\gamma )!}{(\alpha +\beta )!\,(\alpha +\gamma )!}}}$

Und ${\displaystyle {\text{Re}}(c-a-b)>0\,}$ ist äquivalent zu ${\displaystyle {\text{Re}}(\alpha +1+\beta +\gamma )>0\,}$ bzw. ${\displaystyle {\text{Re}}(\alpha +\beta +\gamma )>-1\,}$.

Nach der Formel von Dixon ist ${\displaystyle \sum _{n=0}^{\infty }{\frac {\Gamma (a+n)\,\Gamma (b+n)}{n!\,\Gamma (1+a-b+n)}}\,\left[{\frac {\Gamma (c+n)}{\Gamma (c)}}\,{\frac {\Gamma \left(1+{\frac {a}{2}}-c\right)\,\Gamma (1+a-b-c)}{\Gamma \left(1+a+n-c\right)\,\Gamma \left(1+{\frac {a}{2}}-b-c\right)}}\right]={\frac {\Gamma \left({\frac {a}{2}}\right)\,\Gamma (b)}{2\,\Gamma \left(1+{\frac {a}{2}}-b\right)}}}$.

Für jedes ${\displaystyle n\in \mathbb {N} }$ geht der Ausdruck in eckigen Klammern gegen ${\displaystyle (-1)^{n}\,}$ wenn man ${\displaystyle c\to -\infty \,}$ gehen lässt.

Denn es gilt ${\displaystyle {\frac {\Gamma (c+n)}{\Gamma (c)}}\sim c^{n}\,}$ und ${\displaystyle {\frac {\Gamma \left(1+{\frac {a}{2}}-c\right)\,\Gamma (1+a-b-c)}{\Gamma \left(1+a+n-c\right)\,\Gamma \left(1+{\frac {a}{2}}-b-c\right)}}\sim {\frac {(-c)^{\left(1+{\frac {a}{2}}\right)+(1+a-b)}}{(-c)^{(1+a+n)-\left(1+{\frac {a}{2}}-b\right)}}}={\frac {1}{(-c)^{n}}}}$ für ${\displaystyle c\to -\infty \,}$.

Betrachte die Funktion ${\displaystyle f(z)=\pi \cot \pi z\,\,{\frac {\Gamma (a+z)\,\Gamma (b+z)}{\Gamma (c+z)\,\Gamma (d+z)}}}$.

${\displaystyle \cot \pi z\,}$ besitzt einfache Polstellen bei ${\displaystyle z=0,\pm 1,\pm 2,...\,}$

${\displaystyle \Gamma (a+z)\,}$ bei ${\displaystyle z=-a\,,\;-a-1\,,\;-a-2\,,\;...\,}$

${\displaystyle \Gamma (b+z)\,}$ bei ${\displaystyle z=-b\,,\;-b-1\,,\;-b-2\,,\;...\,}$

Es gibt eine Folge ${\displaystyle (\gamma _{N})\,}$ von Quadraten mit ${\displaystyle {\text{dist}}(\gamma _{N},0)\to \infty \,}$, so dass ${\displaystyle \oint _{\gamma _{N}}f\,dz}$ gegen null geht.

Also muss die Summe aller Residuen von ${\displaystyle f\,}$ gleich null sein.

${\displaystyle \forall n\in \mathbb {Z} }$ ist ${\displaystyle {\text{res}}(f,n)={\frac {\Gamma (a+n)\,\Gamma (b+n)}{\Gamma (c+n)\,\Gamma (d+n)}}}$

und ${\displaystyle \forall n\in \mathbb {Z} ^{\geq 0}}$ ist ${\displaystyle {\text{res}}(f,-a-n)=\underbrace {\pi \cot(-a\pi -n\pi )} _{-\pi \,\cot a\pi }\,{\frac {{\frac {(-1)^{n}}{n!}}\,\Gamma (b-a-n)}{\Gamma (c-a-n)\,\Gamma (d-a-n)}}}$.

Nach der Formel ${\displaystyle \Gamma (x-n)={\frac {(-1)^{n}\,\Gamma (x)\,\Gamma (1-x)}{\Gamma (1-x+n)}}={\frac {(-1)^{n}}{\Gamma (1-x+n)}}\,{\frac {\pi }{\sin \pi x}}}$

ist das ${\displaystyle -\pi \cot a\pi \,{\frac {{\frac {(-1)^{n}}{n!}}\,{\frac {(-1)^{n}}{\Gamma (1+a-b+n)}}\,{\frac {\pi }{\sin(b-a)\pi }}}{{\frac {(-1)^{n}\,\Gamma (c-a)\,\Gamma (1+a-c)}{\Gamma (1+a-c+n)}}\,{\frac {(-1)^{n}\,\Gamma (d-a)\,\Gamma (1+a-d)}{\Gamma (1+a-d+n)}}}}}$

${\displaystyle =-\pi ^{2}{\frac {\cot a\pi }{\sin(b-a)\pi }}\,{\frac {1}{\Gamma (c-a)\,\Gamma (1+a-c)\,\Gamma (d-a)\,\Gamma (1+a-d)}}\,{\frac {\Gamma (1+a-c+n)\,\Gamma (1+a-d+n)}{n!\,\;\Gamma (1+a-b+n)}}}$.

Nach der Gaußschen Formel ${\displaystyle \sum _{n=0}^{\infty }{\frac {\Gamma (a+n)\,\Gamma (b+n)}{n!\,\;\Gamma (c+n)}}={\frac {\Gamma (a)\,\Gamma (b)\,\Gamma (c-a-b)}{\Gamma (c-a)\,\Gamma (c-b)}}}$ ist

${\displaystyle \sum _{n=0}^{\infty }{\frac {\Gamma (1+a-c+n)\,\Gamma (1+a-d+n)}{n!\,\;\Gamma (1+a-b+n)}}={\frac {\Gamma (1+a-c)\,\Gamma (1+a-d)\,\Gamma (c+d-a-b-1)}{\Gamma (c-b)\,\Gamma (d-b)}}}$.

Also ist ${\displaystyle \sum _{n=0}^{\infty }{\text{res}}(f,-a-n)=-{\frac {\pi ^{2}\,\cot a\pi }{\sin(b-a)\pi }}\,{\frac {\Gamma (c+d-a-b-1)}{\Gamma (c-a)\,\Gamma (d-a)\,\Gamma (c-b)\,\Gamma (d-b)}}}$.

Vertauscht man die Rollen von ${\displaystyle a\,}$ und ${\displaystyle b\,}$, so ist

${\displaystyle \sum _{n=0}^{\infty }{\text{res}}(f,-b-n)={\frac {\pi ^{2}\,\cot b\pi }{\sin(b-a)\pi }}\,{\frac {\Gamma (c+d-a-b-1)}{\Gamma (c-a)\,\Gamma (d-a)\,\Gamma (c-b)\,\Gamma (d-b)}}}$.

Somit ist ${\displaystyle \sum _{n=0}^{\infty }{\text{res}}(f,-a-n)+\sum _{n=0}^{\infty }{\text{res}}(f,-b-n)=-\pi ^{2}\,{\frac {\cot a\pi -\cot b\pi }{\sin(b-a)\pi }}\,{\frac {\Gamma (c+d-a-b-1)}{\Gamma (c-a)\,\Gamma (d-a)\,\Gamma (c-b)\,\Gamma (d-b)}}}$,

wobei ${\displaystyle {\frac {\cot a\pi -\cot b\pi }{\sin(b-a)\pi }}={\frac {{\frac {\cos a\pi }{\sin a\pi }}-{\frac {\cos b\pi }{\sin b\pi }}}{\sin(b-a)\pi }}={\frac {1}{\sin a\pi \,\sin b\pi }}\,{\frac {\sin b\pi \,\cos a\pi -\cos b\pi \,\sin a\pi }{\sin(b-a)\pi }}={\frac {1}{\sin a\pi \,\sin b\pi }}}$ ist.

Da die Summe aller Residuen null ergibt, ist also

${\displaystyle \sum _{n=-\infty }^{\infty }{\frac {\Gamma (a+n)\,\Gamma (b+n)}{\Gamma (c+n)\,\Gamma (d+n)}}={\frac {\pi ^{2}}{\sin a\pi \,\sin b\pi }}\,{\frac {\Gamma (c+d-a-b-1)}{\Gamma (c-a)\,\Gamma (d-a)\,\Gamma (c-b)\,\Gamma (d-b)}}}$.

In der Dougall'schen Formel ${\displaystyle \sum _{n=-\infty }^{\infty }{\frac {\Gamma (c+n)\,\Gamma (d+n)}{\Gamma (a+n)\,\Gamma (b+n)}}={\frac {\pi ^{2}}{\sin c\pi \,\sin d\pi }}\,{\frac {\Gamma (a+b-c-d-1)}{\Gamma (a-c)\,\Gamma (a-d)\,\Gamma (b-c)\,\Gamma (b-d)}}}$

für ${\displaystyle {\text{Re}}(c+d-a-b)>1\,}$ ersetze ${\displaystyle \Gamma (c+n)={\frac {(-1)^{n}\,\Gamma (c)\,\Gamma (1-c)}{\Gamma (1-c-n)}}}$ und ${\displaystyle \Gamma (d+n)={\frac {(-1)^{n}\,\Gamma (d)\,\Gamma (1-d)}{\Gamma (1-d-n)}}}$.

${\displaystyle \sum _{n=-\infty }^{\infty }{\frac {\Gamma (c)\,\Gamma (1-c)\,\,\Gamma (d)\,\Gamma (1-d)}{\Gamma (a+n)\,\Gamma (b+n)\,\Gamma (1-c-n)\,\Gamma (1-d-n)}}={\frac {\pi }{\sin c\pi }}\,{\frac {\pi }{\sin d\pi }}\,{\frac {\Gamma (a+b-c-d-1)}{\Gamma (a-c)\,\Gamma (a-d)\,\Gamma (b-c)\,\Gamma (b-d)}}}$

Kürzt man ${\displaystyle \Gamma (c)\,\Gamma (1-c)={\frac {\pi }{\sin c\pi }}}$ und ${\displaystyle \Gamma (d)\,\Gamma (1-d)={\frac {\pi }{\sin d\pi }}}$ heraus und substituiert ${\displaystyle c\mapsto 1-c\,,\,d\mapsto 1-d}$, so gilt

${\displaystyle \sum _{n=-\infty }^{\infty }{\frac {1}{\Gamma (a+n)\,\Gamma (b+n)\,\Gamma (c-n)\,\Gamma (d-n)}}={\frac {\Gamma (a+b+c+d-3)}{\Gamma (a+c-1)\,\Gamma (a+d-1)\,\Gamma (b+c-1)\,\Gamma (b+d-1)}}}$

für ${\displaystyle {\text{Re}}(a+b+c+d)>3\,}$.

${\displaystyle I:=\int _{0}^{1}\int _{0}^{1}s^{\alpha -1}\,(1-s)^{\beta -1}\,t^{\alpha -1}\,(1-t)^{\beta -1}\,|s-t|^{2\gamma }\,ds\,dt}$

ist aus Symmetriegründen

${\displaystyle 2\int _{0}^{1}\int _{0}^{t}s^{\alpha -1}\,(1-s)^{\beta -1}\,t^{\alpha -1}\,(1-t)^{\beta -1}\,(t-s)^{2\gamma }\,ds\,dt}$.

Und das ist nach Substitution ${\displaystyle s=tu\;,\;ds=t\,du}$ gleich

${\displaystyle 2\int _{0}^{1}\int _{0}^{1}t^{\alpha -1}\,u^{\alpha -1}\,(1-tu)^{\beta -1}\,t^{\alpha -1}\,(1-t)^{\beta -1}\,t^{2\gamma }\,(1-u)^{2\gamma }\,t\,du\,dt}$

Wegen ${\displaystyle (1-tu)^{\beta -1}=\sum _{n=0}^{\infty }{n-\beta \choose n}\,(tu)^{n}}$ ist nun

${\displaystyle I=2\sum _{n=0}^{\infty }{n-\beta \choose n}\int _{0}^{1}t^{2\alpha +2\gamma +n-1}\,(1-t)^{\beta -1}\,dt\,\int _{0}^{1}u^{\alpha +n-1}\,(1-u)^{2\gamma }\,du}$

${\displaystyle =2\sum _{n=0}^{\infty }{\frac {\Gamma (1-\beta +n)}{n!\,\Gamma (1-\beta )}}\,{\frac {\Gamma (2\alpha +2\gamma +n)\,\Gamma (\beta )}{\Gamma (2\alpha +\beta +2\gamma +n)}}\,{\frac {\Gamma (\alpha +n)\,\Gamma (1+2\gamma )}{\Gamma (1+\alpha +2\gamma +n)}}}$.

Nach dem Selberg Integral in zwei Variablen ist ${\displaystyle I={\frac {\Gamma (\alpha )\,\Gamma (\beta )}{\Gamma (\alpha +\beta +\gamma )}}\,{\frac {\Gamma (\alpha +\gamma )\,\Gamma (\beta +\gamma )\,\Gamma (1+2\gamma )}{\Gamma (\alpha +\beta +2\gamma )\,\Gamma (1+\gamma )}}}$.

Also ist ${\displaystyle \sum _{n=0}^{\infty }{\frac {\Gamma (1-\beta +n)\,\Gamma (2\alpha +2\gamma +n)\,\Gamma (\alpha +n)}{n!\,\Gamma (2\alpha +2\gamma +n)\,\Gamma (1+\alpha +2\gamma +n)}}={\frac {\Gamma (\alpha )\,\Gamma (1-\beta )\,\Gamma (\alpha +\gamma )\,\Gamma (\beta +\gamma )}{2\,\Gamma (\alpha +\beta +\gamma )\,\Gamma (\alpha +\beta +2\gamma )\,\Gamma (1+\gamma )}}}$.

Substituiert man ${\displaystyle \alpha =c\,,\,\beta =1-b\,,\,\gamma ={\frac {a}{2}}-c}$, so ist

${\displaystyle \sum _{n=0}^{\infty }{\frac {\Gamma (b+n)\,\Gamma (a+n)\,\Gamma (c+n)}{n!\,\Gamma (1-b+a+n)\,\Gamma (1-c+a+n)}}={\frac {\Gamma (c)\,\Gamma (b)\,\Gamma \left({\frac {a}{2}}\right)\,\Gamma \left(1+{\frac {a}{2}}-b-c\right)}{2\,\Gamma \left(1+{\frac {a}{2}}-b\right)\,\Gamma (1+a-b-c)\,\Gamma \left(1+{\frac {a}{2}}-c\right)}}}$.

Es ist ${\displaystyle I:=\int _{0}^{\frac {\pi }{2}}{\frac {x^{2}}{\sin x}}\,dx=2\pi G-{\frac {7}{2}}\zeta (3)}$.

Ersetzt man beim Integranden ${\displaystyle x^{2}\,}$ durch ${\displaystyle \sum _{n=1}^{\infty }{\frac {2^{2n-1}}{n^{2}\,{2n \choose n}}}\,\sin ^{2n}x}$,

so ist ${\displaystyle I:=\sum _{n=1}^{\infty }{\frac {2^{2n-1}}{n^{2}\,{2n \choose n}}}\,\int _{0}^{\frac {\pi }{2}}\sin ^{2n-1}x\,dx}$.

Hierbei ist ${\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2n-1}x\,dx={\frac {{\sqrt {\pi }}\,\,\Gamma (n)}{2\,\Gamma \left(n+{\frac {1}{2}}\right)}}={\frac {2^{2n-1}}{n\,{2n \choose n}}}}$.

Also stimmt ${\displaystyle I\,}$ mit ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}\left({\frac {{\sqrt {\pi }}\,\,\Gamma (n)}{2\,\Gamma \left(n+{\frac {1}{2}}\right)}}\right)^{2}}$ überein.

Aus der Formel ${\displaystyle \int _{0}^{\infty }{\frac {1}{p}}\left({\frac {2}{1+{\sqrt {1+4x}}}}\right)^{p}\cdot x^{s-1}\,dx=\Gamma (s)\cdot {\frac {\Gamma (p-2s)}{\Gamma (p-s-1)}}}$

folgt nach Ramanujans Master Theorem die Formel ${\displaystyle \sum _{k=0}^{\infty }{\frac {\Gamma (p+2k)}{\Gamma (p+k+1)}}\,{\frac {(-x)^{k}}{k!}}={\frac {1}{p}}\left({\frac {2}{1+{\sqrt {1+4x}}}}\right)^{p}}$.

${\displaystyle \sum _{k=0}^{\infty }{\frac {\Gamma (p+2k)}{\Gamma (p+k+1)}}\,{\frac {x^{k}}{k!}}={\frac {1}{p}}\cdot {\frac {2^{p}}{\left(1+{\sqrt {1-4x}}\,\right)^{p}}}}$

Differenziere nach ${\displaystyle x\,}$:

${\displaystyle \sum _{k=1}^{\infty }{\frac {\Gamma (p+2k)}{\Gamma (p+k+1)}}\,{\frac {x^{k-1}}{(k-1)!}}={\frac {2^{p+1}}{{\sqrt {1-4x}}\cdot \left(1+{\sqrt {1-4x}}\right)^{p+1}}}}$

Ersetze ${\displaystyle p\,}$ durch ${\displaystyle p-1}$:

${\displaystyle \sum _{k=1}^{\infty }{\frac {\Gamma (p-1+2k)}{\Gamma (p+k)}}\,{\frac {x^{k-1}}{(k-1)!}}={\frac {2^{p}}{{\sqrt {1-4x}}\cdot \left(1+{\sqrt {1-4x}}\right)^{p}}}}$

Die Reihe ist nach Laufindexverschiebung gleich

${\displaystyle \sum _{k=0}^{\infty }{\frac {\Gamma (p+2k+1)}{\Gamma (p+k+1)}}\,{\frac {x^{k}}{k!}}=\sum _{k=0}^{\infty }{\frac {(2k+p)!}{(k+p)!\cdot k!}}\,x^{k}=\sum _{k=0}^{\infty }{2k+p \choose k}\,x^{k}}$.