72
VIII – Cauchy Theory
D
A
C
B
T
−T
ib'
ib
0
Fig. 8.10.
therefore, does not come as a surprise that the integral only depends on
the sign of b. To justify this directly, we integrate along the contour of the
rectangle ABCD of the above figure; by Cauchy’s theorems, the result is zero,
and it all boils down to showing that, as T tends to +∞, the contributions
from the vertical sides approach 0. However, on these sides,
ζ
−n e(ζ)
=
T
2 + η
2
−n/2 exp(−2πη) ,
where η = Im(ζ) varies between b and b
, and so 0 < m ≤ |e(ζ)| ≤ M < +∞
with constants m and M independent from T . On the other hand, (T
2 +
η
2 )
−n/2 remains between its values at η = b and η = b
; |ζ
−n e(ζ)| is T
−n
on the vertical sides. Their contribution is, therefore, O(T
−n ), qed.
This argument also explains why the value of the integral over the positive
horizontals is different from that on the negative ones: in such a case, there is
pole at ζ = 0 in the interior of the rectangle ABCD, so that, up to a factor
2πi, the integral is equal to the residue of ζ
−n e(ζ) at 0.
Exercise 1. Translate (15’) using the change of variable 2πitz = ζ.
Exercise 2. Applying Poisson’s summation formula to the function x →
(x − z)
−k , show that
Z
1
(z + n) k =
(−2πi)
k
(k − 1)!
n≥1
n
k−1 e
2πinz for k ≥ 2 , Im(z) > 0 .
(8.16)
Recover (16) by differentiating the partial fractions expansion of cotg πz.
Exercise 3. Check that Plancherel’s formula holds for the Fourier transform (14).
(iv) Semi-convergent Fourier transforms. The calculation of the Fourier
transform of a rational function f = p/q assumes that d
◦ (q) − d
◦ (p) = n ≥ 2.
VIII – Cauchy Theory
D
A
C
B
T
−T
ib'
ib
0
Fig. 8.10.
therefore, does not come as a surprise that the integral only depends on
the sign of b. To justify this directly, we integrate along the contour of the
rectangle ABCD of the above figure; by Cauchy’s theorems, the result is zero,
and it all boils down to showing that, as T tends to +∞, the contributions
from the vertical sides approach 0. However, on these sides,
ζ
−n e(ζ)
=
T
2 + η
2
−n/2 exp(−2πη) ,
where η = Im(ζ) varies between b and b
, and so 0 < m ≤ |e(ζ)| ≤ M < +∞
with constants m and M independent from T . On the other hand, (T
2 +
η
2 )
−n/2 remains between its values at η = b and η = b
; |ζ
−n e(ζ)| is T
−n
on the vertical sides. Their contribution is, therefore, O(T
−n ), qed.
This argument also explains why the value of the integral over the positive
horizontals is different from that on the negative ones: in such a case, there is
pole at ζ = 0 in the interior of the rectangle ABCD, so that, up to a factor
2πi, the integral is equal to the residue of ζ
−n e(ζ) at 0.
Exercise 1. Translate (15’) using the change of variable 2πitz = ζ.
Exercise 2. Applying Poisson’s summation formula to the function x →
(x − z)
−k , show that
Z
1
(z + n) k =
(−2πi)
k
(k − 1)!
n≥1
n
k−1 e
2πinz for k ≥ 2 , Im(z) > 0 .
(8.16)
Recover (16) by differentiating the partial fractions expansion of cotg πz.
Exercise 3. Check that Plancherel’s formula holds for the Fourier transform (14).
(iv) Semi-convergent Fourier transforms. The calculation of the Fourier
transform of a rational function f = p/q assumes that d
◦ (q) − d
◦ (p) = n ≥ 2.