is Note that there is no value of z for which Γ ( z ) = 0, hence [Γ ( z )] − 1 is an entire function. An integral representation is given by Using (B.3), it can be shown that Γ ( 1 2 ) = √ π . The argument of the Gamma function can be simplified using the identity In order to construct series expansions of the Gamma function, it is helpful to introduce the Digamma function and the closely related Polygamma function Numerical values of these functions at z = 1 are where γ$_{E}$ ≃ 0 . 577 is the Euler-Mascheroni constant and ζ ( z ) is the Riemann Zeta function. 108 ψ (1) = − γ$_{E}$ , ψ ( n $^{)}$(1) = ( − 1) n n ! ζ ( n + 1) , (B.7) ψ ( n $^{)}$( z ) = d n dz $_{n}$ψ ( z ) . (B.6) ψ ( z ) = 1 Γ ( z ) d dz Γ ( z ) , (B.5) Γ (2 z ) = 2 x − 1 2 √ 2 π Γ ( z ) Γ$^{(}$ z + 1 2$^{)}$ . (B.4) Γ ( z ) = ∫ ∞ 0 dt t z − $^{1}$e − t , Re( z ) > 0 . (B.3) Res Γ (z) |$_{z}$$_{→−}$$_{n}$ = ( − 1) n n ! . (B.2)