Chapter 8 Euler's Gamma function The Gamma function plays an important role in the functional equation for ζ(s) that we will derive in the next chapter. In the present chapter we have collected some properties of the Gamma function. z z log t For t 2 R>0, z 2 C, define t := e , where log t is he ordinary real logarithm. Euler's Gamma function is defined by the integral Z 1 −t z−1 Γ(z) := e t dt (z 2 C; Re z > 0): 0 Lemma 8.1. Γ(z) defines an analytic function on fz 2 C : Re z > 0g. Proof. We prove that Γ(z) is analytic on Uδ;R := fz 2 C; δ < Re z < Rg for every δ; R with 0 < δ < R. This is standard using Theorem 2.29. First, the function −t z−1 e t is continuous, hence measurable on R>0×Uδ;R. Second, for any fixed t 2 R>0, −t t−1 the function z 7! e z is analytic on Uδ;R. Lastly, for z 2 Uδ;R we have δ−1 −t z−1 t for 0 6 t 6 1; je t j 6 M(t) := −t R−1 −t=2 e t 6 Ce for t > 1; where C is some constant. Now we have Z 1 Z 1 Z 1 M(t)dt = tδ−1dt + C e−t=2dt = δ−1 + 2C < 1: 0 0 1 Hence all conditions of Theorem 2.29 are satisfied, and thus, Γ(z) is analytic on Uδ;R. 115 We now show that Γ has a meromorphic continuation to C. Theorem 8.2. There exists a unique meromorphic function Γ on C with the fol- lowing properties: R 1 −t z−1 (i) Γ(z) = 0 e t dt for z 2 C, Re z > 0; (ii) the function Γ is analytic on C n f0; −1; −2;:::g; (iii) Γ has a simple pole with residue (−1)n=n! at z = −n for n = 0; 1; 2;:::; (iv) Γ(z + 1) = zΓ(z) for z 2 C n f0; −1; −2;:::g; (v) Γ(n) = (n − 1)! for n 2 Z>0. R 1 −t z−1 Proof. The function Γ has already been defined for Re z > 0 by 0 e t dt. By Corollary 2.22, Γ has at most one analytic continuation to any larger connected open set, hence there is at most one function Γ with properties (i){(v). We proceed to construct such a function. Let z 2 C with Re z > 0. Then using integration by parts, Z 1 Z 1 (8.1) Γ(z) = e−ttz−1dt = e−tz−1dtz 0 0 h i1 Z 1 Z 1 = z−1e−ttz + z−1 e−ttzdt = z−1 e−ttzdt 0 0 0 = z−1Γ(z + 1): Now by induction on n it follows that 1 (8.2) Γ(z) = · Γ(z + n) for Re z > 0; n = 1; 2;:::: z(z + 1) ··· (z + n − 1) We continue Γ to B := C n f0; −1; −2;:::g as follows. For z 2 B, choose n 2 Z>0 such that Re z +n > 0 and define Γ(z) by the right-hand side of (8.2). This does not depend on the choice of n. For if m; n are any two integers with m > n > −Re z, then by (8.2) with z + n, m − n instead of z; n we have 1 Γ(z + n) = · Γ(z + m); z + n) ··· (z + m − 1) and so 1 1 · Γ(z + n) = · Γ(z + m): z(z + 1) ··· (z + n − 1) z(z + 1) ··· (z + m − 1) 116 So Γ is well-defined on B, and it is analytic on B since the right-hand side of (8.2) is analytic if Re z + n > 0. This proves (ii). We prove (iii). By (8.2) we have 1 lim (z + n)Γ(z) = lim (z + n) Γ(z + n + 1) z→−n z→−n z(z + 1) ··· (z + n) 1 (−1)n = Γ(1) = : (−n)(−n + 1) ··· (−1) n! Hence Γ has a simple pole at z = −n of residue (−1)n=n!. We prove (iv). Both functions Γ(z + 1) and zΓ(z) are analytic on B, and by (8.1), they are equal on the set fz 2 C : Re z > 0g which has limit points in B. So by Corollary 2.21, Γ(z + 1) = zΓ(z) for z 2 B. R 1 −t Identity (v) follows easily by observing that Γ(1) = 0 e dt = 1, and by re- peatedly applying (iv). π Theorem 8.3. We have Γ(z)Γ(1 − z) = for z 2 n . sin πz C Z Proof. We prove that zΓ(z)Γ(1 − z) = πz= sin πz, or equivalently, πz (8.3) Γ(1 + z)Γ(1 − z) = for z 2 A := ( n ) [ f0g; sin πz C Z which implies Theorem 8.3. Notice that by Theorem 8.2 the left-hand side is analytic on A, while by limz!0 πz= sin πz = 1 the right-hand side is also analytic on A. By Corollary 2.19, it suffices to prove that (8.3) holds for z 2 S, where S is any subset 1 of A having a limit point in A. For the set S we take f 2n : n 2 Z>0g; this set has limit point 0 in A. Thus, (8.3), and hence Theorem 8.3, follows once we have proved that π=2n (8.4) Γ(1 + 1 ) · Γ(1 − 1 ) = (n = 1; 2;:::): 2n 2n sin π=2n Notice that Z 1 Z 1 1 1 −s 1=2n −t −1=2n Γ(1 + 2n ) · Γ(1 − 2n ) = e s ds · e t dt 0 0 Z 1 Z 1 = e−s−t(s=t)1=2ndsdt: 0 0 117 Define new variables u = s + t, v = s=t. Then s = uv=(v + 1), t = u=(v + 1). The Jacobian of the substitution (s; t) 7! (u; v) is v u @s @s @(s; t) v + 1 (v + 1)2 @u @v = = 1 u @(u; v) @t @t − @u @v v + 1 (v + 1)2 −uv − u −u = = : (v + 1)3 (v + 1)2 It follows that Z 1 Z 1 1 1 −u 1=2n @(s; t) Γ(1 + 2n ) · Γ(1 − 2n ) = e v · dudv 0 0 @(u; v) Z 1 Z 1 Z 1 Z 1 1=2n −u 1=2n u −u v = e v 2 · dudv = e udu · 2 dv: 0 0 (v + 1) 0 0 (v + 1) In the last product, the first integral is equal to 1, while for the second integral we have, by homework exercise 4, Z 1 1=2n Z 1 v 1=2n 1 2 dv = − v d 0 (v + 1) 0 v + 1 1=2n 1 Z 1 Z 1 v 1 1=2n dw π=2n = − + · dv = 2n = : v + 1 0 0 v + 1 0 w + 1 sin π=2n This implies (8.4), hence Theorem 8.3. 1 p Corollary 8.4. Γ( 2 ) = π. 1 1 Proof. Substitute z = 2 in Theorem 8.3, and use Γ( 2 ) > 0. Corollary 8.5. (i) Γ(z) 6= 0 for z 2 C n f0; −1; −2;:::g. (ii) 1=Γ is analytic on C, and 1=Γ has simple zeros at z = 0; −1; −2;:::. Proof. (i) Recall that Γ(n) = (n − 1)! 6= 0 for n = 1; 2;:::. Further, Γ(z) 6= 0 for z 2 C n Z by Theorem 8.3. (ii) By (i), the function 1=Γ is analytic on C n f0; −1; −2;:::g. Further, at z = 0; −1; −2;:::, Γ has a simple pole, hence 1=Γ is analytic and has a simple zero. 118 We give two other expressions for the Gamma function. Recall that the Euler- Mascheroni constant γ is defined by N X 1 γ := lim − log N: N!1 n n=1 Theorem 8.6. For z 2 Z n f0; −1; −2;:::g we have 1 n! · nz Y ez=n Γ(z) = lim = e−γzz−1 : n!1 z(z + 1) ··· (z + n) 1 + z=n n=1 Proof. We first show that the second and third expression are equal, assuming that either the limit exists, or the product converges. Indeed, 1 z=n N z=n −γz −1 Y e (log N−1− 1 −···− 1 )z −1 Y e e z = lim e 2 N z 1 + z=n N!1 1 + z=n n=1 n=1 N Y 1 N z · N! = lim ez log N z−1 = lim : N!1 1 + z=n N!1 z(z + 1) ··· (z + N) n=1 The remainder of the proof of Theorem 8.6 is a combination of a few lemmas. Define for the moment 1 n! · nz Y ez=n g(z) := lim = e−γzz−1 : n!1 z(z + 1) ··· (z + n) 1 + z=n n=1 Lemma 8.7. g(z) defines an analytic function on C n f0; −1; −2;:::g. Q1 z −z=n Proof. It suffices to prove that h(z) := n=1 1 + n e is analytic on C and that h(z) 6= 0 on B := C n f0; −1; −2;:::g. For this, it is sufficient to prove that h(z) is analytic on D(0;R) = fz 2 C : jzj < Rg and h(z) 6= 0 for z 6= 0; −1; −2;:::.