1550251515-Classical_Complex_Analysis__Gonzalez_

728 Chapter9

so that

J

27rB

fo(z)dz = J=A

r+

Also,

J

fo(z) dz= 27ri Res fo(z) = 27ri lim J Az^2 + 2Bz + C

z=O z->0

'"ft

since the square root must be of the form -ik(k > 0) for x <a. As to the

integrals along 1i and 1t, we have

lim E->0 J fo(z) dz= 0

'"(i

and

lim E->0 J f 0 (z) dz= 0

'"ft

since

lim(z - a)fo(z) = 0

z->a

and lim(z - b)fo(z) = 0
z-tb
Therefore, by taking limits in (9.11-63) as E -t 0, and using the preceding
results, we find that

27rB 1

6

r-x-= 271'V-C + 2 -v'^1 Ax2 + 2Bx + C dx

v-A a x

or,

t .!_ J Ax^2 + 2Bx + C dx = 71' ( _!!__ - r-c)

Ja X J=A

(9.11-64)

(ii) In the case a < 0 < b, x = 0 is a simple pole of the integrand and

the integral must be considered in the sense of a principal value. As before,
we have A < 0 and the equation ab = C /A implies that C > 0. We shall
use the branch off as in case (i), namely, fo(z) for any point z =/= O, and a

contour similar to that of Fig. 9.28, except for an indentation I'd: z = 8ei^8 ,

0 ~ 0 ~ 71', 0 < 8 < min(JaJ, b), on the upper boundary of the cut, and

another 1~ +: z = 8ei^8 , 71' ~ 0 ~ 271', on the lower boundary which will