INTEGRALS
49. MULTIPLY CONNECTED DOMAINS
A domain that is not simply connected (Sec. 48) is said to be multiply connected.
The following theorem is an adaptation of the Cauchy–Goursat theorem to multiply connected domains.
Theorem. Suppose that
(a) Cis a simple closed contour, described in the counterclockwise direction;
(b) Ck(k=1,2, . . . , n)are simple closed contours interior to C, all described in the clockwise direction, that are disjoint and whose interiors have no points in common (Fig. 60).
If a function f is analytic on all of these contours and throughout the multiply connected domain consisting of the points inside Cand exterior to eachCk,then
C
f (z) dz+ n k=1
Ck
f (z) dz=0.
(1)
sec. 49 Multiply Connected Domains 159
O x y
C L1
L2 L3
C2 C1
FIGURE 60
Note that in equation (1), the direction of each path of integration is such that the multiply connected domain lies to theleft of that path.
To prove the theorem, we introduce a polygonal pathL1, consisting of a finite number of line segments joined end to end, to connect the outer contour C to the inner contour C1. We introduce another polygonal path L2 which connects C1 to C2; and we continue in this manner, with Ln+1 connecting Cn toC. As indicated by the single-barbed arrows in Fig. 60, two simple closed contours1 and2 can be formed, each consisting of polygonal pathsLk or−Lk and pieces ofC andCk
and each described in such a direction that the points enclosed by them lie to the left. The Cauchy–Goursat theorem can now be applied tof on1and2, and the sum of the values of the integrals over those contours is found to be zero. Since the integrals in opposite directions along each pathLk cancel, only the integrals along Cand theCk remain; and we arrive at statement (1).
Corollary. LetC1andC2denote positively oriented simple closed contours, whereC1is interior toC2(Fig. 61). If a functionf is analytic in the closed region consisting of those contours and all points between them, then
C2
f (z) dz=
C1
f (z) dz.
(2)
O x y
C2 C1
FIGURE 61
This corollary is known as theprinciple of deformation of paths since it tells us that if C1 is continuously deformed into C2, always passing through points at
which f is analytic, then the value of the integral off overC1 never changes. To verify the corollary, we need only write equation (2) as
C2
f (z) dz+
−C1
f (z) dz=0 and apply the theroem.
EXAMPLE. WhenC is any positively oriented simple closed contour sur- rounding the origin, the corollary can be used to show that
C
dz z =2π i.
This is done by constructing a positively oriented circleC0with center at the origin and radius so small thatC0lies entirely insideC(Fig. 62). Since (see Example 2, Sec. 42)
C0
dz z =2π i
and since 1/zis analytic everywhere except atz=0, the desired result follows.
Note that the radius of C0 could equally well have been so large that C lies entirely inside C0.
x C0
C
O y
FIGURE 62
EXERCISES
1. Apply the Cauchy–Goursat theorem to show that
C
f (z) dz=0
when the contourC is the unit circle|z| =1, in either direction, and when (a)f (z)= z2
z−3; (b)f (z)=z e−z; (c)f (z)= 1 z2+2z+2; (d)f (z)=sechz; (e)f (z)=tanz; ( f)f (z)=Log(z+2).
sec. 49 Exercises 161
2. LetC1denote the positively oriented boundary of the square whose sides lie along the linesx= ±1, y= ±1 and let C2 be the positively oriented circle|z| =4 (Fig. 63).
With the aid of the corollary in Sec. 49, point out why
C1
f (z) dz=
C2
f (z) dz when
(a)f (z)= 1
3z2+1; (b)f (z)= z+2
sin(z/2); (c)f (z)= z 1−ez.
1 4 x
y
C1 C2
FIGURE 63 3. IfC0 denotes a positively oriented circle|z−z0| =R, then
C0
(z−z0)n−1 dz=0 whenn= ±1,±2, . . . , 2π i whenn=0,
according to Exercise 10(b), Sec. 42. Use that result and the corollary in Sec. 49 to show that if C is the boundary of the rectangle 0≤x≤3,0≤y≤2, described in the positive sense, then
C
(z−2−i)n−1dz=0 whenn= ±1,±2, . . . , 2π i whenn=0.
4. Use the following method to derive the integration formula ∞
0
e−x2cos 2bx dx=
√π
2 e−b2 (b >0).
(a) Show that the sum of the integrals ofe−z2 along the lower and upper horizontal legs of the rectangular path in Fig. 64 can be written
x a a + bi –a + bi
–a O
y
FIGURE 64
2 a
0
e−x2dx−2eb2 a
0
e−x2cos 2bx dx
and that the sum of the integrals along the vertical legs on the right and left can be written
ie−a2 b
0
ey2e−i2aydy−ie−a2 b
0
ey2ei2aydy.
Thus, with the aid of the Cauchy–Goursat theorem, show that a
0
e−x2cos 2bx dx=e−b2 a
0
e−x2dx+e−(a2+b2) b
0
ey2sin 2ay dy.
(b) By accepting the fact that∗ ∞
0
e−x2dx=
√π 2 and observing that
b
0
ey2sin 2ay dy ≤
b
0
ey2dy,
obtain the desired integration formula by lettingatend to infinity in the equation at the end of part(a).
5. According to Exercise 6, Sec. 39, the pathC1from the origin to the pointz=1 along the graph of the function defined by means of the equations
y(x)=
x3sin(π/x) when 0< x≤1,
0 whenx=0
is a smooth arc that intersects the real axis an infinite number of times. LetC2 denote the line segment along the real axis fromz=1 back to the origin, and letC3 denote any smooth arc from the origin toz=1 that does not intersect itself and has only its end points in common with the arcsC1 andC2(Fig. 65). Apply the Cauchy–Goursat theorem to show that if a functionf is entire, then
C1
f (z) dz=
C3
f (z) dz and
C2
f (z) dz= −
C3
f (z) dz.
Conclude that even though the closed contourC=C1+C2intersects itself an infinite
number of times,
C
f (z) dz=0.
∗The usual way to evaluate this integral is by writing its square as
∞
0
e−x2dx
∞
0
e−y2dy=∞
0
∞
0
e−(x2+y2)dxdy
and then evaluating this iterated integral by changing to polar coordinates. Details are given in, for example, A. E. Taylor and W. R. Mann, “Advanced Calculus,” 3d ed., pp. 680–681, 1983.
sec. 49 Exercises 163
O 1 x
y
C1
C3 C2
1– 3
1– 2
FIGURE 65
6. LetC denote the positively oriented boundary of the half disk 0≤r≤1,0≤θ ≤π, and letf (z)be a continuous function defined on that half disk by writingf (0)=0 and using the branch
f (z)=√ reiθ/2
r >0,−π
2 < θ <3π 2
of the multiple-valued functionz1/2. Show that
C
f (z) dz=0
by evaluating separately the integrals of f (z) over the semicircle and the two radii which make upC. Why does the Cauchy–Goursat theorem not apply here?
7. Show that if C is a positively oriented simple closed contour, then the area of the region enclosed byC can be written
1 2i
C
z dz.
Suggestion:Note that expression (4), Sec. 46, can be used here even though the functionf (z)=zis not analytic anywhere [see Example 2, Sec. 19].
8. Nested Intervals. An infinite sequence of closed intervalsan≤x≤bn(n=0,1,2, . . .) is formed in the following way. The intervala1≤x≤b1 is either the left-hand or right-hand half of the first intervala0≤x≤b0, and the intervala2≤x≤b2 is then one of the two halves ofa1≤x≤b1, etc. Prove that there is a pointx0which belongs to every one of the closed intervalsan≤x≤bn.
Suggestion: Note that the left-hand end points an represent a bounded nonde- creasing sequence of numbers, sincea0≤an≤an+1< b0; hence they have a limit Aasntends to infinity. Show that the end pointsbnalso have a limitB. Then show thatA=B, and writex0=A=B.
9. Nested Squares. A square σ0:a0≤x≤b0, c0≤y≤d0 is divided into four equal squares by line segments parallel to the coordinate axes. One of those four smaller squaresσ1:a1≤x≤b1, c1≤y≤d1 is selected according to some rule. It, in turn, is divided into four equal squares one of which, calledσ2, is selected, etc. (see Sec.
47). Prove that there is a point(x0, y0) which belongs to each of the closed regions of the infinite sequenceσ0, σ1, σ2, . . . .
Suggestion: Apply the result in Exercise 8 to each of the sequences of closed intervalsan≤x≤bnandcn≤y≤dn(n=0,1,2, . . .).
50. CAUCHY INTEGRAL FORMULA