Euler's Reflection Formula Derivation
📂Functions Euler's Reflection Formula Derivation For non-integer p p p Γ ( 1 − p ) Γ ( p ) = π sin π p
{\Gamma (1-p) \Gamma ( p )} = { {\pi} \over {\sin \pi p } }
Γ ( 1 − p ) Γ ( p ) = sin π p π
Description It is the most famous formula among the formulas using the Gamma function .
A useful result that can be obtained from the reflection formula is Γ ( 1 2 ) = π \Gamma ( { 1 \over 2} ) = \sqrt{\pi} Γ ( 2 1 ) = π 1 2 \frac{1}{2} 2 1
Derivation Weierstrass’s infinite product : 1 Γ ( p ) = p e γ p ∏ n = 1 ∞ ( 1 + p n ) e − p n
{1 \over \Gamma (p)} = p e^{\gamma p } \prod_{n=1}^{\infty} \left( 1 + {p \over n} \right) e^{- {p \over n} }
Γ ( p ) 1 = p e γ p n = 1 ∏ ∞ ( 1 + n p ) e − n p
1 Γ ( p ) ⋅ 1 Γ ( − p ) = p e γ p ∏ n = 1 ∞ ( 1 + p n ) e − p n ⋅ ( − p ) e − γ p ∏ n = 1 ∞ ( 1 − p n ) e p n = − p 2 ∏ n = 1 ∞ ( 1 − p 2 n 2 )
\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*}
Γ ( p ) 1 ⋅ Γ ( − p ) 1 = = p e γ p n = 1 ∏ ∞ ( 1 + n p ) e − n p ⋅ ( − p ) e − γ p n = 1 ∏ ∞ ( 1 − n p ) e n p − p 2 n = 1 ∏ ∞ ( 1 − n 2 p 2 ) Γ ( 1 − p ) = − p Γ ( − p ) { \Gamma ( 1-p )} = -p \Gamma (-p) Γ ( 1 − p ) = − p Γ ( − p ) 1 Γ ( 1 − p ) Γ ( p ) = p ∏ n = 1 ∞ ( 1 − p 2 n 2 )
{ 1 \over {\Gamma (1-p) \Gamma ( p )} } = p \prod_{n=1}^{\infty} \left( 1 - {p^2 \over n^2} \right)
Γ ( 1 − p ) Γ ( p ) 1 = p n = 1 ∏ ∞ ( 1 − n 2 p 2 )
Euler’s representation of the sinc function : sin π x π x = ∏ n = 1 ∞ ( 1 − x 2 n 2 )
{{\sin \pi x} \over {\pi x}} = \prod_{n=1}^{\infty} \left( 1 - {{x^2} \over { n^2}} \right)
π x sin π x = n = 1 ∏ ∞ ( 1 − n 2 x 2 )
By fine-tuning the Euler representation of the sinc function, the desired formula is obtained.
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