hapter Four Integration 4.. Introduction. If : D is simply a function on a real interval D,,thenthe integral tdt is, of course, simply an ordered pair of everyday 3 rd grade calculus integrals: tdt xtdt i ytdt, where t xt iyt. Thus, for example, t 2 it 3 dt 4 3 i 4. Nothing really new here. The excitement begins when we consider the idea of an integral of an honest-to-goodness complex function f : D, whered is a subset of the complex plane. Let s define the integral of such things; it is pretty much a straight-forward extension to two dimensions of what we did in one dimension back in Mrs. Turner s class. Suppose f is a complex-valued function on a subset of the complex plane and suppose a and b are complex numbers in the domain of f. In one dimension, there is just one way to get from one number to the other; here we must also specify a path from a to b. Let be a path from a to b, and we must also require that be a subset of the domain of f. 4.
Note we do not even require that a b; but in case a b, we must specify an orientation for the closed path. We call a path, or curve, closed in case the initial and terminal points are the same, and a simple closed path is one in which no other points coincide. Next, let P be a partition of the curve; that is, P z,z,z 2,,z n is a finite subset of, such that a z, b z n, and such that z j comes immediately after z j as we travel along from a to b. A Riemann sum associated with the partition P is just what it is in the real case: n SP fz j z j, j where z j is a point on the arc between z j and z j,andz j z j z j. (Note that for a given partition P, there are many SP depending on how the points z j are chosen.) If there is a number L so that given any, there is a partition P of such that SP L whenever P P, then f is said to be integrable on and the number L is called the integral of f on. This number L is usually written fzdz. Some properties of integrals are more or less evident from looking at Riemann sums: cfzdz c fzdz for any complex constant c. 4.2
fz gzdz fzdz gzdz 4.2 Evaluating integrals. Now, how on Earth do we ever find such an integral? Let :, be a complex description of the curve. We partition by partitioning the interval, in the usual way: t t t 2 t n. Then a,t,t 2,, b is partition of. (Recall we assume that t for a complex description of a curve.) A corresponding Riemann sum looks like n SP ft j t j t j. j We have chosen the points z j t j,wheret j t j t j. Next, multiply each term in the sumbyindisguise: n SP ft j t j t j t t j t j t j. j j I hope it is now reasonably convincing that in the limit, we have fzdz ft tdt. (We are, of course, assuming that the derivative exists.) Example We shall find the integral of fz x 2 y ixy from a tob i along three different paths, or contours, as some call them. First, let be the part of the parabola y x 2 connecting the two points. A complex description of is t t it 2, t : 4.3
.8.6.4.2.2.4.6.8 x Now, t 2ti, andf t t 2 t 2 itt 2 2t 2 it 3. Hence, fzdz f t tdt 2t 2 it 3 2tidt 2t 2 2t 4 5t 3 idt 5 4 5 4 i Next, let s integrate along the straight line segment 2 joining and i..8.6.4.2.2.4.6.8 x Here we have 2 t t it, t. Thus, 2 t i, and our integral looks like 4.4
fzdz f 2 t 2 tdt 2 t 2 t it 2 idt t it 2t 2 dt 2 7 6 i Finally, let s integrate along 3, the path consisting of the line segment from to together with the segment from to i..8.6.4.2.2.4.6.8 Weshalldothisintwoparts: 3,thelinefromto;and 32, the line from to i. Then we have fzdz fzdz fzdz. 3 3 32 For 3 we have t t, t. Hence, fzdz 3 t 2 dt 3. For 32 we have t it, t. Hence, fzdz 32 t itidt 2 3 2 i. 4.5
Thus, fzdz fzdz fzdz 3 3 32 6 3 2 i. Suppose there is a number M so that fz M for all z. Then where L t dt is the length of. Exercises fzdz ft tdt ft t dt M t dt ML,. Evaluate the integral zdz,where is the parabola y x 2 fromto i. 2. Evaluate z dz, where is the circle of radius 2 centered at oriented counterclockwise. 4. Evaluate fzdz, where is the curve y x 3 from i to i,and fz for y 4y for y. 5. Let be the part of the circle t e it in the first quadrant from a tob i. Findas small an upper bound as you can for z 2 z 4 5dz. 4.6
6. Evaluate fzdz where fz z 2 z and is the path from z toz 2i consisting of the line segment from to together with the segment from to 2i. 4.3 Antiderivatives. Suppose D is a subset of the reals and : D is differentiable at t. Suppose further that g is differentiable at t. Then let s see about the derivative of the composition gt. It is, in fact, exactly what one would guess. First, gt uxt,yt ivxt,yt, where gz ux,y ivx,y and t xt iyt. Then, d gt u dt x dx u dt y dy dt i v x dx v dt y dy dt. The places at which the functions on the right-hand side of the equation are evaluated are obvious. Now, apply the auchy-riemann equations: d gt u dt x dx v dt x u x i v x g t t. dy dt i v x dx i dy dt dt dx u dt x dy dt The nicest result in the world! Now, back to integrals. Let F : D and suppose F z fz in D. Suppose moreover that a and b are in D and that D is a contour from a to b. Then fzdz ft tdt, where :, describes. From our introductory discussion, we know that d dt Ft F t t ft t. Hence, 4.7
fzdz ft tdt d Ftdt F F dt Fb Fa. This is very pleasing. Note that integral depends only on the points a and b and not at all on the path. Wesaytheintegralispath independent. Observe that this is equivalent to saying that the integral of f around any closed path is. We have thus shown that if in D the integrand f is the derivative of a function F, then any integral fzdz for D is path independent. Example Let be the curve y x 2 from the point z i to the point z 3 i 9.Let sfind z 2 dz. This is easy we know that F z z 2,whereFz 3 z3. Thus, z 2 dz 3 i3 3 i 9 26 27 728 287 i 3 Now, instead of assuming f has an antiderivative, let us suppose that the integral of f between any two points in the domain is independent of path and that f is continuous. Assume also that every point in the domain D is an interior point of D and that D is connected. We shall see that in this case, f has an antiderivative. To do so, let z be any point in D, and define the function F by Fz z fzdz, where z is any path in D from z to z. Here is important that the integral is path independent, otherwise Fz would not be well-defined. Note also we need the assumption that D is connected in order to be sure there always is at least one such path. 4.8
Now, for the computation of the derivative of F: Fz z Fz L z fsds, where L z is the line segment from z to z z. Next, observe that L z ds z. Thus, fz z L z fzds, and we have Fz z Fz z fz z L z fs fzds. Now then, z fs fzds z max fs fz : sl z z L z max fs fz : sl z. We know f is continuous at z, andso lim max fs fz : sl z. Hence, z lim z Fz z Fz z fz lim z z L z fs fzds. 4.9
In other words, F z fz, and so, just as promised, f has an antiderivative! Let s summarize what we have shown in this section: Suppose f : D is continuous, where D is connected and every point of D is an interior point. Then f has an antiderivative if and only if the integral between any two points of D is path independent. Exercises 7. Suppose is any curve from to 2i. Evaluate the integral cos z 2 dz. 8. a)let Fz logz, 3 4 argz 5 4. Show that the derivative F z z. b)let Gz logz, 4 argz 7 4. Show that the derivative G z z. c)let be a curve in the right-half plane D z :Rez from i to i that does not pass through the origin. Find the integral z dz. d)let 2 be a curve in the left-half plane D 2 z :Rez from i to i that does not pass through the origin. Find the integral. z dz. 2 9. Let be the circle of radius centered at with the clockwise orientation. Find z dz.. a)let Hz z c, argz. Find the derivative H z. b)let Kz z c, argz 7. Find the derivative 4 4 K z. c)let be any path from to that lies completely in the upper half-plane and does not pass through the origin. (Upper half-plane z :Imz.) Find 4.
Fzdz, where Fz z i, argz.. Suppose P is a polynomial and is a closed curve. Explain how you know that Pzdz. 4.