§ 1. Integrals of Holomorphic Functions
17
D 1 F (z) =
D 1 {f [H(z, t)] DH(z, t)} dt =
(2.15)
=
{f
[H(z, t)] D 1 H(z, t).DH(z, t)+
+ f [H(z, t)] D 1 DH(z, t)} dt =
=
{f
[H(z, t)] DH(z, t).D 1 H(z, t)+
+ f [H(z, t)] DD 1 H(z, t)} dt =
=
D {f [H(z, t)] D 1 H(z, t)} dt = f [H(z, 1)] D 1 H(z, 1) −
−f [H(z, 0)] D 1 H(z, 0)
But since H(z, 0) = μ z (0) = a is independent of z and in particular of x,
D 1 H(z, 0) = 0; and since H(z, 1) = μ z (1) = z = x + iy, D 1 H(z, 1) = 1. So
(15) becomes D 1 F (z) = f (z). If D 1 = d/dx is replaced by D 2 = d/dy, the
calculation is similar except that D 2 f = if
and D 2 H(z, 1) = i; hence
D 2 F (z) = if (z). The function F is, therefore, holomorphic and is a primitive
for f .
This is all formal calculation. To justify it, the theorem on differentiation
under the
sign (Chap. V, n
◦ 9, Theorem 9) needs to be applied. Leaving
aside subtleties unnecessary for the time being, this supposes that the function f [H(z, t)] DH(z, t) integrated in (14) has continuous functions of the
couple (z, t) ∈ G × I as partial derivatives with respect to x and y . Since f
does not present any problems,the derivatives of H(z, t) and DH(z, t) with
respect to x and y must, therefore, exist and be continuous on G × I. The
formula DD i = D i D has also been used; this is justified if H is of class C
2
on
13 G × I, in which case the previous conditions are obviously satisfied.
Calculation (15) and the relation F
= f are, therefore, justified, provided
there is a map
H : G × I −→ G
13 This is problematic since functions of class C
2 have only been defined on an
open Cartesian space; however, I is compact and G is open, so that the product
G × I ⊂ C × R = R
3 , a vertical cylinder having G as base and height 1, is neither
open nor closed in R
3 . The solution is to constrain H to be C
2 on the open set
G×]0, 1[ and H and its derivatives of at most second order to be the restrictions
of functions defined and continuous on G × I to this set. Then derivatives at
points of the form (z, 0) or (z, 1) are well-defined and the relation D1D = DD1
which holds at (z, t) for 0 < t < 1, by passing to the limit, also holds for t = 0
or 1. It would be simpler to assume that H is defined and of class C
2 on G × J,
where J is an open interval containing I. In practice, this does not change the
results in any way.
17
D 1 F (z) =
D 1 {f [H(z, t)] DH(z, t)} dt =
(2.15)
=
{f
[H(z, t)] D 1 H(z, t).DH(z, t)+
+ f [H(z, t)] D 1 DH(z, t)} dt =
=
{f
[H(z, t)] DH(z, t).D 1 H(z, t)+
+ f [H(z, t)] DD 1 H(z, t)} dt =
=
D {f [H(z, t)] D 1 H(z, t)} dt = f [H(z, 1)] D 1 H(z, 1) −
−f [H(z, 0)] D 1 H(z, 0)
But since H(z, 0) = μ z (0) = a is independent of z and in particular of x,
D 1 H(z, 0) = 0; and since H(z, 1) = μ z (1) = z = x + iy, D 1 H(z, 1) = 1. So
(15) becomes D 1 F (z) = f (z). If D 1 = d/dx is replaced by D 2 = d/dy, the
calculation is similar except that D 2 f = if
and D 2 H(z, 1) = i; hence
D 2 F (z) = if (z). The function F is, therefore, holomorphic and is a primitive
for f .
This is all formal calculation. To justify it, the theorem on differentiation
under the
sign (Chap. V, n
◦ 9, Theorem 9) needs to be applied. Leaving
aside subtleties unnecessary for the time being, this supposes that the function f [H(z, t)] DH(z, t) integrated in (14) has continuous functions of the
couple (z, t) ∈ G × I as partial derivatives with respect to x and y . Since f
does not present any problems,the derivatives of H(z, t) and DH(z, t) with
respect to x and y must, therefore, exist and be continuous on G × I. The
formula DD i = D i D has also been used; this is justified if H is of class C
2
on
13 G × I, in which case the previous conditions are obviously satisfied.
Calculation (15) and the relation F
= f are, therefore, justified, provided
there is a map
H : G × I −→ G
13 This is problematic since functions of class C
2 have only been defined on an
open Cartesian space; however, I is compact and G is open, so that the product
G × I ⊂ C × R = R
3 , a vertical cylinder having G as base and height 1, is neither
open nor closed in R
3 . The solution is to constrain H to be C
2 on the open set
G×]0, 1[ and H and its derivatives of at most second order to be the restrictions
of functions defined and continuous on G × I to this set. Then derivatives at
points of the form (z, 0) or (z, 1) are well-defined and the relation D1D = DD1
which holds at (z, t) for 0 < t < 1, by passing to the limit, also holds for t = 0
or 1. It would be simpler to assume that H is defined and of class C
2 on G × J,
where J is an open interval containing I. In practice, this does not change the
results in any way.