68
VIII – Cauchy Theory
and hence f (z) + f (−z) = O(1/z
2 ). In the decomposition of f into simple
elements, the integrals of terms of the form A/(z − a)
r with r ≥ 2 can be
computed as before; computations may even be omitted because they are
zero, either because that is the case of the corresponding residues or simply
because the function 1/(x−a)
r admitting a primitive approaching 0 at infinity
for r ≥ 2, the FT solves the problem without having to invoke Cauchy.
It, therefore, remains to compute the above integral for f (z) = 1/(z − a),
a /
∈ R. A first method consists in observing (Chapter V, § 6, n
◦ 20) that, on
the simply connected open subset G = C − R − , the function 1/z admits as
primitive any uniform branch L(z) of the pseudo-function Log z, for example
the one obtained by setting
L(z) = log |z| + i. Arg z with | Arg z| < π .
As a is not real, the points x − a are in G for x ∈ R, so that L(x − a) can be
chosen as a primitive for 1/(x − a) on R. Hence
R
−R
dx
x − a
= L(x − a)
R
−R
.
The variation of the real part of
L(x − a) = log |x − a| + i. Arg(x − a)
over [−R, R] is equal to log(|R−a|/|R+a|) and tends to 0 since |R−a|/|R+a|
approaches 1 as R increases; the argument of R − a tends to 0 since the halfline with initial point 0 and terminal point R −a approaches the half-line R + ;
finally, the half-line with initial point 0 and terminal point −R−a approaches
R − , but is in the half-plane Im(z) < 0 if Im(a) > 0 and in the half-plane
Im(z) > 0 if Im(a) < 0; the argument of −R − a, therefore, approaches −π
if Im(a) > 0 and +π if Im(a) < 0. So
R
dx/(x − a) =
πi
if Im(a) > 0
−πi
if Im(a) < 0
.
(8.6)
Finally, in the general case, it follows that
R
f (x)dx = πi
Im(a) > 0
Res(f, a) − πi
Im(a) < 0
Res(f, a) .
(8.7)
When d
◦ (q) ≥ d
◦ (p) + 2, the sum of all the residues of f in C are zero and
(7) reduces to (3’) or (3”), as desired. If on the contrary d
◦ (q) = d
◦ (p) + 1, it
is the sum of residues in C and at infinity, which is zero by (5.14’). Then we
for example find
R
f (x)dx = 2πi
Im(a) > 0
Res(f, a) + πi Res(f, ∞) .
(8.8)
VIII – Cauchy Theory
and hence f (z) + f (−z) = O(1/z
2 ). In the decomposition of f into simple
elements, the integrals of terms of the form A/(z − a)
r with r ≥ 2 can be
computed as before; computations may even be omitted because they are
zero, either because that is the case of the corresponding residues or simply
because the function 1/(x−a)
r admitting a primitive approaching 0 at infinity
for r ≥ 2, the FT solves the problem without having to invoke Cauchy.
It, therefore, remains to compute the above integral for f (z) = 1/(z − a),
a /
∈ R. A first method consists in observing (Chapter V, § 6, n
◦ 20) that, on
the simply connected open subset G = C − R − , the function 1/z admits as
primitive any uniform branch L(z) of the pseudo-function Log z, for example
the one obtained by setting
L(z) = log |z| + i. Arg z with | Arg z| < π .
As a is not real, the points x − a are in G for x ∈ R, so that L(x − a) can be
chosen as a primitive for 1/(x − a) on R. Hence
R
−R
dx
x − a
= L(x − a)
R
−R
.
The variation of the real part of
L(x − a) = log |x − a| + i. Arg(x − a)
over [−R, R] is equal to log(|R−a|/|R+a|) and tends to 0 since |R−a|/|R+a|
approaches 1 as R increases; the argument of R − a tends to 0 since the halfline with initial point 0 and terminal point R −a approaches the half-line R + ;
finally, the half-line with initial point 0 and terminal point −R−a approaches
R − , but is in the half-plane Im(z) < 0 if Im(a) > 0 and in the half-plane
Im(z) > 0 if Im(a) < 0; the argument of −R − a, therefore, approaches −π
if Im(a) > 0 and +π if Im(a) < 0. So
R
dx/(x − a) =
πi
if Im(a) > 0
−πi
if Im(a) < 0
.
(8.6)
Finally, in the general case, it follows that
R
f (x)dx = πi
Im(a) > 0
Res(f, a) − πi
Im(a) < 0
Res(f, a) .
(8.7)
When d
◦ (q) ≥ d
◦ (p) + 2, the sum of all the residues of f in C are zero and
(7) reduces to (3’) or (3”), as desired. If on the contrary d
◦ (q) = d
◦ (p) + 1, it
is the sum of residues in C and at infinity, which is zero by (5.14’). Then we
for example find
R
f (x)dx = 2πi
Im(a) > 0
Res(f, a) + πi Res(f, ∞) .
(8.8)