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“(± + β)
Finally, we show the following result upon the asymptotic behavior of the
Gamma function for z ’ ∞:
Theorem 68 (Stirling™s Formula) For |z| ’ ∞ in sectors | arg z| ¤
π ’ µ, for every µ > 0, we have

ez z
“(z) √ ’ 1. (B.12)
2π z z

fn z ’n , with
ˆ
More precisely, there is a formal power series f (z) = 0
constant term f0 = 1, so that

ez z ∼ ˆ
“(z) √ | arg z| < π,
=1 f (z),
2π z z
in the sense of Section 4.5.

Proof: We give a proof especially suited to the theory developed in previous
chapters; the reader may wish to verify that here we only use the de¬nition
of 1-summability and some results on integral equations, and in particular
do not rely on anything for which we have used (B.12).
As was shown above, the Gamma function is a solution of the di¬erence

equation x(z + 1) = z x(z), and putting y(z) = ez z z ’z x(z), we ¬nd
y(z + 1) = a(z) y(z), with
exp[1 ’ z log(1 + 1/z)]
= 1 + a2 z ’2 + a3 z ’3 + . . . , |z| > 1.
a(z) =
1 + 1/z

yn z ’n ; hence
Formally setting y(z) = y (z) =
ˆ 0
∞ n
’m
’n
y(z + 1) = z ym ,
n’m
n=0 m=0

it is not di¬cult to see that, for arbitrary y0 , the power series formally
satis¬es the di¬erence equation if and only if the remaining coe¬cients are
given by the recursion
n’1
’m
’ an+1’m ym , n ≥ 1.
n yn =
n+1’m
m=0
∞ ∞
n n
Set t(u) = 0 u yn+1 /n!, k(u) = 1 u an+1 /n!, with an being the
coe¬cients in the power series expansion of a(z); in particular, observe
230 Appendix B. Functions with Values in Banach Spaces

a1 = 0. Then it is readily seen that formally t(u) satis¬es the integral
equation
u
’u
’ 1) t(u) = y0 k(u) + k(u ’ w) t(w) dw.
(e
0

Since k(u) vanishes at the origin, we can divide this integral equation by
u to eliminate the ¬rst-order zero of the factor e’u ’ 1 at the origin, and
then arguments very similar to the ones used in the proof of Proposition 26
(p. 217) show the existence and uniqueness of a solution t(u) of this integral
equation that is analytic in the complex plane with cuts from ±2πi along
the positive resp. negative imaginary axes, because of the other zeros of
e’u ’ 1, and the function t(u) satis¬es an estimate of the form

|t(u)| ¤ K ec|u| , | arg u| ¤ π/2 ’ µ,

for every µ > 0 and su¬ciently large c, K > 0 depending upon µ. The
power series expansion of t(u) then can be seen to be of the form as above;
hence, by de¬nition the formal series y (z) is 1-summable in all directions
ˆ
d with |d| < π/2, and its sum y(z) satis¬es the above di¬erence equation,

for every choice of y0 . Thus we are left to show that for y0 = 2π we have

y(z) = ez z z ’z “(z), but this follows from Exercise 4. 2


Exercises: In the following exercises, let G be a simply connected region.
1. Show that if f ∈ H(G, E ) has a power series expansion about some
z0 ∈ G in which all coe¬cients vanish, then f is identically zero.

2. Show that if f ∈ H(G, E ) satis¬es f (n) (z0 ) = 0 for some z0 ∈ G and
every n ≥ 0, then f vanishes identically. So in other words, if f (z0 ) =
0 but f (z) ≡ 0, then for some n ≥ 1 we must have f (n) (z0 ) = 0; the
number z0 then is called a root of f , and the minimal n with this
property is called the order of the root of f .

3. Show the Identity Theorem: If f ∈ H(G, E ) has in¬nitely many roots,
accumulating at some z0 ∈ G, then f is identically zero.

4. In this exercise, let x be a positive real number.
∞ ’x(t’log t) dt
(a) Show “(x) = xx e t.
0
(b) For t > 0, check that u = t ’ 1 ’ log t is strictly increasing and
make a corresponding change of variable to show
∞ ∞
’x(t’log t) dt ’x
e’xu f1 (u) du,
e =e
t
1 0

u f1 (u) ’
with a positive continuous function f1 (u) satisfying

1/ 2 as u ’ 0+.
B.3 Holomorphic Continuation 231

(c) For 0 < t < 1, check that u = t ’ 1 ’ log t is strictly decreasing
and make a corresponding change of variable to show
1 ∞
’x(t’log t) dt ’x
e’xu f2 (u) du,
e =e
t
0 0

u f2 (u) ’
with a positive continuous function f2 (u) satisfying

1/ 2 as u ’ 0+.
(d) Use the identities derived above to show


ex x
’ 2π,
“(x)
xx
for (positive real) x ’ ∞, and from this complete the proof of
Stirling™s formula.

5. For arbitrary a ∈ C , use (B.12) to show for |z| ’ ∞ in sectors
| arg z| ¤ π ’ µ, for every µ > 0,

“(z) z a
’ 1. (B.13)
“(a + z)


6. Show
nz n!
’ “(1 + z), n ’ ∞.
(z + 1) · . . . · (z + n)

7. In this and the following exercises we aim at proving (B.10). For this
purpose, let the integral on the right of (B.10) be denoted by M (z).
Show that M (z) is an entire function, having real values for z ∈ R.

8. For » ∈ C and arbitrary real „ , show

’1 dz
eu/z z » = M (») u» ,
2πi z
γ(„ )

with a path of integration γ(„ ) as follows: From the origin along
arg z = „ + (µ + π)/2 to some z1 of modulus r, then along the circle
|z| = r to the ray arg z = „ ’ (µ + π)/2, and back to the origin along
this ray, with arbitrary r > 0 and 0 < µ < π/2.

9. For Re » > 0 and arbitrary real „ , show
∞(„ )
’1
u» e’u/z du = “(1 + ») z » ,
z
0

with integration along the ray arg u = „ .
232 Appendix B. Functions with Values in Banach Spaces

10. For Re » > 0 and arbitrary real „ , show that
∞(„ )
’1 dz
’1
e’u/w du = w» ,
eu/z z »
z
2πi z
0 γ(„ )


and use this and the previous two exercises to conclude (B.10). Com-
pare this to Section 5.5, where an analogous formula is obtained for
a more general situation.

11. For Re z < ’1, show that the path of integration γ in (B.10) can
be replaced with a path from in¬nity to the origin along the ray
arg w = ’π, and back to in¬nity along the ray arg w = π, interpreting
w’z = e’z log w accordingly. Use this to show
π
“(z) “(1 ’ z) = , z ∈ Z. (B.14)
sin πz

In particular, conclude that “(1/2) = π.

12. Use (B.14) and Exercise 6 to show Wallis™s product formula:

(1 ’ z 2 /k 2 ), z ∈ C.
sin(πz) = πz
k=1




B.4 Order and Type of Holomorphic Functions
¯
Let S = {z : |z| ≥ ρ0 , ± ¤ arg z ¤ β} and assume f holomorphic in
¯
the interior of S and continuous up to the boundary. De¬ne M (ρ, f ) =
sup{ f (z) : ± ¤ arg z ¤ β, |z| = ρ}, for ρ ≥ ρ0 , and call

log(log M (ρ, f ))
k = k(f ) = lim sup (B.15)
log ρ
ρ’∞

¯
the exponential order (or for short: the order) of f (in S). By de¬nition we
have 0 ¤ k ¤ ∞. It is easily seen that one can also de¬ne k as the in¬mum
of all κ > 0 for which a constant K exists such that
¯
f (z) ¤ K exp[|z|κ ], z ∈ S, (B.16)

following the convention that the in¬mum of the empty set is ∞. If k is
neither zero nor in¬nite, we refer to

log M (ρ, f )
„ = „ (f ) = lim sup (B.17)
ρk
ρ’∞
B.4 Order and Type of Holomorphic Functions 233

as the type of f and say that f is of ¬nite type if „ < ∞. As for the order,
one can also de¬ne „ as the in¬mum of all c > 0 for which K exists such
that
¯
f (z) ¤ K exp[c|z|k ], z ∈ S. (B.18)
If S = {z : |z| > ρ0 , ± < arg z < β} and f ∈ H(S, E ), we de¬ne its
order as the supremum of all orders of f in closed subsectors of S, and
analogously for the type. In particular, f is either of order smaller than k,
or of order k and ¬nite type, if an estimate (B.18) holds in every closed
subsector of S, with constants c, K that may go o¬ to in¬nity when we
approach the boundary of S.
As is standard for C -valued functions, we say that a function f ∈
H(C , E ) is an entire function. For such an entire f we take M (ρ, f ) as
the maximum modulus of f (z) on the circle of radius ρ (about the origin)
and de¬ne order and type as above. As an example, we consider

Mittag-Leffler™s Function
The (C -valued) function

z n /“(1 + ±n) ,
E± (z) = ± > 0,
n=0

is called Mittag-Le¬„er™s function of index ±.

Using (B.10), one can show

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