Steven R. Dunbar
Department of Mathematics
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University of Nebraska-Lincoln
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Topics in
Probability Theory and Stochastic Processes
Steven R. Dunbar
Stirling’s Formula in Real and Complex Variables
Rating
Mathematicians Only: prolonged scenes of intense rigor.
1
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?
Key Concepts
1. The Gamma function is defined as
Z ∞
Γ(z) =
xz−1 e−x dx .
0
2. Stirling’s Formula in real variables is
z−1
p
z−1
Γ(z) ∼
2π(z − 1)
e
as z → ∞.
3. Gauss’s Formula for the Gamma function is
nz n!
n→∞ z(z + 1)(z + 2) . . . (z + n)
Γ(z) = lim
valid for complex values of z 6= −1, −2, −3, . . ..
4. Several important properties of the Gamma function follow immediately from Gauss’s formula.
2
Vocabulary
1. The Gamma function is defined as
Z ∞
Γ(z) =
xz−1 e−x dx .
0
2. Euler’s constant (also called the Euler-Mascheroni constant) is
!
n
X
1
− log(n) .
γ = lim
n→∞
j
j=1
3. Gauss’s Formula for the Gamma function is
nz n!
Γ(z) = lim
n→∞ z(z + 1)(z + 2) . . . (z + n)
valid for complex values of z 6= −1, −2, −3, . . ..
Mathematical Ideas
Stirling’s Formula in Real Variables
The Gamma function is defined as
Z ∞
Γ(z) =
xz−1 e−x dx .
0
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
p
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
Lemma 1.
Γ(z) =
z−1
e
z−1
Z
p
2π(z − 1)
∞
√
− z−1
where
gz−1 (y) =
y
1+ √
z−1
z−1
e−y
gz−1 (v) dv .
√
z−1
Proof.
In the integral representation of Γ(z)√make the substitution
x =
√
√
x
y z − 1 + z − 1 (or equivalently y = √z−1
− z − 1) with dx = z − 1 dy
to give
Z ∞
Z ∞
√
√
√
z−1 −x
x e dx = √ (y z − 1 + z − 1)z−1 e−(y z−1+z−1) z − 1 dy
0
− z−1
Z ∞
√
√
√
z−1
−(z−1)
z−1 −(y z−1)
= (z − 1)
dy
z − 1e
(y/
z
−
1
+
1)
e
√
− z−1
Z ∞
√
z−1
−(z−1)
= (z − 1)
z − 1e
gz−1 (y) dy .
√
− z−1
Provided that we can show
Z ∞
Z
lim √ gz−1 (y) dy = lim
z→∞
− z−1
z→∞
√
√
y
z−1 −(y z−1)
√
(1
+
)
e
2π
dy
=
√
z−1
− z−1
∞
this is Stirling’s Formula as expressed in (1).
Lemma 2. Let L be a large finite value, then
lim gz−1 (v) = e−v
2 /2
z→∞
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.
4
√
Proof. Note that gz−1 (v) is defined for v > − z − 1. Fix L as a large value,
then
v
v
log(gz−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 > L2 .
Then the conclusion follows from Lemma 3 in Stirling’s Formula Derived
from the Gamma Function.
Alternatively,
z−1
√
y
2
e−y z−1 → e−y /2
1+ √
z−1
if and only if
√
y
−y 2
(z − 1) log 1 + √
−y z−1→−
2
z−1
if and only if
y
y
−y 2
(z − 1) log 1 + √
−√
→−
2
z−1
z−1
if and only if
y
−y 2
y
log 1 + √
→−
.
−√
2(z − 1)
z−1
z−1
√
Now letting u = y/ z − 1, we wish to show that log(1 + u) − u + u2 /2 → 0 as
u → 0. The function log(1 + u) − u + u2 /2 is increasing on (−1, ∞) since its
derivative is 1/(1+u)−1+u ≥ 0. Then the maximum of | log(1+u)−u+u2 /2|
on an interval occurs at the endpoints. If −1 < −a < 0 < a then
max | log(1+x)−x+x2 /2| = max{| log(1−a)+a+a2 /2|, | log(1+a)−a+a2 /2|}.
[−a,a]
Continuity of the functions log(1 − a) + a + a2 /2 and log(1 + a) − a + a2 /2
2
in the neighborhood of√0 shows that
√ log(1 + u) − u + u /2 → 0 as u → 0
uniformly on u ∈ [−L/ z − 1, L/ z − 1].
5
Lemma 3.
Z
lim
z→∞
∞
√
− z−1
y
1+ √
z−1
z−1
e−y
√
z−1
dy =
Proof. Take z to be a large finite value and fix 1 < L <
Z
√
√
2π .
z − 1. Then
∞
√
− z−1
gz−1 (y) dy
Z
=
−L
Z
√
− z−1
L
gz−1 (y) dy +
Z
∞
gz−1 (y) dy +
−L
gz−1 (y) dy
L
Using Lemma 2,
Z
L
Z
L
gz−1 (y) dy →
e−y
2 /2
dy .
−L
−L
For temporary convenience in the proof, set
y
y
h(y) = (z − 1) log(1 + √
)− √
z−1
z−1
so that gz−1 (y) = eh(y) . Note that
√
−y z − 1
h (y) = √
z−1+y
0
so that h0 (y) < h0 (L) < 0 for 0 < L <
→ −y 2 /2 as
R ∞y. Note also that
R ∞ h(y)
z → ∞. Now estimate the upper tail L gz−1 (y) dy = L eh(y) dy by
Z ∞
Z ∞
1
h(y)
e
dy ≤ 0
h0 (y)eh(y) dy
h
(L)
L
L
1 h(L)
= 0
e
h (L)
1
2
→ e−L /2 as z → ∞ .
L
6
R −L
Estimate the lower tail −√z−1 eh(y) dy. Since h0 (−L) < h0 (y) < 0 for
√
− z − 1 < −L < y < 0
Z −L
Z −L
1
h(z)
e
dz ≤ 0
h0 (y)eh(y) dy
√
√
h (−L) − z−1
− z−1
√
1
(eh(−L) − eh(− z−1) )
= 0
h (−L)
1
2
→ e−L /2 as z → ∞ .
L
z−1
√
R∞
y
√
√
Taking z and L sufficiently large makes − z−1 1 + z−1
e−y z−1 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
∞
1
1X
B2n
1
1
.
log(Γ(z + 1)) ∼ log(2π) + (z + ) log(z) − z +
2n−1
2
2
2 n=1 n(2n − 1) z
By exponentiating both sides of this asymptotic limit we can obtain
√
1
1
z −z
+ ... .
Γ(z + 1) ∼ 2πzz e exp
−
12z 360z 2
By expressing the last exponential in a Maclaurin series, we can express this
as
√
1
1
z −z
Γ(z + 1) ∼ 2πzz e
1+
+
+ ... .
12z 288z 2
See [2] for a sketch of this proof of this asymptotic series.
7
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
∞ Y
m=1
z2
1− 2 2
mπ
.
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).
2 · 2 · 4 · 4 · 6 · 6 . . . (2n) · (2n)
π
= .
n→∞ 1 · 3 · 3 · 5 · 5 . . . (2n − 1) · (2n − 1) · (2n + 1)
2
lim
Proof. Substitute z = π/2 in the product expansion of sin(z).
Definition. Euler’s constant (also called the Euler-Mascheroni constant) is
!
n
X
1
γ = lim
− log(n) .
n→∞
j
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 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
∞ Y
1
z −z
γz
=e z
1+
e n
Γ(z)
n
n=1
8
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.
nz n!
n→∞ z(z + 1)(z + 2) . . . (z + n)
Γ(z) = lim
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 z 6= 0, − 12 , −1, 32 , . . .
√
1
Γ(z)Γ(z + ) = 21−2z πΓ(2z)
2
which is known as Legendre’s Duplication Formula.
√
6. Γ(1/2) = π.
Proof. Veech [7, page 193] gives a proof using
P (z) =
∞ Y
1+
n=1
z −z/n
e
n
to show zP (z)P (−z) = sin(πz)/π, then reducing with Gauss’s Formula.
9
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 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].
Problems to Work for Understanding
1. Show that
lim gz−1 (v) = e−v
2 /2
z→∞
as z → ∞, uniformly on [−L, L] follows from Lemma 3 in Stirling’s
Formula Derived from the Gamma Function
10
2. Given > 0 show that there is a value Z so large that
Z
z−1
∞ √
√
y
1+ √
e−y z−1 dy − 2π .
√
− z−1
z−1
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
1
(2n)! √
Γ(n + ) = 2n
π
2
2 n!
√
and in particular that Γ( 21 ) = 12 π.
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.
11
[7] William A. Veech. A Second Course in Complex Analysis. W. A. Benjamin Inc., New York, 1967.
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