Topics in Probability Theory and Stochastic Processes Steven R. Dunbar. Stirling s Formula in Real and Complex Variables

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1 Steven R. Dunbar Department of Mathematics 203 Aver Hall Universit of Nebraska-Lincoln Lincoln, NE Voice: Fax: Topics in Probabilit Theor and Stochastic Processes Steven R. Dunbar Stirling s Formula in Real and Complex Variables Rating Mathematicians Onl: prolonged scenes of intense rigor. 1

2 Section Starter Question Can ou name a function defined as an integral? Can ou name a function defined as a limit? Can ou name a function defined as an infinite product? What is the Hadamard Product Theorem in complex analsis? Ke Concepts 1. The Gamma function is defined as Γ(z) = 2. Stirling s Formula in real variables is 0 x z 1 e x dx. as z. ( ) z 1 z 1 Γ(z) 2π(z 1) e 3. Gauss s Formula for the Gamma function is Γ(z) = lim n n z n! z(z + 1)(z + 2)... (z + n) valid for complex values of z 1, 2, 3, Several important properties of the Gamma function follow immediatel from Gauss s formula. 2

3 Vocabular 1. The Gamma function is defined as Γ(z) = 0 x z 1 e x dx. 2. Euler s constant (also called the Euler-Mascheroni constant) is ( n ) 1 γ = lim n j log(n). j=1 3. Gauss s Formula for the Gamma function is Γ(z) = lim n n z n! z(z + 1)(z + 2)... (z + n) valid for complex values of z 1, 2, 3,.... Mathematical Ideas Stirling s Formula in Real Variables The Gamma function is defined as Γ(z) = 0 x z 1 e x dx. For an integer n, Γ(n) = (n 1)!. See the proof in Lemma 1 in Stirlings Formula Derived from the Gamma Function. Stirling s Formula in real variables is ( ) z 1 z 1 Γ(z) 2π(z 1) (1) e as z. Start with the same change of variables as in Lemma 2 of the section Stirlings Formula Derived from the Gamma Function. 3

4 Lemma 1. ( ) z 1 z 1 Γ(z) = 2π(z 1) g e (v) dv. z 1 where g z 1 () = ( ) z e z 1 z 1 Proof. In the integral representation of Γ(z) make the substitution x = z 1 + z 1 (or equivalentl = x z 1 z 1) with dx = z 1 d to give x z 1 e x dx = 0 ( z 1 + z 1) z 1 e ( z 1+z 1) z 1 d = (z 1) z 1 z 1e (z 1) (/ z 1 + 1) z 1 e ( z 1) d = (z 1) z 1 z 1e (z 1) g z 1 () d. Provided that we can show lim z g z 1 () d = lim (1 + ) z 1 e ( z 1) d = 2π z z 1 this is Stirling s Formula as expressed in (1). Lemma 2. Let L be a large finite value, then as z, uniforml for v [ L, L]. lim g z 1(v) = e v2 /2 z Remark. Compare this result to Lemma 5 of the section Stirling s Formula Derived from the Gamma Function. 4

5 Proof. Note that g z 1 (v) is defined for v > z 1. Fix L as a large value, then [ ] v v log(g z 1 (v)) = (z 1) log(1 + ) z 1 z 1 is defined on the interval [ L, L] as long as z 1 < L, or z 1 > L 2. Then the conclusion follows from Lemma 3 in Stirling s Formula Derived from the Gamma Function. Alternativel, ( ) z e z 1 e 2 /2 z 1 if and onl if if and onl if ( ) (z 1) log 1 + z 1 2 z 1 2 [ ( (z 1) log 1 + z 1 ) ] 2 z 1 2 if and onl if ( ) log z 1 z 1 2(z 1). Now letting u = / z 1, we wish to show that log(1+u) u+u 2 /2 0 as u 0. The function log(1 + u) u + u 2 /2 is increasing on ( 1, ) since its derivative is 1/(1+u) 1+u 0. Then the maximum of log(1+u) u+u 2 /2 on an interval occurs at the endpoints. If 1 < a < 0 < a then max [ a,a] log(1+x) x+x2 /2 = max{ log(1 a)+a+a 2 /2, log(1+a) a+a 2 /2 }. Continuit of the functions log(1 a) + a + a 2 /2 and log(1 + a) a + a 2 /2 in the neighborhood of 0 shows that log(1 + u) u + u 2 /2 0 as u 0 uniforml on u [ L/ z 1, L/ z 1]. 5

6 Lemma 3. ( ) z 1 lim 1 + e z 1 d = 2π. z z 1 Proof. Take z to be a large finite value and fix 1 < L < z 1. Then g z 1 () d = L L g z 1 () d + g z 1 () d + L L g z 1 () d Using Lemma 2, L L g z 1 () d L L For temporar convenience in the proof, set [ h() = (z 1) log(1 + e 2 /2 d. z 1 ) ] z 1 so that g z 1 () = e h(). Note that h () = z 1 z 1 + so that h () < h (L) < 0 for 0 < L <. Note also that h() 2 /2 as z. Now estimate the upper tail g L z 1() d = L eh() d b L e h() d 1 h (L) L = 1 h (L) eh(l) h ()e h() d 1 L e L2 /2 as z. 6

7 Estimate the lower tail L z 1 < L < < 0 eh() d. Since h ( L) < h () < 0 for L L e h(z) 1 dz h ()e h() d h ( L) 1 = h ( L) (eh( L) e h() ) 1 L e L2 /2 as z. Taking z and L sufficientl large makes ( ) z z 1 z 1 e d as close as desired to 2π. Remark. The estimate used here is much less precise than the estimates proved in Lemma 9 in Stirling s Formula Derived from the Gamma Function. Therefore, the results are onl asmptotic limits and not bounds. Remark. Another derivation of the asmptotic limit for Γ(z + 1) starts with the Frullani integral representation of the digamma function. Expanding the integrand in a power series, defining the Bernoulli numbers B n, and then using the definition of the Gamma function as the derivative of the logarithm of the digamma function, one can derive the asmptotic expansion log(γ(z + 1)) 1 2 log(2π) + (z ) log(z) z B 2n 1 n(2n 1) z. 2n 1 B exponentiating both sides of this asmptotic limit we can obtain Γ(z + 1) ( 1 2πzz z e z exp 12z 1 ) 360z B expressing the last exponential in a Maclaurin series, we can express this as Γ(z + 1) ( 2πzz z e z z + 1 ) 288z n=1 See [2] for a sketch of this proof of this asmptotic series. 7

8 Stirling s Formula in Complex Variables This section provides the statements of Stirling s Formula for the complex variable form of the Gamma Function. This section onl sketches the proofs because of the extensive background required in complex variable theor. The bibliograph provides references for the proofs. Theorem 4. The expansion of sin(z) as an infinite product is sin(z) = z (1 z2 m=1 m 2 π 2 Proof. See [6, page 312] for the proof. The article in the Mathworld.com article on also gives references to Edwards 2001, pages 18 and 47; and Borwein et al. 2004, page 5. Corollar 1 (Wallis s Formula). lim n ) (2n) (2n) (2n 1) (2n 1) (2n + 1) = π 2. Proof. Substitute z = π/2 in the product expansion of sin(z). Definition. Euler s constant (also called the Euler-Mascheroni constant) is ( n ) 1 γ = lim n j log(n). j=1 See [6, page 313] for the motivation and derivation, as well as a proof that the limit exists using the Integral Test for convergence of series. See also [7, page 190] for an alternative proof of existence b appealing to general theorems on convergence of product representations of entire functions of order 1. Definition. Define the reciprocal of the Gamma Function as a product 1 Γ(z) = eγz z n=1 ( 1 + z ) e z n n 8

9 Remark. See [6, page 313] or [7, page 192] for the justification of the construction as a meromorphic function with simple poles at the points z = 0, 1, 2,.... Of course, with this definition of the Gamma function, it remains to be shown that the function satisfies the usual integral representation as in the Definition at the beginning of the section. Veech [7] gives the proof in Section 12, pages , Saks and Zgmund [6] gives a similar proof on pages Corollar 2. Γ(z) = lim n n z n! z(z + 1)(z + 2)... (z + n) Saks and Zgmund [6, page 313] sa that this is known as Gauss s Formula for the Gamma function. Veech [7] gives the same proof. Corollar 3. From Gauss s Formula it follows that 1. Γ(z + 1) = zγ(z) 2. Γ(1) = 1 3. Γ(n + 1) = n! 4. For z not an integer, Γ(z)Γ(1 z) = π sin(πz) 5. For z 0, 1 2, 1, 3 2,... Γ(z)Γ(z ) = 21 2z πγ(2z) which is known as Legendre s Duplication Formula. 6. Γ(1/2) = π. Proof. Veech [7, page 193] gives a proof using P (z) = n=1 ( 1 + z ) e z/n n to show zp (z)p ( z) = sin(πz)/π, then reducing with Gauss s Formula. 9

10 Theorem 5 (Stirling s Formula). For 0 < δ < π, as z goes to within the sector π + δ arg z π δ Γ(z) 2πe z z z 1/2. Proof. Saks and Zgmund [6] give a proof on pages using Gauss s Formula and a version of the Euler-Maclaurin summation formula. Remark. Diaconis and Freedman [5] mention three other proofs of the complex variable form of Stirling s Formula for the Gamma function: 1. The cite de Bruijn, [4] who uses the saddlepoint method from asmptotic analsis. 2. The sa that Artin, [3], gives a proof based on the fact that the Gamma Function is the onl function which is log-convex on (0, ), satisfies Γ(z + 1) = zγ(z) and Γ(1) = A third approach due to Lindelof in Ahlfors, [1] uses residue calculus from complex analsis. Sources The proof of the real variable form of Stirling s Formula is adapted from the short article b Diaconis and Freedman, [5]. The results on the complex variable form of the Gamma function and Stirling s Formula are drawn from Saks and Zgmund, [6] and Veech, [7]. Problems to Work for Understanding 1. Show that lim g z 1(v) = e v2 /2 z as z, uniforml on [ L, L] follows from Lemma 3 in Stirling s Formula Derived from the Gamma Function 10

11 2. Given ɛ > 0 show that there is a value Z so large that ( ) z e z 1 d 2π z 1. for z > Z. 3. Prove Γ(z + 1) = zγ(z) b using Gauss s Formula. 4. Prove Γ(1) = 1 b using Gauss s Formula. 5. Show that Γ(n ) = (2n)! π 2 2n n! and in particular that Γ( 1 2 ) = 1 2 π. Reading Suggestion: References [1] L. Ahlfors. Complex Analsis. McGraw Hill, 3rd edition, [2] Larr C. Andrews. Special Functions for Engineers and Applied Mathematicians. MacMillan, [3] E. Artin. The Gamma Function. Holt, Rinehart, Winston, [4] N. G. de Bruijn. Asmptotic Methods in Analsis. Dover, [5] P. Diaconis and D. Freedman. An elmentar proof of Stirling s formula. American Mathematical Monthl, 93: , [6] S. Saks and A. Zgmund. Analtic Functions. Elsevier Publishing,

12 [7] William A. Veech. A Second Course in Complex Analsis. W. A. Benjamin Inc., New York, Outside Readings and Links: I check all the information on each page for correctness and tpographical errors. Nevertheless, some errors ma occur and I would be grateful if ou would alert me to such errors. I make ever reasonable effort to present current and accurate information for public use, however I do not guarantee the accurac or timeliness of information on this website. Your use of the information from this website is strictl voluntar and at our risk. I have checked the links to external sites for usefulness. Links to external websites are provided as a convenience. I do not endorse, control, monitor, or guarantee the information contained in an external website. I don t guarantee that the links are active at all times. Use the links here with the same caution as ou would all information on the Internet. This website reflects the thoughts, interests and opinions of its author. The do not explicitl represent official positions or policies of m emploer. Information on this website is subject to change without notice. Steve Dunbar s Home Page, to Steve Dunbar, sdunbar1 at unl dot edu Last modified: Processed from L A TEX source on Ma 23,