TITLE: A gamma function identity
QUESTION [21 upvotes]: In some of my previous work on mean values of Dirichlet L-functions, I came upon the following identity for the Gamma function:
\begin{equation} 
\frac{\Gamma(a) \Gamma(1-a-b)}{\Gamma(1-b)} 
+ \frac{\Gamma(b) \Gamma(1-a-b)}{ \Gamma(1-a)} 
+ \frac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)} = 
\pi^{\frac12} 
\frac{\Gamma\left(\frac{ 1-a-b}{2}\right) }{\Gamma\left(\frac{a+b}{2}\right)}
\frac{\Gamma\left(\frac{a}{2}\right)}{\Gamma\left(\frac{1-a}{2}\right)}
\frac{\Gamma\left(\frac{b}{2}\right)}{\Gamma\left(\frac{1-b}{2}\right)}.
\end{equation}
As is often the case, once one knows such a formula should be true then it is easy to prove it.  I give my proof below.  My questions are 1) Has this formula been observed before?  I have no idea how to search the literature for such a thing.  2) Is there a better proof? (Of course this is totally subjective, but one thing that would please me would be to avoid trigonometric functions since they do not appear in the formula.)
Proof.
Using
\begin{equation}
\frac{\Gamma(\frac{s}{2})}{\Gamma(\frac{1-s}{2})} = \pi^{-\frac12} 2^{1-s} \cos({\textstyle \frac{\pi s}{2}}) \Gamma(s),
\end{equation}
the right hand side is
\begin{equation}
2 \frac{\cos(\frac{\pi a}{2}) \cos(\frac{\pi b}{2}) \Gamma(a) \Gamma(b)}{\cos(\frac{\pi (a + b)}{2}) \Gamma(a+b)}.
\end{equation}
On the other hand, the left hand side is
\begin{equation}
\frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} 
\left(
 \frac{\Gamma(a+b) \Gamma(1-a-b)}{\Gamma(b) \Gamma(1-b)} 
 + \frac{\Gamma(a+b) \Gamma(1-a-b)}{\Gamma(a) \Gamma(1-a)}
 + 1
\right),
\end{equation}
which becomes after using $\Gamma(s) \Gamma(1-s) = \frac{\pi}{\sin(\pi s)}$,
\begin{equation}
\frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} 
\left(\frac{\sin(\pi a) + \sin( \pi b) + \sin(\pi(a + b))}{\sin(\pi(a+b))} \right).
\end{equation}
Using trig formulas, we get that this is
\begin{equation}
 2 \frac{\Gamma(a)\Gamma(b)}{\Gamma(a + b)} \frac{\sin(\frac{\pi}{2}(a+b)) \cos(\frac{\pi}{2}(a-b)) + \sin(\frac{\pi}{2}(a+b)) \cos(\frac{\pi}{2}(a+b)) }{\sin(\pi(a+b))}
\end{equation}
I think I've run out of space?  The rest is easy trig.

REPLY [22 votes]: A proof of the statement has already been given, so I will just add a small historical remark.
The left hand side of your identity is the Veneziano amplitude in the case of four identical scalar particles. The right hand side corresponds to another crossing symmetric four-point amplitude for scalar particles suggested by Virasoro in "Alternative Constructions of Crossing-Symmetric Amplitudes with Regge Behavior". These are known to coincide for the choice of parameters $a+b+c=1$ giving your identity. I would recommend the article "The birth of string theory" for a quick survey of the Veneziano model and its generalization to N scalar particles and related mathematics.