ガンマ関数
Γ ( − 3 2 ) = 4 π 3 ≈ 2.363 {\displaystyle \Gamma \left(-{\frac {3}{2}}\right)\,={\frac {4{\sqrt {\pi }}}{3}}\approx 2.363\,} Γ ( − 1 2 ) = − 2 π ≈ − 3.545 {\displaystyle \Gamma \left(-{\frac {1}{2}}\right)\,=-2{\sqrt {\pi }}\approx -3.545\,} Γ ( 1 2 ) = π ≈ 1.772 {\displaystyle \Gamma \left({\frac {1}{2}}\right)\,={\sqrt {\pi }}\approx 1.772\,} Γ ( 1 ) = 0 ! = 1 {\displaystyle \Gamma (1)\,=0!=1\,} Γ ( 3 2 ) = π 2 ≈ 0.
g 0 ( z ) = ∫ 0 1 u z − 1 d u = [ u z z ] u = 0 1 = 1 z {\displaystyle g_{0}(z)=\int _{0}^{1}{u^{z-1}}{\rm {d}}u=\left[{\frac {u^{z}}{z}}\right]_{u=0}^{1}={\frac {1}{z}}} g n ( z ) = ∫ 0 1 ( u z z ) ′ ( 1 − u ) n d u = n z ∫ u = 0 1 u z ( 1 − u ) n − 1 d u = n z g n − 1 ( z + 1 ) {\displaystyle g_{n}(z)=\int _{0}^{1}{\left({\frac {u^{z}}{z}}\right)'(1-u)^{n}}{\rm {d}}u={\frac {n}{z}}\int _{u=0}^{1}{u^{z}(1-u)^{n-1}}{\rm {d}}u={\frac {n}{z}}g_{n-1}(z+1)}
これにより
G n ( z ) = n z n ! ∏ k = 0 n ( z + k ) {\displaystyle G_{n}(z)={\frac {n^{z}n!}{\prod _{k=0}^{n}{(z+k)}}}}
を得る。故に
∫ 0 ∞ t z − 1 e − t d t = lim n → ∞ G n ( z ) = lim n → ∞ n z n ! ∏ k = 0 n ( z + k ) {\displaystyle \int _{0}^{\infty }t^{z-1}e^{-t}{\ d}t=\lim _{n\to \infty }G_{n}(z)=\lim _{n\to \infty }{\frac {n^{z}n!}{\prod _{k=0}^{n}{(z+k)}}}}
である。
R z > 0 {\displaystyle \Re {z}>0} とすると δ → 0 {\displaystyle \delta \to 0} で δ z → 0 {\displaystyle \delta ^{z}\to 0} であるから