[Calculation 5] Gauss’s Summation Formula

Edited by Leun Kim

Theorem. (Gauss’s Summation Formula)
For $\text{Re}c>\text{Re}b>0$,
$$
{}_2 F_1 (a,b;c;1) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}
$$
holds.

 
Proof. We remember that the Euler Integral Representation for the hypergeometric function is
$$
_2 F_1 (a,b;c;z)
=
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-tz)^a}\,dt.
$$
Taking the limit $z\to 1$ both sides, we obtain
\begin{eqnarray*}
_2 F_1 (a,b;c;1)
&=&
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\int_0^1 t^{b-1} (1-t)^{c-a-b-1}\,dt\\
&=&
\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\frac{\Gamma(b)\Gamma(c-b-a)}{\Gamma(c-a)}\\
&=&
\frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)},
\end{eqnarray*}
which proves the theorem.$\square$
 

Corollary. (Chu–Vandermonde Identity)
$$
{}_2 F_1 (-n,b;c;1) = \frac{(c-b)_n}{(c)_n}
$$

 
Proof. Taking $a = -n$ in Gauss’s Summation Formula, we have
$$
{}_2 F_1 (-n,b;c;1) = \frac{\Gamma(c)\Gamma(c-b+n)}{\Gamma(c+n)\Gamma(c-b)} = \frac{(c-b)_n}{(c)_n}.
$$
 
References.
[1] en.wikipedia.org/wiki/Hypergeometric_function
[2] Leun Kim, Fundamentals of Hypergeometric Functions.

 
I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.



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