266 CHAPTER 11. FUNCTIONS OF ONE COMPLEX VARIABLE
where γ is the unit circle oriented counter clockwise. Now the integrand has poles of order1 at those points where z
(4z+ z2 +1
)= 0. These points are
0,−2+√
3,−2−√
3.
Only the first two are inside the unit circle. It is also clear the function has simple poles atthese points. Therefore,
res( f ,0) = limz→0
z(
z2 +1z(4z+ z2 +1)
)= 1.
res(
f ,−2+√
3)=
limz→−2+
√3
(z−(−2+
√3)) z2 +1
z(4z+ z2 +1)=−2
3
√3.
It follows ∫π
0
cosθ
2+ cosθdθ =
12i
∫γ
z2 +1z(4z+ z2 +1)
dz =12i
2πi(
1− 23
√3)
= π
(1− 2
3
√3).
Other rational functions of the trig functions will work out by this method also.Sometimes you have to be clever about which version of an analytic function you wish
to use. The following is such an example.
Example 11.5.7 The integral here is∫
∞
0lnx
1+x4 dx.
It is natural to try and use the contour in the following picture in which the small circlehas radius r and the large one has radius R.
x
y
However, this will create problems with the log since the usual version of the log is notdefined on the negative real axis. This difficulty may be eliminated by simply using anotherbranch of the logarithm as discussed above. Leave out the ray from 0 along the negativey axis and use this example to define L(z) on this set. Thus L(z) = ln |z|+ iarg1 (z) wherearg1 (z) will be the angle θ , between −π
2 and 3π
2 such that z = |z|eiθ . Of course, withthis contour, this will end up finding the integral
∫∞
−∞
ln|x|1+x4 dx. Then the function used is
f (z)≡ L(z)1+z4 . Now the only singularities contained in this contour are
12
√2+
12
i√
2,−12
√2+
12
i√
2
and the integrand f has simple poles at these points. Thus res(
f , 12
√2+ 1
2 i√
2)=
limz→ 1
2√
2+ 12 i√
2
(z−(
12
√2+ 1
2 i√
2))
(ln |z|+ iarg1 (z))
1+ z4