§ 3. Some Applications of Cauchy’s Method
93
When f is bounded on R without being integrable, the function
F (z) =
1
2πi
f (t)dt
t − z
= lim F n (z)
is not well-defined in general. u f (x, y) can indeed be defined: if f is real, it
is, up to a factor of 2, the real part of the non-existent function F (z); but its
imaginary part is defined by a divergent integral (12). However, if the integral
that should define F is integrated term by term with respect to z, we now
get a function
G(z) =
1
2πi
f (t)dt
(t − z)
2 =
1
2πi
d
dz
1
t − z
f (t)dt , Im(z) = 0
(11.16)
defined and holomorphic for Im(z) = 0 and satisfying G(z) = F
(z) when
F exists. Failing to use (3’) in order to define the function F that we are
looking for, a primitive for G on Im(z) > 0 could perhaps instead be used as
a substitute. To simplify, f will be assumed to be real in the rest of this n
◦ .
Let us first prove an important result which could have been another
corollary of Theorem 3 in n
◦ 3 related to the existence of primitives of holomorphic functions :
Theorem 10. On a simply connected domain, a real harmonic function is
the real part of a holomorphic function and is unique up to the addition of a
pure imaginary constant.
Uniqueness is obvious. Let u be a harmonic function; suppose there is a
holomorphic function f = u + iv such that u = Re f . Then
f
= D 1 u + iD 1 v and D 1 v = −D 2 u ,
and so f
= D 1 u − iD 2 u. Conversely, starting with a harmonic function u,
Laplace’s equation says precisely that D 1 u − iD 2 u is holomorphic. Now, on
a simply connected domain, this function has a primitive
f (z) =
z
a
[D 1 u(ζ) − iD 2 u(ζ)] dζ ,
(
∗∗ )
where integration is along an arbitrary path connecting a fixed point a to the
point z in G. Setting f = p + iq,
f
= D 1 p − iD 2 p = D 1 u − iD 2 u ,
so that the derivatives of u and p are identical. Hence u = p + c, where c is
a real constant and the function f (z) + c solves the problem, qed.
Coming back to the construction of a primitive for G, we see that a
holomorphic function F 1 exists on the half-plane H
+ : Im(z) > 0 such that
u f (x, y) = F 1 (z) + F 1 (z) .
(11.17)
93
When f is bounded on R without being integrable, the function
F (z) =
1
2πi
f (t)dt
t − z
= lim F n (z)
is not well-defined in general. u f (x, y) can indeed be defined: if f is real, it
is, up to a factor of 2, the real part of the non-existent function F (z); but its
imaginary part is defined by a divergent integral (12). However, if the integral
that should define F is integrated term by term with respect to z, we now
get a function
G(z) =
1
2πi
f (t)dt
(t − z)
2 =
1
2πi
d
dz
1
t − z
f (t)dt , Im(z) = 0
(11.16)
defined and holomorphic for Im(z) = 0 and satisfying G(z) = F
(z) when
F exists. Failing to use (3’) in order to define the function F that we are
looking for, a primitive for G on Im(z) > 0 could perhaps instead be used as
a substitute. To simplify, f will be assumed to be real in the rest of this n
◦ .
Let us first prove an important result which could have been another
corollary of Theorem 3 in n
◦ 3 related to the existence of primitives of holomorphic functions :
Theorem 10. On a simply connected domain, a real harmonic function is
the real part of a holomorphic function and is unique up to the addition of a
pure imaginary constant.
Uniqueness is obvious. Let u be a harmonic function; suppose there is a
holomorphic function f = u + iv such that u = Re f . Then
f
= D 1 u + iD 1 v and D 1 v = −D 2 u ,
and so f
= D 1 u − iD 2 u. Conversely, starting with a harmonic function u,
Laplace’s equation says precisely that D 1 u − iD 2 u is holomorphic. Now, on
a simply connected domain, this function has a primitive
f (z) =
z
a
[D 1 u(ζ) − iD 2 u(ζ)] dζ ,
(
∗∗ )
where integration is along an arbitrary path connecting a fixed point a to the
point z in G. Setting f = p + iq,
f
= D 1 p − iD 2 p = D 1 u − iD 2 u ,
so that the derivatives of u and p are identical. Hence u = p + c, where c is
a real constant and the function f (z) + c solves the problem, qed.
Coming back to the construction of a primitive for G, we see that a
holomorphic function F 1 exists on the half-plane H
+ : Im(z) > 0 such that
u f (x, y) = F 1 (z) + F 1 (z) .
(11.17)