§ 3. Some Applications of Cauchy’s Method
75
Replacing t by −t, we get
(x − a)
−1 e(tx)dx = 2πit
0
+ e(at)
(8.20)
by stipulating, like Fej´ er, that
t
0
+ = 1 if t > 0 , = 1/2 if t = 0 , = 0 if t < 0 .
We could have spared ourselves all these calculations – but we did not
want to – by alluding to Theorem 27 of Chapter VII, § 6, n
◦ 30 : if f is a
regulated and absolutely integrable function on R, then
lim
R
−R
ˆ
f (y)e(xy)dy =
1
2
[f (x+) + f (x−)]
at every point where f has left and right derivatives. In the present case, it
can be applied to the function f defined by the right hand sides of (19): it is
obviously regulated, right and left differentiable everywhere, integrable since
Im(a) > 0, and finally its Fourier transform is 1/(x − a) as shown by a most
simple direct calculation. It then remains to check that
1
2 [f (0+) + f (0−)] =
πi.
9 – Summation Formulas
The residue method allows us to prove several summation formulas. Let us
for example show that if f (z) is an entire function satisfying an inequality
of the form
|f (z)| ≤ M.e
πa|y|
with 0 < a < 1 ,
(9.1)
then
π
f (z)
sin πz
=
Z
(−1)
n f (n)
z − n
= lim
p∞
|n|< p
.
(9.2)
For this, let us consider the meromorphic function
46 g(w) = πf (w)/ sin πw.
Its only singularities are, at worst, simple poles at points n ∈ Z with
Res(g, n) = (−1)
n f (n). For given z /
∈ Z, the function
h(w) = g(w)/(w − z) = πf (w)/(w − z) sin πw
has simple poles at w ∈ Z, with
46 Recall that the only singularities of a meromorphic function on a domain G
(here, C) are necessarily isolated poles. If f = p/q where q has a simple zero at
a and p is holomorphic at a, then Res(f, a) = p(a)/q
(a).
75
Replacing t by −t, we get
(x − a)
−1 e(tx)dx = 2πit
0
+ e(at)
(8.20)
by stipulating, like Fej´ er, that
t
0
+ = 1 if t > 0 , = 1/2 if t = 0 , = 0 if t < 0 .
We could have spared ourselves all these calculations – but we did not
want to – by alluding to Theorem 27 of Chapter VII, § 6, n
◦ 30 : if f is a
regulated and absolutely integrable function on R, then
lim
R
−R
ˆ
f (y)e(xy)dy =
1
2
[f (x+) + f (x−)]
at every point where f has left and right derivatives. In the present case, it
can be applied to the function f defined by the right hand sides of (19): it is
obviously regulated, right and left differentiable everywhere, integrable since
Im(a) > 0, and finally its Fourier transform is 1/(x − a) as shown by a most
simple direct calculation. It then remains to check that
1
2 [f (0+) + f (0−)] =
πi.
9 – Summation Formulas
The residue method allows us to prove several summation formulas. Let us
for example show that if f (z) is an entire function satisfying an inequality
of the form
|f (z)| ≤ M.e
πa|y|
with 0 < a < 1 ,
(9.1)
then
π
f (z)
sin πz
=
Z
(−1)
n f (n)
z − n
= lim
p∞
|n|< p
.
(9.2)
For this, let us consider the meromorphic function
46 g(w) = πf (w)/ sin πw.
Its only singularities are, at worst, simple poles at points n ∈ Z with
Res(g, n) = (−1)
n f (n). For given z /
∈ Z, the function
h(w) = g(w)/(w − z) = πf (w)/(w − z) sin πw
has simple poles at w ∈ Z, with
46 Recall that the only singularities of a meromorphic function on a domain G
(here, C) are necessarily isolated poles. If f = p/q where q has a simple zero at
a and p is holomorphic at a, then Res(f, a) = p(a)/q
(a).