§ 3. Some Applications of Cauchy’s Method
119
special functions. Proving them is a complex logarithm calculation exercise
involving all the pitfalls of the topic.
65 In what follows, set
C − R − = C + .
The idea behind the proof is simple. A naive calculation shows (3) to be
seemingly equivalent to
log Γ (s) =
1
2
log 2π + (s − 1/2) log s − s + o(1) .
(14.4)
However, the formula
Γ (s) = lim n!n
s /s(s + 1) . . . (s + n) ,
(14.5)
which holds for all non-integral s ≤ 0, seems to show that
log Γ (s) = lim
log(n!) + s log n −
n
0
log(s + p)
(14.6)
and (2) shows that
log(n!) = 1/2 log(2π) + (n + 1/2) log n − n + o(1) .
(14.7)
The problem, therefore, appears to lie in the evaluation of the sum of the
log(s + p), which, as we shall see, is made possible by the Euler-MacLaurin
summation formula (Chapter VI, § 2, n
◦ 16). Combining these results, (4)
and hence (3) can be expected to be justified. But complex logarithms, not
to speak of their limits, cannot be used like those of Neper; hence their
meaning will first need to be specified.
Let us start by specifying the meaning of the expression s
s−1/2 occurring
in (3), namely
s
s−1/2 = exp [(s − 1/2) Log s] for s ∈ C + ,
(14.8)
where, on C + ,the Log function is the uniform branch which reduces to the
Neper function on the positive real axis:
Log z = log |z| + i Arg(z) with | Arg(z)| < π .
(14.9)
This function allows the following more general definition
z
w = exp(w Log z) for z ∈ C + , w ∈ C .
For technical reasons, the Log function needs to be extended to all of C
∗
by setting
65 See Serge Lang, Complex Analysis (Springer-New York, 4th. ed., 1999), pp. 422–
428 for an example of a proof where complex logs are used without precaution.
119
special functions. Proving them is a complex logarithm calculation exercise
involving all the pitfalls of the topic.
65 In what follows, set
C − R − = C + .
The idea behind the proof is simple. A naive calculation shows (3) to be
seemingly equivalent to
log Γ (s) =
1
2
log 2π + (s − 1/2) log s − s + o(1) .
(14.4)
However, the formula
Γ (s) = lim n!n
s /s(s + 1) . . . (s + n) ,
(14.5)
which holds for all non-integral s ≤ 0, seems to show that
log Γ (s) = lim
log(n!) + s log n −
n
0
log(s + p)
(14.6)
and (2) shows that
log(n!) = 1/2 log(2π) + (n + 1/2) log n − n + o(1) .
(14.7)
The problem, therefore, appears to lie in the evaluation of the sum of the
log(s + p), which, as we shall see, is made possible by the Euler-MacLaurin
summation formula (Chapter VI, § 2, n
◦ 16). Combining these results, (4)
and hence (3) can be expected to be justified. But complex logarithms, not
to speak of their limits, cannot be used like those of Neper; hence their
meaning will first need to be specified.
Let us start by specifying the meaning of the expression s
s−1/2 occurring
in (3), namely
s
s−1/2 = exp [(s − 1/2) Log s] for s ∈ C + ,
(14.8)
where, on C + ,the Log function is the uniform branch which reduces to the
Neper function on the positive real axis:
Log z = log |z| + i Arg(z) with | Arg(z)| < π .
(14.9)
This function allows the following more general definition
z
w = exp(w Log z) for z ∈ C + , w ∈ C .
For technical reasons, the Log function needs to be extended to all of C
∗
by setting
65 See Serge Lang, Complex Analysis (Springer-New York, 4th. ed., 1999), pp. 422–
428 for an example of a proof where complex logs are used without precaution.