§ 3. Some Applications of Cauchy’s Method
101
with a function
ˆ
f (t) =
f (z)e(−tz)dx
(12.14)
independent of y and that can serve as the Fourier transform of f .
Let us show that the Fourier inversion formula can be applied to (11).
Since the series
f
(r) (z + n) converges, f
(r) (x + iy) is integrable at x and
approaches 0 as |x| increases indefinitely; (11) can, therefore, be computed
by integrating by parts. f being holomorphic, D 1 f = f
and the usual computation shows that
f
(r) (x + iy)e(−tx)dx = (−2πit)
r F (t, y) .
(12.15)
It follows that for all y ∈ I and all r,
F (t, y) = O
|t|
−r
as |t| −→ +∞ ,
(12.16)
a condition more than sufficient to justify the use of the inversion formula.
In view of (12), it can be written
f (z) =
ˆ
f (t)e(tz)dt
and proves that f is the complex Fourier transform of ˆ
f .
Besides, (16) shows that
|F (n, y)| < +∞ for all y, which allows Poisson’s summation formula to be applied (Chap. V, n
◦ 27, theorem 24) to
x → f (x + iy); in view of (12), it can be written
f (z + n) =
ˆ
f (n)e(nz) .
Ultimately, the following formulas are obtained:
ˆ
f (t) =
Re(z)=y
f (z)e(−tz)dz ,
(12.17)
f (z) =
ˆ
f (t)e(tz)dt ,
(12.18)
f (z + n) =
ˆ
f (n)e(nz) .
(12.19)
They obviously assume that a < y < b, a condition that, as seen earlier,
makes the integrals absolutely convergent.
(iv) Holomorphic functions integrable over a half-plane. Let P be the halfplane Im(z) > 0. In what precedes, take I =]0, +∞[ and B = P and instead
of (6), impose the stronger condition
P
|f (z)| ρ(y)dxdy < +∞
(12.20)
101
with a function
ˆ
f (t) =
f (z)e(−tz)dx
(12.14)
independent of y and that can serve as the Fourier transform of f .
Let us show that the Fourier inversion formula can be applied to (11).
Since the series
f
(r) (z + n) converges, f
(r) (x + iy) is integrable at x and
approaches 0 as |x| increases indefinitely; (11) can, therefore, be computed
by integrating by parts. f being holomorphic, D 1 f = f
and the usual computation shows that
f
(r) (x + iy)e(−tx)dx = (−2πit)
r F (t, y) .
(12.15)
It follows that for all y ∈ I and all r,
F (t, y) = O
|t|
−r
as |t| −→ +∞ ,
(12.16)
a condition more than sufficient to justify the use of the inversion formula.
In view of (12), it can be written
f (z) =
ˆ
f (t)e(tz)dt
and proves that f is the complex Fourier transform of ˆ
f .
Besides, (16) shows that
|F (n, y)| < +∞ for all y, which allows Poisson’s summation formula to be applied (Chap. V, n
◦ 27, theorem 24) to
x → f (x + iy); in view of (12), it can be written
f (z + n) =
ˆ
f (n)e(nz) .
Ultimately, the following formulas are obtained:
ˆ
f (t) =
Re(z)=y
f (z)e(−tz)dz ,
(12.17)
f (z) =
ˆ
f (t)e(tz)dt ,
(12.18)
f (z + n) =
ˆ
f (n)e(nz) .
(12.19)
They obviously assume that a < y < b, a condition that, as seen earlier,
makes the integrals absolutely convergent.
(iv) Holomorphic functions integrable over a half-plane. Let P be the halfplane Im(z) > 0. In what precedes, take I =]0, +∞[ and B = P and instead
of (6), impose the stronger condition
P
|f (z)| ρ(y)dxdy < +∞
(12.20)