1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 449

= J P 6..x + Q 6..y = 0

8Lmn

Corollary 7.11 If 6..xQ - 6..yP = 0 on Lm-l,n-1, then

f P 6..x + Q 6..y = 0

JaLmn ·

Lemma 7.4 Let S = [O, 1] x [O, 1] and let AC C be open. Suppose that

h: S --t A defines a continuous mapping of the square S into A, and let

f: A --t C be analytic in A. Consider the boundary {JS of Sas oriented in

the positive sense and decomposed as follows: {JS= {)SB+8SR+8Sr+8SL,

where {)SB stands for the bottom side, {)SR for the right-hand side, 8Sr

stands for the top side, and {)SL for the left-hand side (Fig. 7.10). Then
we have

1. Jh(as) f(z) dz = 0

2. If h(O,r) = h(l,r), 0:::; r:::; 1, then

J f(z) dz+ J f(z) dz= 0

h(8Sn) h(8ST)

Proof If A= C, the conclusions follow from Corollary 7.10 and the remark

preceding Definition 7.8. Assuming now that A f C, let r = d(A', h(S)).

Clearly, r > 0 since A' = C - A is closed and h(S) is compact. By virtue

of the uniform continuity of h on S there is a positive integer N such that

v'2

lh(p 1 ) - h(pz)I < r whenever P1,P2 ES and IP1 - Pzl < N

T y

2§r h

asL! s tasR

(^0) as;^1 x
h(il8 8 )
Fig. 7.10