Steven R. Dunbar
Department of Mathematics
203 Avery Hall
University of Nebraska-Lincoln
Lincoln, NE 68588-0130
http://www.math.unl.edu
Voice: 402-472-3731
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Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar

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Stirling’s Formula in Real and Complex Variables

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Rating

Rating

Mathematicians Only: prolonged scenes of intense rigor.

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Section Starter Question

Section Starter Question

Can you name a function defined as an integral? Can you name a function defined as a limit? Can you name a function defined as an infinite product? What is the Hadamard Product Theorem in complex analysis?

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Key Concepts

Key Concepts

  1. The Gamma function is defined as
    Γ(z) =0xz1ex dx.

  2. Stirling’s Formula in real variables is
    Γ(z) z 1 e z12π (z 1)

    as z .

  3. Gauss’s Formula for the Gamma function is
    Γ(z) = lim n nzn! z(z + 1)(z + 2)(z + n)

    valid for complex values of z 1,2,3,.

  4. Several important properties of the Gamma function follow immediately from Gauss’s formula.

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Vocabulary

Vocabulary

  1. The Gamma function is defined as
    Γ(z) =0xz1ex dx.

  2. Euler’s constant (also called the Euler-Mascheroni constant) is
    γ = lim n j=1n1 j log(n) .

  3. Gauss’s Formula for the Gamma function is
    Γ(z) = lim n nzn! z(z + 1)(z + 2)(z + n)

    valid for complex values of z 1,2,3,.

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Mathematical Ideas

Mathematical Ideas

Stirling’s Formula in Real Variables

The Gamma function is defined as

Γ(z) =0xz1ex 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

Γ(z) z 1 e z12π (z 1) (1)

as z .

Start with the same change of variables as in Lemma 2 of the section Stirlings Formula Derived from the Gamma Function..

Lemma 1.

Γ(z) = z 1 e z12π (z 1)z1g z1(v) dv .

where

gz1(y) = 1 + y z 1 z1eyz1

Proof. In the integral representation of Γ(z) make the substitution x = yz 1 + z 1 (or equivalently y = x z1 z 1) with dx = z 1 dy to give

0xz1ex dx =z1(yz 1 + z 1)z1e(yz1+z1)z 1 dy = (z 1)z1z 1e(z1)z1(yz 1 + 1)z1e(yz1) dy = (z 1)z1z 1e(z1)z1g z1(y) dy .

Provided that we can show

lim zz1g z1(y) dy = lim zz1(1+ y z 1)z1e(yz1) dy = 2π

this is Stirling’s Formula as expressed in (1).

Lemma 2. Let L be a large finite value, then

lim zgz1(v) = ev22

as z , uniformly for v [L,L].

Remark. Compare this result to Lemma 5 of the section Stirling’s Formula Derived from the Gamma Function..

Proof. Note that gz1(v) is defined for v > z 1. Fix L as a large value, then

log(gz1(v)) = (z 1) log(1 + v z 1) v z 1

is defined on the interval [L,L] as long as z 1 < L, or z 1 > L2. Then the conclusion follows from Lemma 3 in Stirling’s Formula Derived from the Gamma Function..

Alternatively,

1 + y z 1 z1eyz1 ey22  if and only if (z 1) log 1 + y z 1 yz 1 y2 2  if and only if (z 1) log 1 + y z 1 y z 1 y2 2  if and only if log 1 + y z 1 y z 1 y2 2(z 1).

Now letting u = yz 1, we wish to show that log(1 + u) u + u22 0 as u 0. The function log(1 + u) u + u22 is increasing on (1,) since its derivative is 1(1 + u) 1 + u 0. Then the maximum of | log(1 + u) u + u22| on an interval occurs at the endpoints. If 1 < a < 0 < a then

max [a,a]| log(1+x)x+x22| = max{| log(1a)+a+a22|,| log(1+a)a+a22|}.

Continuity of the functions log(1 a) + a + a22 and log(1 + a) a + a22 in the neighborhood of 0 shows that log(1 + u) u + u22 0 as u 0 uniformly on u [Lz 1,Lz 1]. □

Lemma 3.

lim zz11 + y z 1 z1eyz1 dy = 2π.

Proof. Take z to be a large finite value and fix 1 < L < z 1. Then

z1g z1(y) dy =z1Lg z1(y) dy +LLg z1(y) dy +Lg z1(y) dy

Using Lemma 2,

LLg z1(y) dy LLey22 dy .

For temporary convenience in the proof, set

h(y) = (z 1) log(1 + y z 1) y z 1

so that gz1(y) = eh(y). Note that

h(y) = yz 1 z 1 + y

so that h(y) < h(L) < 0 for 0 < L < y. Note also that h(y) y22 as z . Now estimate the upper tail Lg z1(y) dy =Leh(y) dy by

Leh(y) dy 1 h(L)Lh(y)eh(y) dy = 1 h(L)eh(L) 1 LeL22   as z .

Estimate the lower tail z1Leh(y) dy. Since h(L) < h(y) < 0 for z 1 < L < y < 0

z1Leh(z) dz 1 h(L)z1Lh(y)eh(y) dy = 1 h(L)(eh(L) eh(z1)) 1 LeL22   as z .

Taking z and L sufficiently large makes z11 + y z1 z1eyz1 dy 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 only asymptotic limits and not bounds.

Remark. Another derivation of the asymptotic 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 Bn, and then using the definition of the Gamma function as the derivative of the logarithm of the digamma function, one can derive the asymptotic expansion

log(Γ(z + 1)) 1 2 log(2π) + (z + 1 2) log(z) z + 1 2 n=1 B2n n(2n 1) 1 z2n1.

By exponentiating both sides of this asymptotic limit we can obtain

Γ(z + 1) 2π zzzez exp 1 12z 1 360z2 + .

By expressing the last exponential in a Maclaurin series, we can express this as

Γ(z + 1) 2π zzzez 1 + 1 12z + 1 288z2 + .

See [2] for a sketch of this proof of this asymptotic series.

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 only sketches the proofs because of the extensive background required in complex variable theory. The bibliography provides references for the proofs.

Theorem 4. The expansion of sin(z) as an infinite product is

sin(z) = z m=11 z2 m2π2 .

Proof. See [6, page 312] for the proof. The article in the Mathworld.com article on http://mathworld.wolfram.com/Sine.html. also gives references to Edwards 2001, pages 18 and 47; and Borwein et al. 2004, page 5. □

Corollary 1 (Wallis’s Formula).

lim n 2 2 4 4 6 6(2n) (2n) 1 3 3 5 5(2n 1) (2n 1) (2n + 1) = π 2.

Proof. Substitute z = π2 in the product expansion of sin(z). □

Definition. Euler’s constant (also called the Euler-Mascheroni constant) is

γ = lim n j=1n1 j log(n) .

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 by 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) = eγzz n=11 + z nez n

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 195-198, Saks and Zygmund [6] gives a similar proof on pages 414-415.

Corollary 2.

Γ(z) = lim n nzn! z(z + 1)(z + 2)(z + n)

Saks and Zygmund [6, page 313] say that this is known as Gauss’s Formula for the Gamma function. Veech [7] gives the same proof.

Corollary 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 z0,1 2,1, 3 2,
    Γ(z)Γ(z + 1 2) = 212zπΓ(2z)

    which is known as Legendre’s Duplication Formula.

  6. Γ(12) = π.

Proof. Veech [7, page 193] gives a proof using

P(z) = n=11 + z nezn

to show zP(z)P(z) = sin(πz)π, then reducing with Gauss’s Formula. □

Theorem 5 (Stirling’s Formula). For 0 < δ < π, as z goes to within the sector π + δ arg z π δ

Γ(z) 2πezzz12.

Proof. Saks and Zygmund [6] give a proof on pages 421-424 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. They cite de Bruijn, [4] who uses the saddlepoint method from asymptotic analysis.
  2. They say that Artin, [3], gives a proof based on the fact that the Gamma Function is the only function which is log-convex on (0,), satisfies Γ(z + 1) = zΓ(z) and Γ(1) = 1.
  3. A third approach due to Lindelof in Ahlfors, [1] uses residue calculus from complex analysis.

Sources

The proof of the real variable form of Stirling’s Formula is adapted from the short article by Diaconis and Freedman, [5]. The results on the complex variable form of the Gamma function and Stirling’s Formula are drawn from Saks and Zygmund, [6] and Veech, [7].

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Problems to Work

Problems to Work for Understanding

  1. Show that
    lim zgz1(v) = ev22

    as z , uniformly on [L,L] follows from Lemma 3 in Stirling’s Formula Derived from the Gamma Function.

  2. Given ϵ > 0 show that there is a value Z so large that
    z11 + y z 1 z1eyz1 dy 2π.

    for z > Z.

  3. Prove Γ(z + 1) = zΓ(z) by using Gauss’s Formula.
  4. Prove Γ(1) = 1 by using Gauss’s Formula.
  5. Show that
    Γ(n + 1 2) = (2n)! 22nn!π

    and in particular that Γ(1 2) = 1 2π.

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Books

Reading Suggestion:

References

[1]   L. Ahlfors. Complex Analysis. McGraw Hill, 3rd edition, 1979.

[2]   Larry C. Andrews. Special Functions for Engineers and Applied Mathematicians. MacMillan, 1985.

[3]   E. Artin. The Gamma Function. Holt, Rinehart, Winston, 1964.

[4]   N. G. de Bruijn. Asymptotic Methods in Analysis. Dover, 1981.

[5]   P. Diaconis and D. Freedman. An elmentary proof of Stirling’s formula. American Mathematical Monthly, 93:123–126, 1986.

[6]   S. Saks and A. Zygmund. Analytic Functions. Elsevier Publishing, 1971.

[7]   William A. Veech. A Second Course in Complex Analysis. W. A. Benjamin Inc., New York, 1967.

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Links

Outside Readings and Links:

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Last modified: Processed from LATEX source on May 23, 2011