Cauchy's Integral Formula

Transcription

1 Cauchy's Integral Formula Thursday, October 24, :57 PM Last time we discussed functions defined by integration over complex contours, and let's look at a particularly important example: Where C is a fixed simple closed contour and f is a function which is Analytic in a domain containing the contour C and its interior. We will assume the orientation of C is counterclockwise. Our objective will be to characterize in detail the behavior of the function F(z) so defined. We know from the results from last time that F(z) will be analytic whenever because then the integrand will satisfy the conditions of the proposition concerning integrals with respect to a parameter. When then analyticity can (and generally does) fail because of the singularity arising from the denominator being zero for some value of the integration variable Let's now examine in more detail by considering three different cases for where z is: A. If z is in the exterior of the contour C, then the integrand is a function which is analytic on and inside the contour C, so by the Cauchy-Goursat theorem, the integral is zero: F(z) = 0 when z is in the exterior of the contour C. B. If z is in the interior of the contour C, then there is a singularity of the integrand inside the contour so we can't simply say the integral is zero; Cauchy-Goursat theorem doesn't apply in that way. But we can apply Cauchy-Goursat theorem in the form of principle of deformation of contours: If C can be deformed continuously into contour C' (by a homotopy) over a region over which g remains analytic in the interior. (OK to be nonanalytic on contours so long as continuous.) Proof is to consider the oriented boundary of the region R between these two contours as being the union of C and C' running backward. New Section 2 Page 1

2 Apply Cauchy-Goursat theorem to region R: Let's apply the principle of deformation of contours to shrink the given contour to the contour a circle around the point z, with radius, oriented counterclockwise (in the same direction as C). Claim that by doing this calculation, we will show that F(z) = f(z) for z inside C. Proof of the claim: If one didn't have an idea ahead of time what the answer would be, one could compute it by evaluating this integral parametrically. Then use analyticity of f to argue how the resulting integral should behave as In the third case, we will do a calculation which is close to how this calculation would go. For variety, we'll provide a "softer" analytical argument. Proceeding with this alternative, we will take a common approach in analysis in establishing equalities by first establishing bounds on the difference between the two quantities and proving that the bound on the difference can be made arbitrarily small. Here we apply another common technique in analysis, which is when comparing an integral involving a function to an evaluation of the function is to express the function as an integral, usually by multiplying by a suitable integral whose value is 1. New Section 2 Page 2

3 Intuition behind this manipulation is that since is analytic over the small loop, it should be approximately constant, namely approximately f(z), over that small loop. Now that we have expressed the desired difference as an integral, let's apply standard contour integral inequalities to estimate this difference: New Section 2 Page 3

4 Because f is analytic, and therefore in particular continuous at z. This establishes the classical Cauchy Integral formula, but we will present in a somewhat generalized form by proceeding to consider the third case: C. When then the contour integral is not well-defined because of a divergent singularity which would make the parametric integral not defined under usual calculus rules. (1/x singularity) However, it is very useful to make sense out of such integrals, and this can be done systematically through the notion of a Cauchy prinicpal value integral. This rigorously encodes the intuition about certain kinds of singularities having cancelling divergences upon integration. The definition of a Cauchy principal value integral for a function which has a singularity at some point is: New Section 2 Page 4

5 And if the contour integral has multiple singularities, then the principal value integral is defined similarly but with exclusions simultaneously about all singularities. If the contour integral has no singularities, then the principal value integral is same as original integral. This is not inherently a complex-valued notion; it applies perfectly well in real analysis, but not as much emphasized. Let's see now what the behavior of: New Section 2 Page 5

6 And let's assume that the contour is smooth at z (no cusps or corners). is not a closed contour, so can't use Cauchy-Goursat theorem tricks on it. But we can make a slight adjustment to get a closed contour by including the approximate semicircular contour If the curve is smooth at z, then the angles at which this contour terminates (intersects the given contour C) will approach a difference of because the contour C will look more and more like the tangent line to the curve at z, and the intersection of a circle with a line through the center of the circle will be apart. (This statement, and therefore the final answer, would be modified if the contour C had a corner or a cusp at z.) Consider now: New Section 2 Page 6

7 By Cauchy-Goursat theorem because integrand is analytic inside the contour, and continuous on it. The second integral is over a simple small contour; let's evaluate its small limit explicitly by introducing the parametric integral. (Could have done the earlier calculation for z inside the contour C similarly). New Section 2 Page 7

8 Justify bringing the limit inside the integral by using the fact that f is analytic at z, so has a neighborhood of z over which f is continuous, and therefore by looking at a subdisc of this neighborhood, f is bounded over a closed disk of the form. And then we can apply the bounded convergence theorem to bring the limit inside the integral because the integrand is uniformly bounded for and the integration domain is also uniformly bounded (by say Added after lecture: That was quite ugly. A cleaner way to conduct the endgame of this calculation: By the same argument (and intuition) as in Case 2, New Section 2 Page 8

9 New Section 2 Page 9