§ 3. Some Applications of Cauchy’s Method
81
As f (s) and Γ (s) are bounded on 1 ≤ Re(s) ≤ 2, because of their functional equation, they are bounded at infinity on the entire strip a ≤ Re(s) ≤
a + 1 ≤ 2; the same, therefore, holds for g, and as g does not have poles,
it is globally bounded on that strip. Hence h(s) is bounded on the strip
0 ≤ Re(s) ≤ 1.
The pseudo periodicity of h then shows that h is bounded on C, and
so is constant, and hence is zero since g(1) = f (1) − f (1)Γ (1) = 0. Since
g(s)g(1 − s) = 0 for all s, g(s) = 0, and so
f (s) = f (1)Γ (s) ,
qed.
Having done this, we can return to (10.5.9), which can also be written
Γ (s) = π
−1/2 2
s−1 Γ (s/2)Γ [(s + 1)/2] ,
(10.5.10)
and observe that the function f (s) = 2
s Γ (s/2)Γ [(s + 1)/2] satisfies Wielandt’s
assumptions, which is immediate. Hence f (s) = f (1)Γ (s) = 2π
1/2 Γ (s), and
the formula follows. See Chap. XII, n
◦ 1 also.
(ii) Fourier transform of e
−x x
s−1
+ . To show that (10.4) remains valid for
Re(w) > 0, it suffices – analytic extension – to check that both sides are
holomorphic functions of w on this half-plane. Setting
w
−s = exp [−sL(w)] = |w|
−s e
−is Arg(w)
(10.6)
where L(w) = log |w| + i Arg(w) is a uniform branch of the pseudo-function
Log(w) on Re(w) > 0 (§ 2, n
◦ 4, (i)), it is the case of the right hand side. As
our aim is to find the usual function for w real > 0, we should choose
|Arg(w)| < π/2 .
As for function (3), theorem 9 on integrals depending holomorphically on a
parameter can be applied to it. The only non-obvious condition is the existence of a function p H (t) ∈ L
1 (R + ) for every compact subset H ⊂ {Re(w) >
0} such that
exp(−wt)t
s−1
≤ |p H (t)|
for all w ∈ H and t > 0. But the distance from a compact subset of an open
set U to the boundary of U is > 0. H is therefore contained in a half-plane
Re(w) ≥ a with a > 0; then
exp(−wt)t
s−1
≤ exp(−at)t
Re(s)−1 = p H (t) .
81
As f (s) and Γ (s) are bounded on 1 ≤ Re(s) ≤ 2, because of their functional equation, they are bounded at infinity on the entire strip a ≤ Re(s) ≤
a + 1 ≤ 2; the same, therefore, holds for g, and as g does not have poles,
it is globally bounded on that strip. Hence h(s) is bounded on the strip
0 ≤ Re(s) ≤ 1.
The pseudo periodicity of h then shows that h is bounded on C, and
so is constant, and hence is zero since g(1) = f (1) − f (1)Γ (1) = 0. Since
g(s)g(1 − s) = 0 for all s, g(s) = 0, and so
f (s) = f (1)Γ (s) ,
qed.
Having done this, we can return to (10.5.9), which can also be written
Γ (s) = π
−1/2 2
s−1 Γ (s/2)Γ [(s + 1)/2] ,
(10.5.10)
and observe that the function f (s) = 2
s Γ (s/2)Γ [(s + 1)/2] satisfies Wielandt’s
assumptions, which is immediate. Hence f (s) = f (1)Γ (s) = 2π
1/2 Γ (s), and
the formula follows. See Chap. XII, n
◦ 1 also.
(ii) Fourier transform of e
−x x
s−1
+ . To show that (10.4) remains valid for
Re(w) > 0, it suffices – analytic extension – to check that both sides are
holomorphic functions of w on this half-plane. Setting
w
−s = exp [−sL(w)] = |w|
−s e
−is Arg(w)
(10.6)
where L(w) = log |w| + i Arg(w) is a uniform branch of the pseudo-function
Log(w) on Re(w) > 0 (§ 2, n
◦ 4, (i)), it is the case of the right hand side. As
our aim is to find the usual function for w real > 0, we should choose
|Arg(w)| < π/2 .
As for function (3), theorem 9 on integrals depending holomorphically on a
parameter can be applied to it. The only non-obvious condition is the existence of a function p H (t) ∈ L
1 (R + ) for every compact subset H ⊂ {Re(w) >
0} such that
exp(−wt)t
s−1
≤ |p H (t)|
for all w ∈ H and t > 0. But the distance from a compact subset of an open
set U to the boundary of U is > 0. H is therefore contained in a half-plane
Re(w) ≥ a with a > 0; then
exp(−wt)t
s−1
≤ exp(−at)t
Re(s)−1 = p H (t) .