682 Chapter9
where Lemma 9.1 has been used in passing from the fourth to the fifth
integral. Since e > 0 is arbitrary the conclusion follows.
Lemma 9.3 Let -y: z = Rei^8 , a~ 0 ~ /3, and suppose that:
- f(z) is continuous on "Y for all R ~ Ro
2. zf(z) ~Las R-+ oo for z E -y, L being a constant
Then
lim j f(z) dz= iL(/3 -a)
R-+oo(9.10-3)
'YProof For every e > 0 we have zf(z) = L + T/• with IT/I < e, provided that
R is large enough, say R >Re. Hence if R > max(Ro,Re), we have
j f(z)dz= j L~T/ dz= 1:(L+T1)idO
'Y 'Y= iL(/3 - a) + i lp T/ dO
But
Thus
Hm {P T/dO = 0
R->oo}OL
and it follows thatli~ j f(z) dz= iL(/3 -a)
R-+oo
'YCorollary 9.4 If f(z) is continuous on "Y for R ~ R 0 , and if f is regular
at oo having a zero of order 2:: 2 at oo, then
lim j f(z) dz= 0
R-+oo
(9.10-4)
'YProof Under the assumptions the function g(z) = f(l/z) is regular at
z = 0 and has at this point a zero of order two or greater. Hence