70
VIII – Cauchy Theory
with Re(w) > 0. The function g(z) = exp(−2πitz)f (z) has simple poles at
iw and −iw, and clearly
44
Res(g, iw) = e
2πtw /2iw, Res(g, −iw) = −e
−2πtw /2iw .
As Im(iw) > 0 and Im(−iw) < 0, multiplying by 2πi, we get
e(−tx)
x 2 + w 2 dx =
πe
−2πtw /w if t ≥ 0
πe
2πtw /w
if t ≤ 0 ,
if Re(w) > 0 ;
in other words
e(−tx)
x 2 + w 2 dx = πe
−2πw|t| /w if Re(w) > 0 .
(8.12)
Note that |e
−2πw|t|
| = e
−2π|t| Re(w) approaches 0 exponentially as |t| increases
indefinitely since Re(w) > 0, so that ˆ
f ∈ L
1 (R); in other words, f ∈ F
1 (R).
Fourier’s inversion formula can, therefore, be applied, and so
e
−2πw|t|+2πitx dt = w/π
x
2 + w
2
. Re(w) > 0 ,
This formula easy to check directly: integrate over t > 0 and t < 0 taking
into account that e
ct has e
ct /c as primitive for any c ∈ C
∗ .
To obtain an explicit form of ˆ
f (t) in the general case, observe that any
rational fraction is the sum of a polynomial and of a linear combination of
functions of the form
f (x) = (x − a)
−n ,
(8.13)
where n is an integer ≥ 1 and a is a constant; its Fourier transform is welldefined only if its decomposition does not have any polynomial terms and if
its poles are not real. Hence if an explicit formula is found for the Fourier
transform of (13) for a /
∈ R, the decomposition into simple elements will give
the result in the general case. In this section, we will suppose that n ≥ 2 and
that Im(a) > 0, the case Im(a) < 0 being similar.
The only pole of the function
g(z) = (z − a)
−n e(−tz)
being a, it is already possible to conclude that ˆ
f (t) = 0 for t ≥ 0. For t < 0,
the residue at a of
44 If ϕ has a simple pole at a and if ψ is holomorphic and non-zero at a, then
ϕ(z)ψ(z) =
c(z − a)
−1 + . . .
[ψ(a) + . . .] ,
and so Res(ϕψ, a) = ψ(a) Res(ϕ, a).
VIII – Cauchy Theory
with Re(w) > 0. The function g(z) = exp(−2πitz)f (z) has simple poles at
iw and −iw, and clearly
44
Res(g, iw) = e
2πtw /2iw, Res(g, −iw) = −e
−2πtw /2iw .
As Im(iw) > 0 and Im(−iw) < 0, multiplying by 2πi, we get
e(−tx)
x 2 + w 2 dx =
πe
−2πtw /w if t ≥ 0
πe
2πtw /w
if t ≤ 0 ,
if Re(w) > 0 ;
in other words
e(−tx)
x 2 + w 2 dx = πe
−2πw|t| /w if Re(w) > 0 .
(8.12)
Note that |e
−2πw|t|
| = e
−2π|t| Re(w) approaches 0 exponentially as |t| increases
indefinitely since Re(w) > 0, so that ˆ
f ∈ L
1 (R); in other words, f ∈ F
1 (R).
Fourier’s inversion formula can, therefore, be applied, and so
e
−2πw|t|+2πitx dt = w/π
x
2 + w
2
. Re(w) > 0 ,
This formula easy to check directly: integrate over t > 0 and t < 0 taking
into account that e
ct has e
ct /c as primitive for any c ∈ C
∗ .
To obtain an explicit form of ˆ
f (t) in the general case, observe that any
rational fraction is the sum of a polynomial and of a linear combination of
functions of the form
f (x) = (x − a)
−n ,
(8.13)
where n is an integer ≥ 1 and a is a constant; its Fourier transform is welldefined only if its decomposition does not have any polynomial terms and if
its poles are not real. Hence if an explicit formula is found for the Fourier
transform of (13) for a /
∈ R, the decomposition into simple elements will give
the result in the general case. In this section, we will suppose that n ≥ 2 and
that Im(a) > 0, the case Im(a) < 0 being similar.
The only pole of the function
g(z) = (z − a)
−n e(−tz)
being a, it is already possible to conclude that ˆ
f (t) = 0 for t ≥ 0. For t < 0,
the residue at a of
44 If ϕ has a simple pole at a and if ψ is holomorphic and non-zero at a, then
ϕ(z)ψ(z) =
c(z − a)
−1 + . . .
[ψ(a) + . . .] ,
and so Res(ϕψ, a) = ψ(a) Res(ϕ, a).