66
VIII – Cauchy Theory
π =
μ
f (ζ)dζ =
R
−R
dx
1 + x 2 + O(1/R) .
As the integral over [−R, R] approaches the integral sought, it is equal to π
as expected.
It may be wondered why we choose to integrate over a half-circle rather
than over other curves. The most probable reason is that, for the last two
thousand five hundred years, not to go further back to homo erectus fascinated by the Moon and the Sun, the circle is justifiably an object of adoration
for mathematicians. But we might as well integrate over the upper or lower
part of the square bounded by the lines Re(z) = R or −R and Im(z) = 0 or
R. The main point is that its length should be O(R) as R increases indefinitely, where R denotes the minimum distance from the origin to the chosen
contour points; it is even possible to go until a length of the order of o(R
2 ).
The method generalizes to integrals of rational fractions f (x) = p(x)/q(x),
where q is assumed not to have any real roots and, to start with, that d
◦ (q) −
d
◦ (p) = n ≥ 2 so as to ensure that f ∈ L
1 (R). f is integrated along the
same contour μ as above; choosing R sufficiently large for the roots of q in
the upper half plane to be in the interior of μ, we get
2πi
Im(a) > 0
Res(f, a) ,
(8.1)
a sum extended to the roots of q in the upper half plane. Besides, the integral
over μ is the sum of the integral of f over [−R, R] and of the integral along
the half-circle of radius R. If d
◦ (q) − d
◦ (p) = n, then
p(z)/q(z) ∼ c/z
n
for large |z|
(8.2)
with a constant c = 0. For large R, the integral along the half-circle is,
therefore, πR.O(R
−n ) = O(R
1−n ) and tends to 0 since n ≥ 2. Whence the
final result:
f (x)dx = 2πi
Im(a) > 0
Res(f, a) .
(8.3’)
Instead of integrating along the above contour, its symmetric with respect
to the real axis would do as well; as it is followed clockwise,we get
f (x)dx = −2πi
Im(a) < 0
Res(f, a) .
(8.3”)
Comparing these results shows that
a∈ C
Res(p/q, a) = 0 if d
◦ (q) − d
◦ (p) ≥ 2 ,
(8.4)
which has already been shown in (5.14).
VIII – Cauchy Theory
π =
μ
f (ζ)dζ =
R
−R
dx
1 + x 2 + O(1/R) .
As the integral over [−R, R] approaches the integral sought, it is equal to π
as expected.
It may be wondered why we choose to integrate over a half-circle rather
than over other curves. The most probable reason is that, for the last two
thousand five hundred years, not to go further back to homo erectus fascinated by the Moon and the Sun, the circle is justifiably an object of adoration
for mathematicians. But we might as well integrate over the upper or lower
part of the square bounded by the lines Re(z) = R or −R and Im(z) = 0 or
R. The main point is that its length should be O(R) as R increases indefinitely, where R denotes the minimum distance from the origin to the chosen
contour points; it is even possible to go until a length of the order of o(R
2 ).
The method generalizes to integrals of rational fractions f (x) = p(x)/q(x),
where q is assumed not to have any real roots and, to start with, that d
◦ (q) −
d
◦ (p) = n ≥ 2 so as to ensure that f ∈ L
1 (R). f is integrated along the
same contour μ as above; choosing R sufficiently large for the roots of q in
the upper half plane to be in the interior of μ, we get
2πi
Im(a) > 0
Res(f, a) ,
(8.1)
a sum extended to the roots of q in the upper half plane. Besides, the integral
over μ is the sum of the integral of f over [−R, R] and of the integral along
the half-circle of radius R. If d
◦ (q) − d
◦ (p) = n, then
p(z)/q(z) ∼ c/z
n
for large |z|
(8.2)
with a constant c = 0. For large R, the integral along the half-circle is,
therefore, πR.O(R
−n ) = O(R
1−n ) and tends to 0 since n ≥ 2. Whence the
final result:
f (x)dx = 2πi
Im(a) > 0
Res(f, a) .
(8.3’)
Instead of integrating along the above contour, its symmetric with respect
to the real axis would do as well; as it is followed clockwise,we get
f (x)dx = −2πi
Im(a) < 0
Res(f, a) .
(8.3”)
Comparing these results shows that
a∈ C
Res(p/q, a) = 0 if d
◦ (q) − d
◦ (p) ≥ 2 ,
(8.4)
which has already been shown in (5.14).