426 Chapter^7= J f(g(())g'(() d(
"'!'Theorem 7.4 Let t = h( T) be a continuous one-to-one mapping of the
interval [a', ,8'] onto the interval [a, ,8], such that a == h( a'), ,8 = h(,8').
Let"(: z = z(t), a:::; t :::; ,8, be an arc of class C^1 and 'Y': z == z(h(r)),
a' ::::; T :::; ,8^1 • Then
J f(z) dz= J f(z) dz (7.8-6)
"'! "'!'Under the conditions above, 'Y' is a reparametrization of "(, and (7.8-6)
means that the integral is invariant under a reparametrization of the path.
Proof We have/3'
J f(z)dz = i, f[z(h(r))]z'(h(r))h'(r)dr
"'!'= l/3 f[z(t)]z'(t) dt
= J f(z)dz
"'!Remark This theorem is also valid when 'Y is a rectifiable arc. The
corresponding proof is left to the reader.
Theorem 7.5 Let f(z) be continuous in a region R. Then the integral
of f along a rectifiable path with graph contained in R depends only on
the endpoints of the path iff the integral over any closed rectifiable curve
with graph contained in R is zero.Proof Suppose that the rectifiable arcs 'Yl and "( 2 have the same endpoints.
Then "(1 + (-"(2) is a close.cl rectifiable curve, and if the integral over any
such curve is zero, on applying Theorem 7.2-2 and 3, we obtainJ f(z) dz= J f(z) dz+ J f(z) dz
"'11 +(-"'12)= J f(z) dz -J f(z) dz== 0