§ 3. Some Applications of Cauchy’s Method
77
The integral along BC is, therefore, up to a constant, bounded above by
(2p + 1)pe
π(a−1)p and tends to 0 since a < 1. A similar argument holds for
the contribution from side DA.
Next, let us investigate the contribution from AB. As w = p − 1/2 + iv,
sin πw = ± cosh πv ,
whence | sin πw| ≥ 1/2e
−π|v| and
|h(w)| ≤ 2e
−π|v|
|f (p − 1/2 + iv)| / |p − 1/2 + iv − z| ≤
≤ c 3 e
π(a−1)|v| / |p − 1/2 − Re(z)|
for all v. Integrating from v = −p to v = p, the upper bound
c 4 |p − 1/2 − Re(z)|
−1
p
−p
e
π(a−1)|v| dv
is obtained. It tends to 0 like the first factor since the extended integral over
R converges for a < 1.
Therefore, the contribution from AB and obviously that from CD too
both tend to 0, giving formula (2). The series extended to Z needs to be
interpreted as the limit of its partial symmetric sums since it may very well
not converge unconditionally, for instance in the case of the function f (z) = 1.
In this case, the partial fraction expansion of the function 1/ sin z :
π
sin πz
=
Z
(−1)
n
z − n
=
1
z
+ 2z
n≥1
(−1)
n
z 2 − n 2 .
(9.7’)
Most authors reject the first series or interpret it as the limit of its symmetric
sums, but in fact taken separately its “ positive ” and “ negative ” parts are
convergent though not absolutely. Indeed,
1/(z − n) = −1/n + z/n(z − n) = −1/n + O(1/n
2 )
for large|n|, so that the sum extended to n > 0 is the sum of an absolutely
convergent series and of the alternating series
(−1)
n+1 /n, and so converges.
(7’) can also be written as
π
sin πz
=
1
z
+
n =0
(−1)
n
1
z − n
+
1
n
with, this time, an absolutely convergent series.
Replacing z by
1
2 (1 − z),
π
2 cos πz/2
=
Z
(−1)
n
z + 2n + 1
,
(9.7”)