§ 1. Integrals of Holomorphic Functions
15
point a in G, we consider the uniform branch t → L(t) from Log z along μ
which takes a given value in the set Log a at t = 0, then the value of this
branch should only depend on the endpoint z = μ(1) of μ.
As we will see, the problem of primitives has a similar solution. First, if f
has a primitive F on G, integral (13) only depends on z. Conversely, suppose
that for fixed a, the value of integral (13) is, for all z, independent of the
path μ; we can then talk unequivocally of the function F (z). The function F
is then a global primitive for f .
Indeed, consider an arbitrary point b ∈ G and let D ⊂ G be an open disc
centered at b. Then, formula (2) adapted to the point b gives a primitive F D
of f on D, and F D (z) − F D (b) =
f (ζ)dζ where integration is along the line
segment [b, z]. Since a constant can be added to F D , we may assume that
F D (b) = F (b). To calculate F (z) at a point z ∈ D, f must be integrated
along an arbitrary path connecting a to z in G; for example, we can choose
a path connecting a to b in G, then a path from b to z in D; by definition,
integration along the arc connecting a to b gives F (b), and, as seen above,
the arc connecting b to z, for example the radius, gives F D (z) − F D (b) =
F D (z) − F (b); adding, we find F (z) = F D (z) on D. It follows that F is
holomorphic and satisfies F
= f on D, and hence globally on G since b is
arbitrary. As a result :
Theorem 1. A holomorphic function f on a domain G has primitive on G
if and only if its integral along any admissible path in G depends only on the
latter’s endpoints.
In particular, the integral of f along a closed path, i.e. such that μ(0) =
μ(1), is zero. In fact this condition is sufficient for ensuring the existence of
a primitive. Indeed, if
μ 1 , μ 2 : [0, 1] −→ G
are two paths connecting a given point a to the same point z, we get a
closed path [0, 1] −→ G by following first the path [0, 1/2] −→ G given by
t → μ 1 (2t), then the path : [1/2, 1] −→ G given by t → μ 2 (2 − 2t); Clearly,
the integral of f along the first path is equal to the integral along μ 1 and
the integral along the second path is the opposite of the integral along the
first one. The integral along the total path:
11 [0, 1] −→ G is, therefore, the
difference between the integrals along μ 1 and μ 2 . As a result, these are equal
for all μ 1 and μ 2 . Hence the result follows from Theorem 1 : A holomorphic
function f on a domain G has primitive on G if and only if its integral along
any admissible closed path in G is zero.
For another proof of theorem 1, take an open disc D ⊂ G centered at z.
To go from a to a point z + h ∈ D, we can follow a path connecting a ` a z and
then the radius [z, z + h], i.e. the path t → z + th; then F (z + h) − F (z) is
11 which may not be C
1 even if that is the case of μ1 and μ2. Hence it is necessary
to include paths that are. . . piecewise admissible or at least C
1 .
15
point a in G, we consider the uniform branch t → L(t) from Log z along μ
which takes a given value in the set Log a at t = 0, then the value of this
branch should only depend on the endpoint z = μ(1) of μ.
As we will see, the problem of primitives has a similar solution. First, if f
has a primitive F on G, integral (13) only depends on z. Conversely, suppose
that for fixed a, the value of integral (13) is, for all z, independent of the
path μ; we can then talk unequivocally of the function F (z). The function F
is then a global primitive for f .
Indeed, consider an arbitrary point b ∈ G and let D ⊂ G be an open disc
centered at b. Then, formula (2) adapted to the point b gives a primitive F D
of f on D, and F D (z) − F D (b) =
f (ζ)dζ where integration is along the line
segment [b, z]. Since a constant can be added to F D , we may assume that
F D (b) = F (b). To calculate F (z) at a point z ∈ D, f must be integrated
along an arbitrary path connecting a to z in G; for example, we can choose
a path connecting a to b in G, then a path from b to z in D; by definition,
integration along the arc connecting a to b gives F (b), and, as seen above,
the arc connecting b to z, for example the radius, gives F D (z) − F D (b) =
F D (z) − F (b); adding, we find F (z) = F D (z) on D. It follows that F is
holomorphic and satisfies F
= f on D, and hence globally on G since b is
arbitrary. As a result :
Theorem 1. A holomorphic function f on a domain G has primitive on G
if and only if its integral along any admissible path in G depends only on the
latter’s endpoints.
In particular, the integral of f along a closed path, i.e. such that μ(0) =
μ(1), is zero. In fact this condition is sufficient for ensuring the existence of
a primitive. Indeed, if
μ 1 , μ 2 : [0, 1] −→ G
are two paths connecting a given point a to the same point z, we get a
closed path [0, 1] −→ G by following first the path [0, 1/2] −→ G given by
t → μ 1 (2t), then the path : [1/2, 1] −→ G given by t → μ 2 (2 − 2t); Clearly,
the integral of f along the first path is equal to the integral along μ 1 and
the integral along the second path is the opposite of the integral along the
first one. The integral along the total path:
11 [0, 1] −→ G is, therefore, the
difference between the integrals along μ 1 and μ 2 . As a result, these are equal
for all μ 1 and μ 2 . Hence the result follows from Theorem 1 : A holomorphic
function f on a domain G has primitive on G if and only if its integral along
any admissible closed path in G is zero.
For another proof of theorem 1, take an open disc D ⊂ G centered at z.
To go from a to a point z + h ∈ D, we can follow a path connecting a ` a z and
then the radius [z, z + h], i.e. the path t → z + th; then F (z + h) − F (z) is
11 which may not be C
1 even if that is the case of μ1 and μ2. Hence it is necessary
to include paths that are. . . piecewise admissible or at least C
1 .