Derivation of Euler's Reflection Formula
Formula
For non-integer $p$, $$ {\Gamma (1-p) \Gamma ( p )} = { {\pi} \over {\sin \pi p } } $$
Explanation
Among the formulas involving the gamma function, this is the most famous one.
A useful result obtained from the reflection formula is $ \Gamma ( { 1 \over 2} ) = \sqrt{\pi}$. Perhaps that is why? It is said that the name “reflection formula” comes from the fact that it reflects about $\frac{1}{2}$.
Derivation
Weierstrass’s infinite product: $$ {1 \over \Gamma (p)} = p e^{\gamma p } \prod_{n=1}^{\infty} \left( 1 + {p \over n} \right) e^{- {p \over n} } $$
$$ \begin{align*} {{1} \over {\Gamma (p)}} \cdot { 1 \over { \Gamma ( -p )}} =& p e^{\gamma p } \prod_{n=1}^{\infty} \left( 1 + {p \over n} \right) e^{- {p \over n} } \cdot (-p) e^{- \gamma p } \prod_{n=1}^{\infty} \left( 1 - {p \over n} \right) e^{ {p \over n} } \\ =& -p^2 \prod_{n=1}^{\infty} \left( 1 - {p^2 \over n^2} \right) \end{align*} $$ Meanwhile, since ${ \Gamma ( 1-p )} = -p \Gamma (-p)$, $$ { 1 \over {\Gamma (1-p) \Gamma ( p )} } = p \prod_{n=1}^{\infty} \left( 1 - {p^2 \over n^2} \right) $$
Euler’s representation of the sinc function: $$ {{\sin \pi x} \over {\pi x}} = \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { n^2}} \right) $$
Rearranging the equation using Euler’s representation of the sinc function yields the desired formula.
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