Chapter 8. Euler s Gamma function

Transcription

1 Chapter 8 Euler s Gamma fuctio The Gamma fuctio plays a importat role i the fuctioal equatio for ζ(s) that we will derive i the ext chapter. I the preset chapter we have collected some properties of the Gamma fuctio. For t R >, z C, defie t z : e z log t, where log t is he ordiary real logarithm. Euler s Gamma fuctio is defied by the itegral Γ(z) : e t t z dt (z C, Re z > ). Lemma 8.. Γ(z) defies a aalytic fuctio o {z C : Re z > }. Proof. We prove that Γ(z) is aalytic o U δ,r : {z C, δ < Re z < R} for every δ, R with < δ < R. This is stadard usig Theorem First, the fuctio e t t z is cotiuous, hece measurable o R > U δ,r. Secod, for ay fixed t R >, the fuctio z e t z t is aalytic o U δ,r. Lastly, for z U δ,r we have { t e t t z δ for t, M(t) : e t t R Ce t/2 for t >, where C is some costat. Now we have M(t)dt t δ dt + C e t/2 dt δ + 2C <. Hece all coditios of Theorem 2.29 are satisfied, ad thus, Γ(z) is aalytic o U δ,r. 5

2 We ow show that Γ has a meromorphic cotiuatio to C. Theorem 8.2. There exists a uique meromorphic fuctio Γ o C with the followig properties: (i) Γ(z) e t t z dt for z C, Re z > ; (ii) the fuctio Γ is aalytic o C \ {,, 2,...}; (iii) Γ has a simple pole with residue ( ) /! at z for,, 2,...; (iv) Γ(z + ) zγ(z) for z C \ {,, 2,...}; (v) Γ() ( )! for Z >. Proof. The fuctio Γ has already bee defied for Re z > by e t t z dt. By Corollary 2.22, Γ has at most oe aalytic cotiuatio to ay larger coected ope set, hece there is at most oe fuctio Γ with properties (i) (v). We proceed to costruct such a fuctio. Let z C with Re z >. The usig itegratio by parts, (8.) Γ(z) e t t z dt [ z e t t z ] z Γ(z + ). e t z dt z + z e t t z dt z e t t z dt Now by iductio o it follows that (8.2) Γ(z) Γ(z + ) for Re z >,, 2,.... z(z + ) (z + ) We cotiue Γ to B : C \ {,, 2,...} as follows. For z B, choose Z > such that Re z + > ad defie Γ(z) by the right-had side of (8.2). This does ot deped o the choice of. For if m, are ay two itegers with m > > Re z, the by (8.2) with z +, m istead of z, we have ad so Γ(z + ) Γ(z + m), z + ) (z + m ) z(z + ) (z + ) Γ(z + ) Γ(z + m). z(z + ) (z + m ) 6

3 So Γ is well-defied o B, ad it is aalytic o B sice the right-had side of (8.2) is aalytic if Re z + >. This proves (ii). We prove (iii). By (8.2) we have lim (z + )Γ(z) z lim Γ(z + + ) z z(z + ) (z + ) ( ) Γ(). ( )( + ) ( )! Hece Γ has a simple pole at z of residue ( ) /!. We prove (iv). Both fuctios Γ(z + ) ad zγ(z) are aalytic o B, ad by (8.), they are equal o the set {z C : Re z > } which has limit poits i B. So by Corollary 2.2, Γ(z + ) zγ(z) for z B. Idetity (v) follows easily by observig that Γ() e t dt, ad by repeatedly applyig (iv). Theorem 8.3. We have Γ(z)Γ( z) π si πz for z C \ Z. Proof. We prove that zγ(z)γ( z) πz/ si πz, or equivaletly, (8.3) Γ( + z)γ( z) πz si πz for z A : (C \ Z) {}, which implies Theorem 8.3. Notice that by Theorem 8.2 the left-had side is aalytic o A, while by lim z πz/ si πz the right-had side is also aalytic o A. By Corollary 2.9, it suffices to prove that (8.3) holds for z S, where S is ay subset of A havig a limit poit i A. For the set S we take { : Z 2 >}; this set has limit poit i A. Thus, (8.3), ad hece Theorem 8.3, follows oce we have proved that (8.4) Γ( + 2 ) Γ( 2 ) Notice that Γ( π/2 si π/2 7 e s s /2 ds (, 2,...). e t t /2 dt e s t (s/t) /2 dsdt.

4 Defie ew variables u s + t, v s/t. The s uv/(v + ), t u/(v + ). The Jacobia of the substitutio (s, t) (u, v) is It follows that (s, t) (u, v) s u t u Γ( + 2 ) Γ( 2 ) s v t v uv u (v + ) 3 u (v + ) 2. e u v /2 u (v + ) dudv 2 v u v + (v + ) 2 v + u (v + ) 2 e u v /2 (s, t) (u, v) dudv e u udu v /2 (v + ) 2 dv. I the last product, the first itegral is equal to, while for the secod itegral we have, by homework exercise 4, v /2 (v + ) dv 2 [ ] v /2 + v + This implies (8.4), hece Theorem 8.3. Corollary 8.4. Γ( 2 ) π. ( ) v /2 d v + v + dv/2 Proof. Substitute z 2 i Theorem 8.3, ad use Γ( 2 ) >. dw w 2 + π/2 si π/2. Corollary 8.5. (i) Γ(z) for z C \ {,, 2,...}. (ii) /Γ is aalytic o C, ad /Γ has simple zeros at z,, 2,.... Proof. (i) Recall that Γ() ( )! for, 2,.... Further, Γ(z) for z C \ Z by Theorem 8.3. (ii) By (i), the fuctio /Γ is aalytic o C \ {,, 2,...}. Further, at z,, 2,..., Γ has a simple pole, hece /Γ is aalytic ad has a simple zero. 8

5 We give two other expressios for the Gamma fuctio. Recall that the Euler- Mascheroi costat γ is defied by ( N ) γ : lim log N. N Theorem 8.6. For z Z \ {,, 2,...} we have! z Γ(z) lim z(z + ) (z + ) e γz z e z/ + z/. Proof. We first show that the secod ad third expressio are equal, assumig that either the limit exists, or the product coverges. Ideed, e γz z lim N ez log N z e z/ + z/ lim N e(log N 2 N )z z N + z/ lim N N e z/ + z/ N z N! z(z + ) (z + N). The remaider of the proof of Theorem 8.6 is a combiatio of a few lemmas. Defie for the momet! z g(z) : lim z(z + ) (z + ) e γz z e z/ + z/. Lemma 8.7. g(z) defies a aalytic fuctio o C \ {,, 2,...}. Proof. It suffices to prove that h(z) : {( + z ) } e z/ is aalytic o C ad that h(z) o B : C \ {,, 2,...}. For this, it is sufficiet to prove that h(z) is aalytic o D(, R) {z C : z < R} ad h(z) for z,, 2,.... We proceed to show that there is a sequece {M } such that (8.5) ( + z )e z/ M for z D(, R),, M <. The the ifiite product defiig h is poitwise absolutely coverget, which implies that h(z) wheever ay of the factors i the product is ; that is, 9

6 wheever z,, 2,.... Further, by Corollary 2.27, the ifiite product defies a aalytic fuctio o D(, R). We ow show that there is a sequece {M } with (8.5). Notice that ( z + e ) z/ ( + z ) ( z + z2 / 2 z3 / 3 + ) 2! 3! Hece for z C with z < R we have 2 z2 + ( 2 2! ) z 3 3! ( 3 3! ) z 4 4! + 4 ( + ) z e z/ ( z ) 2 k2 R2 2 er. ( k2 (k )! k! k (k )! z k 2 )( z ) k ( z ) 2 k2 Clearly, R2 e R / 2 coverges. This proves Lemma 8.7. (k 2)! z k 2 To prove that Γ(z) g(z) for z B : C \ {,, 2,...}, it suffices to prove that Γ(z) g(z) for s aubset of B with a limit poit i B. For this subset, we take R >. So it suffices to prove that! x (8.6) Γ(x) lim x(x + ) (x + ) for x R >. Lemma 8.8. We have! x x(x + ) (x + ) Proof. By substitutig s t/, the itegral becomes The rest is left as a exercise. x ( s) s x ds. 2 ( t ) t x dx.

7 Lemma 8.9. For every iteger 2 ad every real t with t we have Proof. This is equivalet to e t ( t ) e t t2 2. t2 et( t ) ( t, 2). Recall that if f, g are cotiuously differetiable, real fuctios with f() g() ad f (x) g (x) for x A, say, the f(x) g(x) for x A. From this observatio, oe easily deduces that + x e x, x e x, ( x) r rx for x, r. This implies o the oe had, for 2, t, o the other had e t( t e t( t ) e t (e t/ ), ) ( t ) ( t ) ( t 2 ) t We prove (8.6) ad complete the proof of Theorem 8.6. We have for x >, by the itegral expressio for Γ(x) for x > ad by Lemma 8.8,! x Γ(x) lim x(x + ) (x + ) { lim e t t x dt ( lim e t ( t ) ) t x dt. Now Lemma 8.9 implies! x Γ(x) lim x(x + ) (x + ) lim } ( t ) t x dt lim e t t x+ dt lim Γ(x + 2). e t t2 tx dt 2

8 We deduce some cosequeces. Corollary 8.. We have ) si πz πz ( z2 2 for z C. Proof. For z C we have by Theorem 8.3, Corollary 8.5 ad Theorem 8.6, si πz π Γ(z)Γ( z) π( z) e γz z πz π Γ(z)( z)γ( z) ( (e z/ + z )) ( z )( + z ) πz ( z)e γz ( z2 2 ). ( (e z/ z )) Recall that the Beroulli umbers B ( ) are give by z e z B! z ( z < 2π). Corollary 8.. We have B, B 2, B 3 B 5 ad ζ(2) ( ) 2 2 B 2 (2)! π2 for, 2,.... Proof. Let z C with < z <. The si πz ad so, by takig the logarithmic derivative of si πz, (8.7) si πz si πz π cos πz si πz π(eπiz + e πiz )/2 (e πiz e πiz )/2i πi + z 2πiz e 2πiz πi + B z! (2πi) z. 22

9 We obtai aother expressio for the logarithmic derivative of si πz by applyig Corollary 2.28 to the product idetity from Corollary 8.. Note that for z C with z < we have z 2 / 2 < 2 ad that 2 coverges. Hece the logarithmic derivative of the ifiite product is the ifiite sum of the logarithmic derivatives of the factors, i.e., (8.8) si πz si πz (πz) πz z + + z 2 ( z 2 / 2 ) z 2 / 2 2z/ 2 z 2 / 2 z 2 z 2 k z 2k+ 2k+2 z 2 ζ(2k + 2)z 2k+. k ( k ) ( z 2 ) k 2 (by absolute covergece) Now Corollary 8. easily follows by comparig the coefficiets of the Lauret series i (8.7) ad (8.8). We fiish with aother importat cosequece of Theorem 8.6, the so-called duplicatio formula. Corollary 8.2. We have π Γ(2z) 2 2z Γ(z)Γ(z + 2 ) for z C, z, 2,, 3 2, 2,.... Proof. Let A be the set of z idicated i the lemma. We show that the fuctio F (z) : 2 2z Γ(z)Γ(z + )/Γ(2z) is costat o A. Substitutig z gives that the 2 2 costat is 2 π, ad the Corollary 8.2 follows. Let z A. To get ice cacellatios i the umerator ad deomiator, we use 23

10 the expressios Γ(z) lim! z z(z + ) (z + ) lim Γ(z + ) lim! z+/2 2 (z + /2)(z + 3/2) (z + + /2) (2 + )! (2 + ) 2z Γ(2z) lim 2z(2z + ) (2z ) 2 +! z 2z(2z + 2) (2z + 2), 2 +! z+/2 lim (2z + )(2z + 3) (2z ), (take i Theorem 8.6 the limit over the odd itegers). Thus, F (z) 22z Γ(z)Γ(z + ) 2 Γ(2z) { 2 2z (!) 2 2z+/2 lim 2z(2z + ) (2z ) { (!) 2 } lim (2 + )! } 2z(2z + ) (2z ) (2 + )! (2 + ) 2z sice lim 2 2z 2z lim (2 + ) 2z e2z log(2/(2+)). This shows that ideed F (z) is costat. Substitutig z, we get 2 F (z) F ( 2 ) 2Γ( 2 )Γ() Γ() 2 π. Remark. More geerally, oe ca derive the multiplicatio formula of Legedre- Gauss, (2π) ( )/2 Γ(z) z /2 Γ(z)Γ(z + ) Γ(z + ) for every iteger 2. The idea of the proof is similar to that of Theorem 8.2 (exercise). 24