680 Chapter9- Show that j (z 2 z~:) 2 dz= 7ritsint, c+: z = 2ei^8 , o::::; ()::::; 27r.
c+
12. Let f be analytic in C except for a finite number of isolated singularities
bi, b 2 , ••• , bm, and let C be a simple closed contour enclosing all thosesingularities. If z E Int C, z =f:. bk ( k = 1, ... , m ), show that
f(z) = - f Res f(() - Res J(()
k=l C=b" ( - z C=oo ( - z9.10 Some Useful Lemmas
We propose to apply the residue theorem to the evaluation of certain im-
proper real integrals. The following lemmas will be found to be of much
help in carrying out those evaluations.Lemma 9.1 (Jordan's Inequality). If 0 ::::; ()::::; %7r, then
(^2) () < sin () < ()
7r - - (9.10-1)
Proof The proof of this simple inequality was proposed in Exercises 7.2(6).
There it was suggested to the reader to consider the behavior of g( 0) =
sin()/() (0 =f:. 0), g(O) = 1, over the interval (0,^1 / 2 7r]. As an alternative we
now offer the following geometrical argument.
In (9.10-1) equality holds for()= 0. For()= %7r we have 1 =sin %7r <
(^1) /
2 7r. For^0 < () <
(^1) /
2 7r the double inequality follows by noting that y = ()
is the equation of the tangent line to the graph of y =sin() at (0, 0), while
y = 20/7r is the equation of the chord joining the points (0, 0) and (1/ 2 7r, 1)
(Fig. 9.9), and that the graph of y =sin() is concave down over (0, %7r)
since y" = - sin() < 0 on that interval. Hence the graph of y = sin() lies
below the tangent y =()and above the chord y = 20/7r, i.e.,
so that- <^20 sm. () < , () 0 < < ()^11
7r^2 7r-^20 ::::; sm. () ::::; f) ,
7r
Lemma 9.2 (Jordan's Lemma). Let 'Y be the circular arc z = Rei^8 ,
0 ::::; a ::::; () ::::; (J ::::; 7r, and suppose that:
1. f ( z) is continuous on 'Y for all R 2:: Ro
- lf(z)I::::; 0 as R -t oo for z E 'Y