304 CHAPTER 8 • RESIDUE THEORY
s
s: g(t)
tt/4- rrJ4
F igure 8.3 Graph of g (t) = J
1+
31
0082
t dt = -Arct~ (^2 cot t).problem: The integrand 1+ 3 ~os" t is a continuous function for all t, but the
function g has a discontinuity at 1r. This condition appears to be a violation
of the fundamental t heorem of calculus, which asserts that the integral of a
continuous function must be differentiable and hence continuous. The problem
is that g (t) is not an antiderivative of 1+ 3 ~s> t for all t in the interval {O, 27r].
Oddly, it is the antiderivative at all points except 0, 7r, and 27r, which you can
verify by computing g' (t) and showing that it equal~ 1+ 3 ~ 05 :i t whenever g (t) is
defined.
The integration algorithm used by computer algebra systems here (the Risch-
Norman algorithm) gives the antiderivative g (t) = -Arct~(^2 cott), and we must
take great care in using this information.
We get the proper value of the integral by using g ( t) on the open subintervals
(0, 7r) and ( 7r, 27r) where it is continuous and taking appropriate limits:
(2
" dt (" dt 12
" __ d_t -
lo 1 + 3 cos^2 t =lo 1+3cos^2 t + ,, 1 + 3cos^2 t
L
t ~ Lt ~
= lim + Jim ----
,_,,,_o-+ • 1+3cos^2 t ,_,.,,_1'"+ - • 1 + 3cos^2 t= Jim g (t) - lim g (s) + Jim g (t) - lim g (s)t-""- s -o+ t-2K- s-11'+
7r -71" 1T -11"
= - - - + - - - = 7r.
4 4 4 4- EXAMPLE 8.12 E I va uate f, 0 2 ,, 5 _cos 4 cos8 w dO ·
Solution For values of z that lie on the unit circle C1 (0), we have