LECTURE-13 : GENERALIZED CAUCHY S THEOREM
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1 LECTURE-3 : GENERALIZED CAUCHY S THEOREM VED V. DATAR The aim of this lecture to prove a general form of Cauchy s theorem applicable to multiply connected domains. We end with computations of some real integrals using Cauchy s theorem. Recall that a curve γ : [a, b] C is called closed if γ(a) = γ(b), and is called simple if γ is injective on [a, b). For this lecture, we restrict attention to open sets Ω C with finitely many boundary components, each of which is a simple piecewise smooth closed curve. The boundary is said to be positively oriented if the domain in always on the left as the boundary is traversed. Note that a Jordan curve is said to be positively oriented if the interior always lies to the left as the curve is traversed. As a result of this, if our domain is multiply connected, then boundary components corresponding to holes will have opposite orientation from their natural orientation as independent Jordan curves (see figure). We not state the theorem that generalizes all the versions of Cauchy s theorems we have seen until now. Theorem 0. (Generalized Cauchy s theorem). Let Ω C with finitely many boundary components, each of which is a simple piecewise smooth closed curve, and let f : Ω C be a holomorphic function which extends continuously to the closure Ω. Then f(z) dz = 0, Ω where the boundary Ω is positively oriented. Outline of a proof of Generalized Cauchy s theorem We first state an extension for Cauchy s theorem for simply connected domains. Since the proof is rather technical, we only offer a brief overview of the proof, indicating where the technicalities lie. Lemma 0.. Let U be a simply connected domain with U a simply, closed curve. If f : U C is a holomorphic function which extends continuously to the boundary, f(z) dz = 0. U We have already shown that f has a primitive on U and hence the integral on any closed curve lying inside U is zero. The problem, is that we want to calculate the integral on the boundary of U. The way to prove this is to
2 construct a closed, simple curve U, for all ε > 0 small, which is close enough to U so that f(z) dz f(z) dz ε. U If a can be constructed close enough to U, then the integral estimate follows from continuity of f on the closure U. Constructing such a approximating curve which is simple and closed is non trivial. If U were smooth, such a curve is constructed by perturbing U a little bit along the inner normal. It still requires some proof that the resulting does not have self intersections. But in our case the boundary might be only piecewise smooth. So extra care must be taken to round off the corners. Assuming the above lemma, let the boundary components of Ω be labeled C, C,, C n so that the domain Ω is given by the intersection of the interior region to C and the exterior regions to C,, C n. The idea, as in the figure below, is to cut the domain into a union of two simply connected domains but introducing polygonal paths L 0, L,, L n, where Each L i intersects exactly two boundary components. L 0 connects C to C, and L n connects C n to C. For j n, L j connects C j to C j+. Figure. Figure from Brown-Churchill. The integrals over the common boundaries are taken with opposite orientations, and hence cancel. With the cross cuts, we can write Ω = Ω Ω where both Ω and Ω are simply connected. Let Γ and Γ denote their boundaries with positive orientation (as in the figure). We then notice that the common part of the boundaries, namely L 0, C, L,, L n, C n, L n have opposite orientations
3 in Γ and Γ. With this observation, we can write f(z) dz = f(z) dz + f(z) dz, Ω Γ Γ since the integrals on the shared parts cancel. But then the two integrals on the right are zero by the extension of Cauchy s theorem for simply connected regions (i.e. Lemma 0.), and this completes the proof of the generalized Cauchy s theorem. Some applications Index of a closed simple curve. Give any closed curve γ and a point a / γ we can define the index as the following integral n(γ, a) = πi z a dz. Then we have the following observation. Proposition 0.. If γ is a simple closed curve, then {, if a is in the interior of γ n(γ, a) = 0, otherwise. Proof. We have already seen that the statement is true if γ is a circle traversed once. For a general γ, suppose a is in the interior. Then since the interior is open, there exists a small disc D r (a) which lies completely inside the interior of γ. But then the intersection of the interior of γ and the exterior of the disc forms an open set which we call Ω. Let the boundary of the disc, with positive orientation, be denoted by C. Then the boundary Ω, with the positive orientation, is given by γ C. So by Cauchy s theorem, since /(z a) is holomorphic in Ω and continuous in Ω, 0 = πi Ω z a dz = πi γ γ z a πi C z a, showing that in this case n(γ, a) =, since the index n(c, a) =. The same argument shows that when a is in the exterior, n(γ, a) = 0. A real variable integral. We will now apply Cauchy s theorem to compute a real variable integral. Later in the course, once we prove a further generalization of Cauchy s theorem, namely the residue theorem, we will conduct a more systematic study of the applications of complex integration to real variable integration. For now, let us compute cos x 0 x dx. This is an improper integral, so by definition 0 cos x x dx = lim lim R ε 0 3 R ε cos x x dx,
4 if the limits exist. Near x = 0, cos x x (which follows for instance from the Taylor expansion), and so near zero, ( cos x)/x is bounded, and so by the comparison principle the improper integral is convergent. Near x =, cos x remains bounded, and so the integrand is comparable to /x, and hence by the p-test, the integral is also convergent near infinity. So all we need is the compute the integral on the right and then take the limits ε ε and R. Since cos x and x are even functions we can write R cos x ( ε cos x R cos x ) ε x dx = R x dx + ε x dx. Consider now the function f(z) = eiz z. This is a holomorphic function on C. Moreover, on the real line ie. when z is a real number the real part of this function is precisely the function that we are looking to integrate. Now consider a contour Γ = γ + γ + γ R where Clearly γ, γ are paths along the real axis, from ( R, 0) to ( ε, 0) and (ε, 0) to (R, 0) respectively. is the semi circle from (0, ε) to ( ε, 0) with positive orientation (that is, going anti-clockwise). γ R is the big semi circle from (R, 0) to ( R, 0) with positive orientation. f(z) dz + f(z) dz = γ γ ε R e ix R e ix x + ε x, which is the integral we want to evaluate. By Cauchy s theorem f(z) dz + γ f(z) dz = γ f(z) dz f(z) dz. γ R So it is enough to evaluate the integrals on the right. On γ R we claim that eiz z R. To see this, for z γ R, we can write z = x+iy with y > 0. So e iz = e y <, and hence by triangle inequality e iz < which proves the claim since z = R on γ R. Using this we can estimate that e γr iz dz R len(γ R) = π R 0 z 4
5 as R. So the contribution on γ R goes to zero as R goes to infinity. Next, on, using the power series definition of exponential, we can write So we can write e iz = + iz iz e iz z + higher order terms. = i z + E(z), where E(z) < on for all small ε > 0. But then clearly, E(z) dz len( ) 0 as ε 0. Hence dz lim f(z) dz = i ε 0 z = π. This shows that cos x x dx = π. Department of Mathematics, UC Berkeley address: vvdatar@berkeley.edu o 5
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