122
VIII – Cauchy Theory
These preliminary explanations allow us to return to
Γ (s) = lim [n!n
s /s(s + 1) . . . (s + n)] .
First of all, by (10), (11) and (17),
Log Γ (s) = lim
log(n!) + s log n −
n
0
L(s + p) + 2k n πi
(14.20)
for properly chosen k n ∈ Z, and by Stirling,
log(n!) =
1
2
log(2π) + (n + 1/2) log n − n + o(1) .
(14.21)
To evaluate the sum of the L(s + p) for given s ∈ C + , set f (x) = L(s +
x) = Log(s + x) for x > 0; we get a C
∞ function such that f
(x) = (s +
x)
−1 , f
(x) = −(s + x)
−2 since Log z is holomorphic on C + and has 1/z as
derivative. Instead of referring the reader to Chapter VI, § 2, n
◦ 16 for the
general Euler-MacLaurin formula, let us introduce the functions
P 1 (x) = x − 1/2, P 2 (x) =
1
2
x
2
− x
.
(14.22)
So P
1 = 1 and P
2 = P 1 . As P 2 (0) = P 2 (1) = 0, integrating twice by parts
immediately show that
p+1
p
f (x)dx =
1
0
f (x + p)dx =
=
1
2
[f (x + p) + f (x + p + 1)] +
1
0
f
(x + p)P 2 (x)dx .
Setting
P
∗
2 (x) = P 2 (x − [x])
(14.23)
to be a function with period 1 equal to P 2 on (0, 1), transform the last integral
into that of the function f
(x)P
∗
2 (x) on (p, p + 1). Summing from p = 0 to
p = n − 1,
n
0
f (x)dx = −
1
2
[f (0) + f (n)] +
n
0
f (x + p) +
n
0
f
(x)P
∗
2 (x)dx
follows. Integrating by parts f (x) = Log(s + x), we get
n
0
f (x)dx = (s + n) Log(s + n) − s Log s − n ,
and so