1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 489

• [! J(()d( J f(()d( l


• 'Y ((-z-h)n((-z) - 'Y ((-z)n+l

• [! J(()d( J f(()d( l +

'Y (( - z - h)n-1(( - z)2 - 'Y (( - z)n+l ...

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!(() d( J f(() d( l

+ 'Y ((-z-h)((-z)n+1 - 'Y ((-z)n+l

• J f(()d(
  n (( - z)n+l (7.20-5)
  'Y
  By the continuity property just proved, each of the expressions in brackets
  in (7.20-5) tends to zero as h --t 0. For instance, the difference in the first
  bracket can be written as

J

f(()d(/((-z) -1 f(()d(/((-z)

((-z-h)n 'Y ((-z)n

'Y

and since fi(() = f(()/(( -z) is continuous along 'Yi that difference tends

to zero ash --t 0. For the second bracket let h(() = !(()/((-z)2, and so
on. Hence, taking lill1:its in (7.20-5) as h --t O, we get

By letting F 1 (z) = F(z) we obtain, applying (7.20-6) repeatedly,

F'(z) = 1 · F2(z)

F"(z) = 1·F~(z)=1·2 · Fa(z)

F'"(z) = 1·2·F~(z)=1 · 2 · 3 · F 4 (z)

p(n)(z) = 1 · 2 · 3 · · · n · Fn+1(z)

'J f(()d(

=n. ((-z)n+l

'Y

The first formula above is the same as (7.20-1,).

(7.20-6)

(7.20-7)

Theorem 7.29 The function F(z), as well as all its derivatives, tend to
zero as z --t oo in Ro.