§ 3. Some Applications of Cauchy’s Method
127
Thus a priori the permutation of the signs
and
seems suspect. To justify
it, replace (3’) by the identity
f (x) = x − x
3 + . . . + (−1)
n x
2n+1 + (−1)
n+1 x
2n+2 f (x)
which is, up to a factor x, just the relation
1/(1 − q) = 1 + q + . . . + q
n + q
n+1 /(1 − q)
for q = −x
2 . Hence the contribution ϕ
− (s) from the interval (0, 1) is equal
to
0 ≤ p ≤ n
(−1)
p
s + 2p + 1
+ (−1)
n+1 ϕ
− (s + 2n + 2)
for all n > 0. This presupposes that Re(s) > −1 in order for the integral at
p = 0 and hence for the following ones to be convergent, but in fact holds for
all s ∈ C by analytic extension since ϕ
− (s) is obviously meromorphic on all
of the plane. However, for any s ∈ C,
ϕ
− (s + 2n + 2) =
1
0
x
s+2n+2
1 + x 2 d
∗ x for Re(s) + 2n + 2 > 0 ,
hence for large n. As n increases, the function x
s+2n+2 /(1+x
2 ) converges to 0
everywhere on ]0, 1[ while remaining, in modulus, ≤ 1 for large n. Therefore,
the integral tends to 0 (dominated convergence with respect to the measure
d
∗ x), and once again the series is convergent and the relation
ϕ
− (s) =
p≥0
(−1)
p
s + 2p + 1
holds for all s.
To deal with the contribution ϕ
+ (s) from the interval (1, +∞) to the
calculation of ϕ(s), note that ϕ
+ (s) = ϕ
− (−s), and so
ϕ
+ (s) =
p≤−1
(−1)
p
s + 2p + 1
.
For | Re(s)| < 1,
ϕ(s) =
+∞
0
x
s
1 + x 2 dx =
Z
(−1)
p
s + 2p + 1
= π/2 cos(πs/2)
again holds thanks to (9.7”).
Third Proof. The reader may be happy with these proofs, but this § is
supposed to present applications of Cauchy’s formula. So here is another way