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: Consequences The integral formula has broad applications. First, it implies that a function which is holomorphic in an open set is in fact infinitely differentiable there. Furthermore, it is an analytic function, meaning that it can be represented as a power series. The proof of this uses the dominated convergence theorem and the geometric series applied to The formula is also used to prove the residue theorem, which is a