1.1. HOMOGENEOUS DISTRIBUTIONS 3 (2) The inverse of the Gamma function 1 Γ(z) defined on the half-plane Re(z) 0 extends to a holomorphic map at z = 0 and (1.4) 1 Γ(z) = z + γz2 + o(z2). In particular, ( 1 Γ ) (0) = 1. Proof. (1) The derivative of Γ at 1 reads Γ (1) = ∂zΓ(1 + z)|z=0 = ∂z ∞ 0 tz e−t dt |z=0 = ∞ 0 log t e−t dx = −γ. The derivative at k ∈ N − {1} reads Γ (k) = lim z→0 Γ (k + z) − Γ (k) z = lim z→0 (k + z − 1) · · · (z + 1) · Γ (z + 1) − (k − 1)! z = (k − 1)! ⎛ ⎝ k−1 j=1 1 j − γ⎠ ⎞ = Γ(k) ⎛ ⎝ k−1 j=1 1 j − γ⎠ ⎞ , so that for k ≥ 2, Γ (k) Γ(k) = k−1 j=1 1 j − γ. (2) It follows from (1.2) that 1 Γ(z) = z Γ(z + 1) = z Γ(1) + Γ (1)z + o(z) = z + γz2 + o(z2). Exercise 1.8. Show that (1.5) Γ(z) ∼0 1 z − γ, where by f(z) ∼0 g(z) we mean that limz→0 f(z) g(z) = 1. Hint: Formula (1.5) follows from (1.2) setting b = z, k = 1 or equivalently from (1.4). Exercise 1.9. Show that2 Γ(z) Γ z + 1 2 = 21−2z √ π Γ(2z). Exercise 1.10. Show that for two complex numbers a and b with positive real parts, we have Γ(a)Γ(b) = B(a, b)Γ(a + b), where B(a, b) := 1 0 ta−1(1 − t)b−1dt is the Beta function. 2 I thank Stephan Weinzierl for pointing out this formula to me.
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