112
VIII – Cauchy Theory
To prove (ii), first note that Γ f (s) is bounded on any strip 0 < a ≤ Re(s) ≤
b < +∞ due to the simple fact that the integral converges for s = a and
s = b. Having said this, let us integrate by parts for Re(s) > 0 :
sΓ f (s) =
+∞
0
f (x)sx
s−1 dx = f (x)x
s
+∞
0
−
+∞
0
f
(x)x
s+1 d
∗ x =
= −Γ
f (s + 1) ;
the part fully integrated is zero because (1) f is continuous for x ≥ 0, where
this is a large inequality, and x
s approaches 0 as x tends to 0, (2) f is a
rapidly decreasing function at infinity. The previous relation generalizing the
formula sΓ (s) = Γ (s + 1) can be integrated since f ∈ S + implies f
∈ S +
and leads to
s(s + 1) . . . (s + n − 1)Γ f (s) = (−1)
n Γ f (n) (s + n) .
(13.14)
By analytic extension, this result holds for all s. If s remains in a vertical
strip of finite width and if n is chosen to be sufficiently large, then s + n
remains in a strip 0 < a ≤ Re(s) ≤ b < +∞, in which the right hand side is
bounded. As
s(s + 1) . . . (s + n − 1) ∼ s
n
for large |s|
and given n, assertion (ii) follows.
(b) Inversion formula. To prove this, it suffices to check conditions (a), (b)
and (c) of section (i). The continuity of f and the convergence of
f (x)x
s d
∗ x
for Re(s) > 0 are obvious since f ∈ S(R + ); integral (4) involved in the inversion formula converges for all non-integral negative σ because of assertion (ii)
of the theorem.
(c) The function f associated to a function ϕ satisfying (i) and (ii). By
(ii), t → ϕ(σ + it) is a rapidly decreasing function at infinity for all σ ∈ R;
it can, therefore, be integrated over any vertical Re(s) = σ = 0, −1, . . . and
integral (4) can be computed. We show that it is independent of σ on any
interval [a, b] which does not contain an integer ≤ 0.
Let us integrate ϕ(s)x
−s over a rectangle bounded by the verticals Re(s) = a
and Re(s) = b > a and by the horizontals Im(s) = ±T .
x
−s
≤ x
−a + x
−b ,
on the horizontal sides. This constant is independent of T . As for the factor
ϕ(s), it is O(T
−n ) for all n since the function s
n ϕ(s) is bounded at infinity
on the vertical strip a ≤ Re(s) ≤ b. The contributions from these sides,
therefore, clearly tend to 0 as T −→ +∞. Setting ψ(s) = ϕ(s)x
−s , at the
limit,
VIII – Cauchy Theory
To prove (ii), first note that Γ f (s) is bounded on any strip 0 < a ≤ Re(s) ≤
b < +∞ due to the simple fact that the integral converges for s = a and
s = b. Having said this, let us integrate by parts for Re(s) > 0 :
sΓ f (s) =
+∞
0
f (x)sx
s−1 dx = f (x)x
s
+∞
0
−
+∞
0
f
(x)x
s+1 d
∗ x =
= −Γ
f (s + 1) ;
the part fully integrated is zero because (1) f is continuous for x ≥ 0, where
this is a large inequality, and x
s approaches 0 as x tends to 0, (2) f is a
rapidly decreasing function at infinity. The previous relation generalizing the
formula sΓ (s) = Γ (s + 1) can be integrated since f ∈ S + implies f
∈ S +
and leads to
s(s + 1) . . . (s + n − 1)Γ f (s) = (−1)
n Γ f (n) (s + n) .
(13.14)
By analytic extension, this result holds for all s. If s remains in a vertical
strip of finite width and if n is chosen to be sufficiently large, then s + n
remains in a strip 0 < a ≤ Re(s) ≤ b < +∞, in which the right hand side is
bounded. As
s(s + 1) . . . (s + n − 1) ∼ s
n
for large |s|
and given n, assertion (ii) follows.
(b) Inversion formula. To prove this, it suffices to check conditions (a), (b)
and (c) of section (i). The continuity of f and the convergence of
f (x)x
s d
∗ x
for Re(s) > 0 are obvious since f ∈ S(R + ); integral (4) involved in the inversion formula converges for all non-integral negative σ because of assertion (ii)
of the theorem.
(c) The function f associated to a function ϕ satisfying (i) and (ii). By
(ii), t → ϕ(σ + it) is a rapidly decreasing function at infinity for all σ ∈ R;
it can, therefore, be integrated over any vertical Re(s) = σ = 0, −1, . . . and
integral (4) can be computed. We show that it is independent of σ on any
interval [a, b] which does not contain an integer ≤ 0.
Let us integrate ϕ(s)x
−s over a rectangle bounded by the verticals Re(s) = a
and Re(s) = b > a and by the horizontals Im(s) = ±T .
x
−s
≤ x
−a + x
−b ,
on the horizontal sides. This constant is independent of T . As for the factor
ϕ(s), it is O(T
−n ) for all n since the function s
n ϕ(s) is bounded at infinity
on the vertical strip a ≤ Re(s) ≤ b. The contributions from these sides,
therefore, clearly tend to 0 as T −→ +∞. Setting ψ(s) = ϕ(s)x
−s , at the
limit,