The Gamma function Michael Taylor. Abstract. This material is excerpted from 18 and Appendix J of [T].

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1 The Gamma fuctio Michael Taylor Abstract. This material is excerpted from 8 ad Appedix J of [T]. The Gamma fuctio has bee previewed i , arisig i the computatio of a atural Laplace trasform: 8. ft = t z = Lfs = Γz s z, for Re z >, with 8. Γz = e t t z dt, Re z >. Here we develop further properties of this special fuctio, begiig with the followig crucial idetity: 8.3 Γz + = = = z Γz, e t t z dt d dt e t t z dt for Re z >, where we use itegratio by parts. The defiitio 8. clearly gives 8.4 Γ =, so we deduce that for ay iteger k, 8.5 Γk = k Γk = = k!. While Γz is defied i 8. for Re z >, ote that the left side of 8.3 is well defied for Re z >, so this idetity exteds Γz to be meromorphic o {z : Re z > }, with a simple pole at z =. Iteratig this argumet, we exted Γz to be meromorphic o C, with simple poles at z =,,,.... Havig such a meromorphic cotiuatio of Γz, we establish the followig idetity.

2 Propositio 8.. For z C \ Z we have 8.6 ΓzΓ z = π si πz. Proof. It suffices to establish this idetity for < Re z <. I that case we have 8.7 ΓzΓ z = = = e s+t s z t z ds dt e u v z + v du dv + v v z dv, where we have used the chage of variables u = s + t, v = t/s. With v = e x, the last itegral is equal to e x e xz dx, which is holomorphic o < Re z <. We wat to show that this is equal to the right side of 8.6 o this strip. It suffices to prove idetity o the lie z = / + iξ, ξ R. The 8.8 is equal to the Fourier itegral 8.9 cosh x e ixξ dx. This was evaluated i 6; by 6.3 it is equal to 8. π cosh πξ, ad sice 8. π si π + iξ = π cosh πξ, the demostratio of 8.6 is complete. Corollary 8.. The fuctio Γz has o zeros, so /Γz is a etire fuctio. For our ext result, we begi with the followig estimate:

3 3 Lemma 8.3. We have 8. e t the latter iequality holdig provided 4. t t e t, t, Proof. The first iequality i 8. is equivalet to the simple estimate e y y for y. To see this, deote the fuctio by fy ad ote that f = while f y = e y for y. As for the secod iequality i 8., write 8.3 log t = log t X = t + 3 = t X, t + t +. 4 We have t/ = e t X ad hece, for t, e t t = e X e t Xe t, usig the estimate x e x for x as above. It is clear from 8.3 that X t / if t /. O the other had, if t / ad 4 we have t / ad hece e t t /e t. We use 8. to obtai, for Re z >, Γz = lim = lim z Repeatedly itegratig by parts gives t t z dt s s z ds. 8.4 Γz = lim z zz + z + which yields the followig result of Euler: Propositio 8.4. For Re z >, we have 8.5 Γz = lim z zz + z +, s z+ ds, Usig the idetity 8.3, aalytically cotiuig Γz, we have 8.5 for all z C other tha,,,.... I more detail, we have Γz = Γz + z = lim z+ z z + z + z + +,

4 4 for Re z > z. We ca rewrite the right side as z zz + z + + = + z + zz + z + + z+, + ad / + z+ as. This exteds 8.5 to {z : Re z > }, ad iteratively we get further extesios. We ca rewrite 8.5 as 8.6 Γz = lim z z + z + z + z. To work o this formula, we defie Euler s costat: 8.7 γ = lim log. The 8.6 is equivalet to 8.8 Γz = lim e γz e z+/+ +/ z + z + z + z, which leads to the followig Euler product expasio. Propositio 8.5. For all z C, we have 8.9 Γz = z eγz = + z e z/. We ca combie 8.6 ad 8.9 to produce a product expasio for si πz. I fact, it follows from 8.9 that the etire fuctio /ΓzΓ z has the product expasio 8. ΓzΓ z = z z. = Sice Γ z = zγ z, we have by 8.6 that 8. si πz = πz z. = For aother proof of this result, see 3, Exercise.

5 Here is aother applicatio of 8.6. If we take z = /, we get Γ/ = π. Sice 8. implies Γ/ >, we have 8. Γ = π. Aother way to obtai 8. is the followig. A chage of variable gives 8.3 e x dx = e t t / dt = Γ It follows from.6 that the left side of 8.3 is equal to π/, so we agai obtai 8.. Note that applicatio of 8.3 the gives, for each iteger k, 8.4 Γ k + = π / k k 3. Oe ca calculate the area A of the uit sphere S R by relatig Gaussia itegrals to the Gamma fuctio. To see this, ote that the argumet givig.6 yields 8.5 e x dx = e dx x = π /. R O the other had, usig spherical polar coordiates to compute the left side of 8.4 gives e x dx = A e r r dr 8.6 R = A e t t / dt, where we use t = r. Recogizig the last itegral as Γ/, we have 8.7 A = π/ Γ/. More details o this argumet are give at the ed of Appedix C.. 5 Exercises. Use the product expasio 8.9 to prove that 8.8 d dz Γ z Γz = z +. =

6 6 Hit. Go from 8.9 to log Γz = log z + γz + [ log + z z ], = ad ote that d dz Γ z Γz = d log Γz. dz. Let γ = log +. Show that γ ad that < γ <. Deduce that γ = lim γ exists, as asserted i Usig / zt z = t z log t, show that has Laplace trasform 4. Show that 8.9 yields f z t = t z log t, Re z > Lf z s = Γ z Γz log s s z, Re s >. 8.9 Γz + = zγz = e γz Use this to show that = + z e z/, z <. 8.3 Γ = d dz zγz z= = γ. 5. Usig Exercises 3 4, show that ad that ft = log t = Lfs = log s + γ, s γ = log te t dt. 6. Show that γ = γ a γ b, with γ a = e t t dt, γ b = e t t dt.

7 Cosider how to obtai accurate umerical evaluatios of these quatities. Hit. Split the itegral for γ i Exercise 5 ito two pieces. Itegrate each piece by parts, usig e t = d/dte t for oe ad e t = d/dte t for the other. See Appedix J for more o this. 7. Use the Laplace trasform idetity 8. for f z t = t z o t, give Re z > plus the results of Exercises 5 6 of 5 to show that 8.3 Bz, ζ = ΓzΓζ, Re z, Re ζ >, Γz + ζ where the beta fuctio Bz, ζ is defied by 8.3 Bz, ζ = s z s ζ ds, Re z, Re ζ >. The idetity 8.3 is kow as Euler s formula for the beta fuctio. 8. Show that, for ay z C, whe z, we have + z e z/ = + w with log + w = log + z/ z/ satisfyig log + w z. Show that this estimate implies the covergece of the product o the right side of 8.9, locally uiformly o C. More ifiite products 9. Show that = 4 = π. Hit. Take z = / i 8... Show that, for all z C, 8.34 cos πz = odd z.

8 8 Hit. Use cos πz/ = si π/z ad 8. to obtai 8.35 cos πz = π z = z 4. Use u = u + u to write the geeral factor i this ifiite product as + z = ad obtai from 8.35 that cos πz = π + z 4 z + = 4 Deduce 8.34 from this ad Show that odd + z, z + z si πz πz = cos πz cos πz 4 cos πz 8. Hit. Make use of 8. ad A. The Legedre duplicatio formula The Legedre duplicatio formula relates Γz ad ΓzΓz + /. Note that each of these fuctios is meromorphic, with poles precisely at {, /,, 3/,,... }, all simple, ad both fuctios are owhere vaishig. Hece their quotiet is a etire holomorphic fuctio, ad it is owhere vaishig, so 8.37 Γz = e Az ΓzΓ z +, with Az holomorphic o C. We seek a formula for Az. We will be guided by 8.9, which implies that 8.38 Γz = zeγz = + z e z/,

9 9 ad via results give i 8B 8.39 ΓzΓz + / = z z + e γz e γz+/{ = + z e z/ + z + / e z+//}. Settig z + / = z + + = + z +, + ad 8.4 e z+// = e z/+ e z[/ /+] e /, we ca write the ifiite product o the right side of 8.39 as 8.4 { + z e z/ + = { + = e /} e z/+} z + e z[/ /+]. = Hece 8.43 ΓzΓz + / = zeγz e γ/ ez + ze z 8.4 = ze γz e γ/ ez = 4 { + z k k= e z/k} { + e /} e z[/ /+]. = Now, settig z = / i 8.9 gives 8.44 Γ/ = eγ/ + e /, = so takig 8.38 ito accout yields 8.45 ΓzΓz + / = Γ/Γz = Γ/Γz e z e αz, = e z[/ /+]

10 where 8.46 α = + = = = log. Hece e αz = z, ad we get 8.47 Γ Γz = z ΓzΓ z +. This is the Legedre duplicatio formula. Recall that Γ/ = π. A equivalet formulatio of 8.47 is z z π / Γz = z / Γ Γ. This geeralizes to the followig formula of Gauss, z z π / Γz = z / Γ Γ Γ valid for = 3, 4,.... z +, 8B. Covergece of ifiite products Here we record some results regardig the covergece of ifiite products, which have arise i this sectio. We look at ifiite products of the form a k. k= Disregardig cases where oe or more factors + a k vaish, the covergece of M k= + a k as M amouts to the covergece 8.5 lim M N + a k =, uiformly i N > M. k=m I particular, we require a k as k. To ivestigate whe 8.5 happes, write N + a k = + a M + a M+ + a N 8.5 k=m = + a j + a j a j + + a M a N, j j <j

11 where, e.g., M j < j N. Hece 8.53 N + a k a j + a j a j + + a M a N j j <j N = + a k k=m k=m = b MN, the last idetity defiig b MN. Our task is to ivestigate whe b MN as M, uiformly i N > M. To do this, we ote that 8.54 ad use the facts 8.55 log + b MN = log = N k=m N + a k k=m log + a k, x = log + x x, x = log + x x. Assumig a k ad takig M so large that k M a k /, we have 8.56 ad hece N k=m 8.57 lim M b MN =, a k log + b MN N k=m a k, uiformly i N > M k a k <. Cosequetly, 8.58 a k < = k = + a k coverges k= + a k coverges. k= Aother cosequece of 8.57 is the followig: 8.59 If + a k for all k, the a k < + a k. We ca replace the sequece a k of complex umbers by a sequece f k of holomorphic fuctios, ad deduce from the estimates above the followig. k=

12 Propositio 8.6. Let f k : Ω C be holomorphic. If 8.6 f k z < o Ω, the we have a coverget ifiite product k f k z = gz, k= ad g is holomorphic o Ω. If z Ω ad + f k z for all k, the gz. Aother cosequece of estimates leadig to 8.57 is that if also g k : Ω C ad g k z < o Ω, the 8.6 { } + f k z + g k z = + f k z + g k z. k= k= k= To make cotact with the Gamma fuctio, ote that the ifiite product i 8.9 has the form 8.6 with f k z = To see that 8.6 applies, ote that + z e z/k. k 8.64 e w = w + Rw, w Rw C w. Hece z e z/k = + z z z k k k + R k = z k + z z R. k k Hece 8.63 holds with 8.66 f k z = z k + z z R, k k so 8.67 f k z C z k which yields 8.6. for k z,

13 3 J. Euler s costat Here we say more about Euler s costat, itroduced i 8.7, i the course of producig the Euler product expasio for /Γz. The defiitio J. γ = lim k log + k= of Euler s costat ivolves a very slowly coverget sequece. I order to produce a umerical approximatio of γ, it is coveiet to use other formulas, ivolvig the Gamma fuctio Γz = e t t z dt. Note that J. Γ z = log te t t z dt. Meawhile the Euler product formula /Γz = ze γz = + z/e z/ implies J.3 Γ = γ. Thus we have the itegral formula J.4 γ = log te t dt. To evaluate this itegral umerically it is coveiet to split it ito two pieces: J.5 γ = = γ a γ b. log te t dt log te t dt We ca apply itegratio by parts to both the itegrals i 5, usig e t = d/dte t o the first ad e t = d/dte t o the secod, to obtai J.6 γ a = e t t dt, γ b = e t t dt. Usig the power series for e t ad itegratig term by term produces a rapidly coverget series for γ a : J.7 γ a = k= k. k k!

14 4 Before producig ifiite series represetatios for γ b, we ote that the chage of variable t = s m gives e sm J.8 γ b = m ds, s which is very well approximated by the itegral over s [, if m =, for example. To produce ifiite series for γ b, we ca break up [, ito itervals [k, k + ad take t = s + k, to write J.9 γ b = k= e k k β k, β k = e t + t/k dt. Note that < β k < /e for all k. For k we ca write J. β k = jαj, α j = k j= t j e t dt. Oe coveiet way to itegrate t j e t is the followig. Write J. E j t = The j l= t l l!. J. hece J.3 so J.4 E j t = E j t, d Ej te t = E j t E j t e t = tj dt j! e t, t j e t dt = j!e j te t + C. I particular, J.5 α j = t j e t dt = j! e = j! e = e l=j+ j l= l! l! j + + j + j + +.

15 5 To evaluate β as a ifiite series, it is coveiet to write J.6 e β = = e t dt t j j= = log + j! j= t j dt j j j! j. To summarize, we have γ = γ a γ b, with γ a give by the coveiet series J.7 ad J.7 γ b = k= j= e k k k jαj + log + j= j j j! j, with α j give by J.5. We ca reverse the order of summatio of the double series ad write J.8 γ b = with j ζ j α j + log + j= J.9 ζ j = Note that k= j= e k k j+. j j j! j. J. < ζ j < j+ e k < j+3, k= while J.5 readily yields < α j < /ej. So oe ca expect 5 digits of accuracy by summig the first series i J.8 over j 5 ad the secod series over j 3, assumig the igrediets α j ad ζ j are evaluated sufficietly accurately. It suffices to sum J.9 over k 4 j/3 to evaluate ζ j to sufficiet accuracy. Note that the quatities α j do ot have to be evaluated idepedetly. Say you are summig the first series i J.8 over j 5. First evaluate α 5 usig terms i J.5, ad the evaluate iductively α 49,..., α usig the idetity J. α j = je + α j j,

16 6 equivalet to α j = jα j /e, which follows by itegratio by parts of tj e t dt. If we sum the series J.7 for γ a over k ad either sum the series J.8 as described above or have Mathematica umerically itegrate J.8, with m =, to high precisio, we obtai J. γ , which is accurate to 5 digits. We give aother series for γ. This oe is more slowly coverget tha the series i J.7 ad J.8, but it makes clear why γ exceeds / by a small amout, ad it has other iterestig aspects. We start with J.3 γ = γ, γ = + = dx x = log +. Thus γ is the area of the regio J.4 Ω = { x, y : x +, x y }. This regio cotais the triagle T with vertices, /, +, /, ad +, / +. The regio Ω \ T is a little sliver. Note that J.5 Area T = δ = ad hece, + J.6 δ =. = Thus J.7 γ = γ δ + γ δ + γ 3 δ 3 +. Now J.8 γ δ = 3 4 log, while, for, we have power series expasios J.9 γ = δ = 3 + 4,

17 7 the first expasio by log + z = z z / + z 3 /3, ad the secod by J.3 δ = + = +, ad the expasio + z = z + z. Hece we have J.3 γ = γ δ , or, with J.3 we have J.33 γ = 3 4 log + ζk = k, [ζ3 ] 3 [ζ4 ] +, 4 a alteratig series from the third term o. We ote that J.34 The estimate J.35 implies 3 log.56858, 4 [ζ3 ].33676, 6 [ζ4 ].588, 4 3 [ζ5 ].783. J.36 < k < k + x k dx k [ζk ] < k, so the series J.33 is geometrically coverget. If k is eve, ζk is a kow ratioal multiple of π k. However, for odd k, the values of ζk are more mysterious. Note that to get ζ3 to 6 digits by summig J.3 oe eeds to sum over 8. O a.3 GHz persoal computer, a C program does this i 4 secods. Of course, this is vastly slower tha summig J.7 ad J.8 over the rages discussed above. Referece [T] M. Taylor, Itroductio to Complex Aalysis. website. Notes, available o this