274 CHAPTER 11. FUNCTIONS OF ONE COMPLEX VARIABLE
semicircles, find, using the method of residues∫
γeiz
z dz. The result should depend on rand R. Show the contour integral over γR converges to 0 as R→∞. Then find the limitof the contour integral over γr and show it is πi in the limit as r→ 0. Then obtainlimits as r→ 0 and R→ ∞ and show that 2i
∫∞
−∞
sinxx ≡ 2i limR→∞
∫ R−R
sinxx dx = πi.
This is another way to get∫
∞
0sinx
x dx = π
2 .
17. Find the following improper integral.∫
∞
−∞
cosx1+x4 dx Hint: Use upper semicircle con-
tour and consider instead∫
∞
−∞
eix
1+x4 dx. This is because the integral over the semicirclewill converge to 0 as R→∞ if you have eiz but this won’t happen if you use cosz be-cause cosz will be unbounded. Just write down and check and you will see why thishappens. Thus you should use eiz
1+z4 and take real part. I think the standard calculustechniques will not work for this horrible integral.
18. Find∫
∞
−∞
cos(x)
(1+x2)2 dx. Hint: Do the same as above replacing cosx with eix.
19. Consider the following contour.
x
The small semicircle has radius r and is centered at (1,0). The large semicircle hasradius R and is centered at (0,0). Use the method of residues to compute
limr→0
(lim
R→∞
∫ R
r
x1− x3 dx+
∫ r
−R
x1− x3 dx
)This is called the Cauchy principal value for
∫∞
−∞
x1−x3 dx. The integral makes no
sense in terms of a real honest integral. The function has a pole on the x axis. How-ever, you can define such a Cauchy principal value. Rather than belabor this issue,I will illustrate with this example. These principal value integrals occur because ofcancelation. They depend on a particular way of taking a limit. They are not asmathematically respectable but are certainly important. They are in that general areaof finding something by taking a certain kind of symmetric limit.
20. Find∫ 2π
0cos(θ)
1+sin2(θ)dθ .
21. Find∫ 2π
0dθ
2−sinθ.
22. Find∫ π/2−π/2
dθ
2−sinθ.
23. Suppose you have a function f (z) which is the quotient of two polynomials in whichthe degree of the top is two less than the degree of the bottom and all zeros of thedenominator have imaginary part nonzero, and you consider the contour.
xThen define
∫γR
f (z)eiszdz in which s is real and positive. Explain why the integralmakes sense and why the part of it on the semicircle converges to 0 as R→ ∞. Usethis to find
∫∞
−∞
eisx
k2+x2 dx, k > 0.