In mathematics, Cauchy's integral formula, named after Augustin-Louis Cauchy, is a central statement in complex analysis. It expresses the fact that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk, and it provides integral formulas for all derivatives of a holomorphic function. Cauchy's formula shows that, in complex analysis, "differentiation is equivalent to integration": complex differentiation, like integration, behaves well under uniform limits – a result that does not hold in real analysis.
Theorem
Let be an open subset of the complex plane , and suppose the closed disk defined as
is completely contained in . Let be a holomorphic function, and let be the circle, oriented counterclockwise, forming the boundary of . Then for every in the interior of ,
The proof of this statement uses the Cauchy integral theorem and like that theorem, it only requires to be complex differentiable. Since can be expanded as a power series in the variable
it follows that holomorphic functions are analytic, i.e. they can be expanded as convergent power series. In particular is actually infinitely differentiable, with
This formula is sometimes referred to as Cauchy's differentiation formula.
The theorem stated above can be generalized. The circle can be replaced by any closed rectifiable curve in which has winding number one about . Moreover, as for the Cauchy integral theorem, it is sufficient to require that be holomorphic in the open region enclosed by the path and continuous on its closure.
Note that not every continuous function on the boundary can be used to produce a function inside the boundary that fits the given boundary function. For instance, if we put the function , defined for , into the Cauchy integral formula, we get zero for all points inside the circle. In fact, giving just the real part on the boundary of a holomorphic function is enough to determine the function up to an imaginary constant — there is only one imaginary part on the boundary that corresponds to the given real part, up to addition of a constant. We can use a combination of a Möbius transformation and the Stieltjes inversion formula to construct the holomorphic function from the real part on the boundary. For example, the function has real part . On the unit circle this can be written . Using the Möbius transformation and the Stieltjes formula we construct the function inside the circle. The term makes no contribution, and we find the function . This has the correct real part on the boundary, and also gives us the corresponding imaginary part, but off by a constant, namely .
Proof sketch
By using the Cauchy integral theorem, one can show that the integral over (or the closed rectifiable curve) is equal to the same integral taken over an arbitrarily small circle around . Since is continuous, we can choose a circle small enough on which is arbitrarily close to . On the other hand, the integral
over any circle centered at . This can be calculated directly via a parametrization (integration by substitution) where and is the radius of the circle.
Letting gives the desired estimate
Example
Let
and let be the contour described by (the circle of radius 2).
To find the integral of around the contour , we need to know the singularities of . Observe that we can rewrite as follows:
where and .
Thus, has poles at and . The moduli of these points are less than 2 and thus lie inside the contour. This integral can be split into two smaller integrals by Cauchy–Goursat theorem; that is, we can express the integral around the contour as the sum of the integral around and where the contour is a small circle around each pole. Call these contours around and around .
Now, each of these smaller integrals can be evaluated by the Cauchy integral formula, but they first must be rewritten to apply the theorem. For the integral around , define as . This is analytic (since the contour does not contain the other singularity). We can simplify to be:
and now
Since the Cauchy integral formula says that:
we can evaluate the integral as follows:
Doing likewise for the other contour:
we evaluate
The integral around the original contour then is the sum of these two integrals:
An elementary trick using partial fraction decomposition: