1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 431

and it follows that

1

z+h

F(z+h)-F(z)=(-y') z J(()d(

1

z+h

=hf(z)+(-y') z [f(()-J(z)]d(

and

F(z + h) - F(z) = f(z) + ]:_ ('Y') 1z+h [!(() - f(z)] d(

h h z

The segment 1' is defined by ( = z +ht, 0 :::; t :=:; 1, and since f is

continuous, for every € > 0 there is a 8 > 0 such that If(() - f(z)I < €

whenever IC -zl = lhtl :::; lhl < 8. By Darboux's inequality we have

1

1 ('Y')

1

z+h I 1

h z [!(() -f(z)] d( :::; ThJclhl = €

so that

IF(z+hi-F(z) -f(z)I <E

provided that lhl < 8. This shows that F'(z) = f(z).
Now suppose that F' ( z) = f ( z) for every z E R, and consider the integral

('Y) 1:

1

F' ( () d(

along any rectifiable path 1: z = z(t), a:::; t:::; /3, with graph contained in
R. Let P be any partition of the interval [a, /3], and let

{zo = (0,(1, ... ,(n = z1}

be the corresponding subdivision induced on 1· By Definition 7.6 we have

(-y) 1z

1

F'(()d( = lim tF^1 ((k-1)((k - (k-1)

zo IPl->O k=l

where F'(() has been evaluated at the points (k-l for convenience. Ifwe let

F((k) -F((k-1) _ F'(r ) _

r r <,k-1 - T/k

<,k - <,k-1

(7.8-10)

by Theorem 6.38 we have that for every € > 0 there is a 8 > 0 such that

l(k - (k-11 < 8 implies that l77kl < € (k = 1, ... , n). Of course, by the

uniform continuity of z = z(t) on [a,/3], there exists a 8' > 0 such that

\Pl < 8' implies that (k - (k-1 \ < 8.