Cauchy`s Integral Formula

Transcription

Cauchy`s Integral Formula
Cauchy's Integral Formula
Thursday, October 24, 2013
1:57 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.
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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.
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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:
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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:
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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:
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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:
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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).
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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,
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