Singularities/Residues/Applications 725Hence
1 1lfo(z)J = JI+ z2JJl + zJlf2JI - zJl/2 < el/2
which givesase-tO
Similarly,~f,(z)dz ~o as e -t^0
Thus if we let R -t oo, then e -t 0 in (9.11-61), we get
0 _ 271" 21
1
dx
--+- v'2 -1 (1 + x^2 )v'l - x^2
or
1
(^1) dx 71"
-1 (1 + x^2 )Vf=X2 = v'2
- To evaluate
l
b v A+ 2B - + C lb 1
a X^2 X dx = a -v' X A(x - a)(x - b) dx
where the coefficients A, B, C, are real and A < 0, by assuming that (i)~O < a < b and (ii) a < 0 < b. The condition A < 0 is necessary for the
square root to be real over the interval [a, b].(i) In this case both a and bare positive, and since a+b = -2B/A, ab=
C /A, it follows that B > 0 and C < 0. Also, the assumption concerning
the roots of the trinomial implies B^2 - AC > 0.
Passing to the complex plane, we let f(z) = y'~(A_/_z_^2 -)(-z---a)_(_z ___ b_) and
slit the plane along the segment joining the branch points a and b. As
before, the value of the square root to be taken at each point of the cut
plane is to be specified. Of course, this is to be done so that the resulting
function is analytic <C - [a, b] and continuous on its closure. Proceeding
as in Example 1, let
z = reiw f:. 0