1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

6.5 • INTEGRAL REPRESENTATIONS FOR ANALYTIC FUNCTIONS 237

f (z) = ex~~;z) and use Theorem 6.10 to conclude that

r exp(i?rz)dz 1 r J(z)dz

lct 2z^2 - 5z + 2 = 2 let co) z - ~

= ~ (27ri) f (~)

.exp (if) 211'

= ?ri 1 = -.

2 -2 3

We now state a general result that shows how to accomplish differentiation
under the integral sign. The proof is presented in some advanced texts. See,
for instance, Rolf Nevanlinna and V. Paatero, Introduction to Complex Analysis
(Reading, Mass.: Addison-Wesley, 1969), Section 9.7.

We now generalize Theorem 6.10 to give an integral representation for the
nth derivative, j(n) (z). We use Leibniz's rule in the proof and note that this
method of proof is a mnemonic device for remembering Theorem 6 .12.