130
VIII – Cauchy Theory
To show that the contributions from the circular arcs tend to 0, use the
lemma from the introduction to this § : in U = C − R + and for all s ∈ C,
|z
s
| | |z|
Re(s)
in U .
(15.9)
As |1 + z
2
|
−1 = O(R
−2 ) over the large circle, the integral is O
R
Re(s)−1
and tends to 0 since Re(s) < 1.
Over the small circle, |z
s
| = O
r
Re(s)
and (1+z
2 )
−1
∼ 1 . Therefore, the
integral is O
r
1+Re(s)
, and so we reach the same conclusion since Re(s) >
−1.
Ultimately, taking (4) into account, we get
(1 − e
2πis )ϕ(s) =
μ
= πe
πis/2 (1 − e
πis )
for | Re(s)| < 1, and so
ϕ(s) = π/2 cos(πs/2) ,
(15.10)
which ends the third proof.
This method shows how to calculate the Mellin transform of a rational
function f (x) = p(x)/q(x) without poles in R + . As f is finite at x = 0,
the integral converges in the neighbourhood of 0, at least for Re(s) > 0.
If d
◦ (q) − d
◦ (p) = n, then at infinity, f (x) x
−n and hence f (x)x
s−1
x
s−n−1 , so that the integral converges for Re(s) < n. The Mellin transform
is, therefore, a priori defined on the vertical strip 0 < Re(s) < n. As Γ f (s) is
obtained by integrating f (x)x
s−1 with respect to the measure dx,
[1 − exp(2πis)] Γ f (s) = 2πi
Res
z
s−1 f (z), a
,
(15.11)
the sum being extended to all the poles of f . If
f (z) =
A k / (z − a k )
only has simple poles, then Res
z
s−1 f (z), a k
= A k a
s−1
k .
For example, for f (z) = (1 + z)
−1 the Mellin integral converges for 0 <
Re(s) < 1 and
(−1)
s−1 = exp [πi(s − 1)] = − exp(πis)
needs to be calculated.
+∞
0
x
s
1 + x
d
∗ x = π/ sin πs for 0 < Re(s) < 1
(15.12)
immediately follows. For f (z) = (z −a)
−k with a /
∈ R + and k > 1, the residue
of z
s f (z) is the coefficient of (z − a)
k−1 in the Taylor series for z
s at a. Now,
by Newton, for |z − a| < |a|,