110
VIII – Cauchy Theory
converges, which is obvious by our formal calculation since it is proportional
to that of Riemann.
Hence the function
Γ f (s) = π
−s/2 Γ (s/2)ζ(s) = ξ(s)
can be analytically extended to all of C, with simple poles at s = 0 and
s = 1 coming from the first two terms of the asymptotic expansion of f
at 0; the residues are equal to 1 at s = 1 and to −1 at s = 0. As the
function 1/Γ (s/2) is entire and has simples zeros at 0, −2, −4, . . ., the function
ζ(s) = π
s/2 Γ f (s)/Γ (s/2) is meromorphic on C. As Γ (1/2) = π
1/2
= 0,
the simple pole of Γ f at s = 1 spreads to ζ(s), with a residue equal to
π
−1/2 π
1/2 = 1. As 1/Γ (s/2) vanishes at s = 0, the pole of Γ f at this point is
neutralized by the zero of the function 1/Γ . Therefore, the function ζ(s) has a
unique singularity in C : a simple pole at s = 1. It vanishes at s = −2, −4, . . .,
as well as at several other points less obvious at first.
It also satisfies a functional equation which can be deduced from that of
the Jacobi function. Indeed,
f (1/x) = θ
1/x
2
− 1 = xθ
x
2
− 1 = x [f (x) + 1] − 1 ,
i.e.
f (1/x) = xf (x) + x − 1 .
Hence, for Re(s) > 1,
Γ
−
f (s) =
1
0
f (x)x
s d
∗ x =
+∞
1
f (1/x)x
−s d
∗ x
=
+∞
1
f (x)x
1−s + x
1−s
− x
−s
d
∗ x .
Each of these three function occurring in the last integral is integrable over
[1, +∞[ with respect to d
∗ x, where Re(s) > 0 : this is obvious for the last
two, and the the first one is integrable for all s since it decreases rapidly at
infinity. In conclusion,
Γ
−
f (s) = Γ
+
f (1 − s) + 1/(s − 1) − 1/s
for Re(s) > 1, and so, by analytic extension, for all s ∈ C. As Γ f = Γ
+
f + Γ
−
f ,
Γ f (s) = Γ
+
f (s) + Γ
+
f (1 − s) + 1/(s − 1) − 1/s ,
which proves that
ξ(s) = ξ(1 − s) .