Cauchy's integral formula

Template:Short description Template:Distinguish Template:Complex analysis sidebar 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.

TheoremEdit

Let Template:Math be an open subset of the complex plane Template:Math, and suppose the closed disk Template:Math defined as <math display="block">D = \bigl\{z:|z - z_0| \leq r\bigr\}</math> is completely contained in Template:Math. Let Template:Math be a holomorphic function, and let Template:Math be the circle, oriented counterclockwise, forming the boundary of Template:Math. Then for every Template:Math in the interior of Template:Math, <math display="block">f(a) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\,dz.\,</math>

The proof of this statement uses the Cauchy integral theorem and like that theorem, it only requires Template:Math to be complex differentiable. Since <math>1/(z-a)</math> can be expanded as a power series in the variable <math>a</math> <math display="block">\frac{1}{z-a} = \frac{1+\frac{a}{z}+\left(\frac{a}{z}\right)^2+\cdots}{z}</math> it follows that holomorphic functions are analytic, i.e. they can be expanded as convergent power series. In particular Template:Math is actually infinitely differentiable, with <math display="block">f^{(n)}(a) = \frac{n!}{2\pi i} \oint_\gamma \frac{f(z)}{\left(z-a\right)^{n+1}}\,dz.</math>

This formula is sometimes referred to as Cauchy's differentiation formula.

The theorem stated above can be generalized. The circle Template:Math can be replaced by any closed rectifiable curve in Template:Math which has winding number one about Template:Math. Moreover, as for the Cauchy integral theorem, it is sufficient to require that Template:Math 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 Template:Math, defined for Template:Math, 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 Template:Math has real part Template:Math. On the unit circle this can be written Template:Math. Using the Möbius transformation and the Stieltjes formula we construct the function inside the circle. The Template:Math term makes no contribution, and we find the function Template:Math. This has the correct real part on the boundary, and also gives us the corresponding imaginary part, but off by a constant, namely Template:Math.

Proof sketchEdit

By using the Cauchy integral theorem, one can show that the integral over Template:Math (or the closed rectifiable curve) is equal to the same integral taken over an arbitrarily small circle around Template:Math. Since Template:Math is continuous, we can choose a circle small enough on which Template:Math is arbitrarily close to Template:Math. On the other hand, the integral <math display="block">\oint_C \frac{1}{z-a} \,dz = 2 \pi i,</math> over any circle Template:Math centered at Template:Math. This can be calculated directly via a parametrization (integration by substitution) Template:Math where Template:Math and Template:Math is the radius of the circle.

Letting Template:Math gives the desired estimate <math display="block">\begin{align} \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-a} \,dz - f(a) \right | &= \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)-f(a)}{z-a} \,dz \right | \\[1ex] &= \left | \frac{1}{2\pi i}\int_0^{2\pi}\left(\frac{f\bigl(z(t)\bigr)-f(a)}{\varepsilon e^{it}}\cdot\varepsilon e^{it} i\right )\,dt\right | \\[1ex] &\leq \frac{1}{2 \pi} \int_0^{2\pi} \frac{ \left|f\bigl(z(t)\bigr) - f(a)\right| } {\varepsilon} \,\varepsilon\,dt \\[1ex] &\leq \max_{|z-a|=\varepsilon} \left|f(z) - f(a)\right| ~~ \xrightarrow[\varepsilon\to 0]{} ~~ 0. \end{align}</math>

ExampleEdit

File:ComplexResiduesExample.png

Surface of the real part of the function Template:Math and its singularities, with the contours described in the text.

Let <math display="block">g(z) = \frac{z^2}{z^2+2z+2},</math> and let Template:Math be the contour described by Template:Math (the circle of radius 2).

To find the integral of Template:Math around the contour Template:Math, we need to know the singularities of Template:Math. Observe that we can rewrite Template:Math as follows: <math display="block">g(z) = \frac{z^2}{(z-z_1)(z-z_2)}</math> where Template:Math and Template:Math.

Thus, Template:Math has poles at Template:Math and Template:Math. 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 Template:Math and Template:Math where the contour is a small circle around each pole. Call these contours Template:Math around Template:Math and Template:Math around Template:Math.

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 Template:Math, define Template:Math as Template:Math. This is analytic (since the contour does not contain the other singularity). We can simplify Template:Math to be: <math display="block">f_1(z) = \frac{z^2}{z-z_2}</math> and now <math display="block">g(z) = \frac{f_1(z)}{z-z_1}.</math>

Since the Cauchy integral formula says that: <math display="block">\oint_C \frac{f_1(z)}{z-a}\, dz=2\pi i\cdot f_1(a),</math> we can evaluate the integral as follows: <math display="block">

 \oint_{C_1} g(z)\,dz
=\oint_{C_1} \frac{f_1(z)}{z-z_1}\,dz
=2\pi i\frac{z_1^2}{z_1-z_2}.

</math>

Doing likewise for the other contour: <math display="block">f_2(z) = \frac{z^2}{z-z_1},</math> we evaluate <math display="block">

 \oint_{C_2} g(z)\,dz
=\oint_{C_2} \frac{f_2(z)}{z-z_2}\,dz
=2\pi i\frac{z_2^2}{z_2-z_1}.

</math>

The integral around the original contour Template:Math then is the sum of these two integrals: <math display="block">\begin{align}

    \oint_C g(z)\,dz

&{}= \oint_{C_1} g(z)\,dz

  + \oint_{C_2} g(z)\,dz \\[.5em]

&{}= 2\pi i\left(\frac{z_1^2}{z_1-z_2}+\frac{z_2^2}{z_2-z_1}\right) \\[.5em] &{}= 2\pi i(-2) \\[.3em] &{}=-4\pi i. \end{align}</math>

An elementary trick using partial fraction decomposition: <math display="block">

 \oint_C g(z)\,dz
=\oint_C \left(1-\frac{1}{z-z_1}-\frac{1}{z-z_2}\right) \, dz
=0-2\pi i-2\pi i
=-4\pi i

</math>

ConsequencesEdit

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 result for meromorphic functions, and a related result, the argument principle. It is known from Morera's theorem that the uniform limit of holomorphic functions is holomorphic. This can also be deduced from Cauchy's integral formula: indeed the formula also holds in the limit and the integrand, and hence the integral, can be expanded as a power series. In addition the Cauchy formulas for the higher order derivatives show that all these derivatives also converge uniformly.

The analog of the Cauchy integral formula in real analysis is the Poisson integral formula for harmonic functions; many of the results for holomorphic functions carry over to this setting. No such results, however, are valid for more general classes of differentiable or real analytic functions. For instance, the existence of the first derivative of a real function need not imply the existence of higher order derivatives, nor in particular the analyticity of the function. Likewise, the uniform limit of a sequence of (real) differentiable functions may fail to be differentiable, or may be differentiable but with a derivative which is not the limit of the derivatives of the members of the sequence.

Another consequence is that if Template:Math is holomorphic in Template:Math and Template:Math then the coefficients Template:Math satisfy Cauchy's estimate<ref>Template:Harvnb</ref> <math display="block">|a_n|\le r^{-n} \sup_{|z|=r}|f(z)|.</math>

From Cauchy's estimate, one can easily deduce that every bounded entire function must be constant (which is Liouville's theorem).

The formula can also be used to derive Gauss's Mean-Value Theorem, which states<ref>Template:WolframAlpha</ref> <math display="block">f(z) = \frac{1}{2\pi} \int_{0}^{2\pi} f(z + r e^{i\theta}) \, d\theta.</math>

In other words, the average value of Template:Math over the circle centered at Template:Math with radius Template:Math is Template:Math. This can be calculated directly via a parametrization of the circle.

GeneralizationsEdit

Smooth functionsEdit

A version of Cauchy's integral formula is the Cauchy–Pompeiu formula,<ref>Template:Harvnb</ref> and holds for smooth functions as well, as it is based on Stokes' theorem. Let Template:Math be a disc in Template:Math and suppose that Template:Math is a complex-valued Template:Math function on the closure of Template:Math. Then<ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref><ref>Template:Harvnb</ref><ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> <math display="block">f(\zeta) = \frac{1}{2\pi i}\int_{\partial D} \frac{f(z) \,dz}{z-\zeta} - \frac{1}{\pi}\iint_D \frac{\partial f}{\partial \bar{z}}(z) \frac{dx\wedge dy}{z-\zeta}.</math>

One may use this representation formula to solve the inhomogeneous Cauchy–Riemann equations in Template:Math. Indeed, if Template:Math is a function in Template:Math, then a particular solution Template:Math of the equation is a holomorphic function outside the support of Template:Math. Moreover, if in an open set Template:Math, <math display="block">d\mu = \frac{1}{2\pi i}\varphi \, dz\wedge d\bar{z}</math> for some Template:Math (where Template:Math), then Template:Math is also in Template:Math and satisfies the equation <math display="block">\frac{\partial f}{\partial\bar{z}} = \varphi(z,\bar{z}).</math>

The first conclusion is, succinctly, that the convolution Template:Math of a compactly supported measure with the Cauchy kernel <math display="block">k(z) = \operatorname{p.v.}\frac{1}{z}</math> is a holomorphic function off the support of Template:Math. Here Template:Math denotes the principal value. The second conclusion asserts that the Cauchy kernel is a fundamental solution of the Cauchy–Riemann equations. Note that for smooth complex-valued functions Template:Math of compact support on Template:Math the generalized Cauchy integral formula simplifies to <math display="block">f(\zeta) = \frac{1}{2\pi i}\iint \frac{\partial f}{\partial \bar{z}}\frac{dz\wedge d\bar{z}}{z-\zeta},</math> and is a restatement of the fact that, considered as a distribution, Template:Math is a fundamental solution of the Cauchy–Riemann operator Template:Math.<ref>Template:Harvnb</ref>

The generalized Cauchy integral formula can be deduced for any bounded open region Template:Math with Template:Math boundary Template:Math from this result and the formula for the distributional derivative of the characteristic function Template:Math of Template:Math: <math display="block"> \frac {\partial \chi_X}{\partial \bar z}= \frac{i}{2} \oint_{\partial X} \,dz,</math> where the distribution on the right hand side denotes contour integration along Template:Math.<ref>Template:Harvnb</ref> Template:Math proof\left(\chi_X\right), \varphi\right\rangle & =-\int_X \frac{\partial \varphi}{\partial \bar{z}} \mathrm{~d}(x, y) \\ & =-\frac{1}{2} \int_X\left(\partial_x \varphi+\mathrm{i} \partial_y \varphi\right) \mathrm{d}(x, y) . \end{aligned} </math> then traverse <math>\partial X</math> in the anti-clockwise direction. Fix a point <math>p \in \partial X</math> and let <math>s</math> denote arc length on <math>\partial X</math> measured from <math>p</math> anti-clockwise. Then, if <math>\ell</math> is the length of <math>\partial X,[0, \ell] \ni s \mapsto(x(s), y(s))</math> is a parametrization of <math>\partial X</math>. The derivative <math>\tau=\left(x'(s), y'(s)\right)</math> is a unit tangent to <math>\partial X</math> and <math>\nu:=\left(-y'(s), x'(s)\right)</math> is the unit outward normal on <math>\partial X</math>. We are lined up for use of the divergence theorem: put <math>V=(\varphi, \mathrm{i} \varphi) \in \mathcal{D}(X)^2</math> so that <math>\operatorname{div} V=\partial_x \varphi+\mathrm{i} \partial_y \varphi</math> and we get

<math>

\begin{aligned} -\frac{1}{2} \int_X\left(\partial_x \varphi+\mathrm{i} \partial_y \varphi\right) \mathrm{d}(x, y) & =-\frac{1}{2} \int_{\partial X} V \cdot \nu \mathrm{d} S \\ & =-\frac{1}{2} \int_0^{\ell}\left(\varphi \nu_1+\mathrm{i} \varphi \nu_2\right) \mathrm{d} s \\ & =-\frac{1}{2} \int_0^{\ell} \varphi(x(s), y(s))\left(y'(s)-\mathrm{i} x'(s)\right) \mathrm{d} s \\ & =\frac{1}{2} \int_0^{\ell} \mathrm{i} \varphi(x(s), y(s))\left(x'(s)+\mathrm{i} y'(s)\right) \mathrm{d} s \\ & =\frac{\mathrm{i}}{2} \int_{\partial X} \varphi \mathrm{d} z \end{aligned} </math> Hence we proved <math> \frac {\partial \chi_X}{\partial \bar z}= \frac{i}{2} \oint_{\partial X} \,dz</math>. }} Now we can deduce the generalized Cauchy integral formula: Template:Math proof^1(X)</math> and since <math>z_0 \in X</math> this distribution is locally in <math>X</math> of the form "distribution times Template:Math function", so we may apply the Leibniz rule to calculate its derivatives:

<math>\frac{\partial u}{\partial \bar{z}} =\frac{\partial}{\partial \bar{z}}\left(\frac{1}{\pi\left(z-z_0\right)}\right) \chi_X+\frac{1}{\pi\left(z-z_0\right)} \frac{\partial}{\partial \bar{z}}\left(\chi_X\right)</math>

Using that Template:Math is a fundamental solution of the Cauchy–Riemann operator Template:Math, we get <math>\frac{\partial}{\partial \bar{z}}\left(\frac{1}{\pi\left(z-z_0\right)}\right)=\delta_{z_0}</math>:

<math>\frac{\partial u}{\partial \bar{z}}=\delta_{z_0}+\frac{1}{\pi\left(z-z_0\right)} \frac{\partial}{\partial \bar{z}}\left(\chi_X\right)

</math> Applying <math>\frac{\partial u}{\partial \bar{z}}</math> to <math>\phi \in \mathcal{D}(X)</math>:

<math>\begin{aligned}

\left\langle\frac{\partial}{\partial \bar{z}}\left(\frac{\chi_X}{\pi\left(z-z_0\right)}\right), \phi\right\rangle & =\phi\left(z_0\right)+\left\langle\frac{1}{\pi\left(z-z_0\right)} \frac{\partial}{\partial \bar{z}}\left(\chi_X\right), \phi\right\rangle \\ & =\phi\left(z_0\right)+\left\langle\frac{\partial}{\partial \bar{z}}\left(\chi_X\right), \frac{\phi}{\pi\left(z-z_0\right)}\right\rangle \\ & =\phi\left(z_0\right)+\frac{\mathrm{i}}{2} \int_{\partial X} \frac{\phi(z)}{\pi\left(z-z_0\right)} \mathrm{d} z \end{aligned}</math> where <math> \frac {\partial \chi_X}{\partial \bar z}= \frac{i}{2} \oint_{\partial X} \,dz</math> is used in the last line.

Rearranging, we get

<math>\phi(z_0)={\frac {1}{2\pi i}}\int _{\partial X}{\frac {\phi(z)\,dz}{z-z_0 }}-{\frac {1}{\pi }}\iint _X{\frac {\partial \phi}{\partial {\bar {z}}}}(z){\frac {dx\wedge dy}{z-z_0 }}.</math>

as desired. }}

Several variablesEdit

In several complex variables, the Cauchy integral formula can be generalized to polydiscs.<ref>Template:Harvnb</ref> Let Template:Math be the polydisc given as the Cartesian product of Template:Math open discs Template:Math: <math display="block">D = \prod_{i=1}^n D_i.</math>

Suppose that Template:Math is a holomorphic function in Template:Math continuous on the closure of Template:Math. Then <math display="block">f(\zeta) = \frac{1}{\left(2\pi i\right)^n}\int\cdots\iint_{\partial D_1\times\cdots\times\partial D_n} \frac{f(z_1,\ldots,z_n)}{(z_1-\zeta_1)\cdots(z_n-\zeta_n)} \, dz_1\cdots dz_n</math> where Template:Math.

In real algebrasEdit

The Cauchy integral formula is generalizable to real vector spaces of two or more dimensions. The insight into this property comes from geometric algebra, where objects beyond scalars and vectors (such as planar bivectors and volumetric trivectors) are considered, and a proper generalization of Stokes' theorem.

Geometric calculus defines a derivative operator Template:Math under its geometric product — that is, for a Template:Math-vector field Template:Math, the derivative Template:Math generally contains terms of grade Template:Math and Template:Math. For example, a vector field (Template:Math) generally has in its derivative a scalar part, the divergence (Template:Math), and a bivector part, the curl (Template:Math). This particular derivative operator has a Green's function: <math display="block">G\left(\mathbf r, \mathbf r'\right) = \frac{1}{S_n} \frac{\mathbf r - \mathbf r'}{\left|\mathbf r - \mathbf r'\right|^n}</math> where Template:Math is the surface area of a unit Template:Math-ball in the space (that is, Template:Math, the circumference of a circle with radius 1, and Template:Math, the surface area of a sphere with radius 1). By definition of a Green's function, <math display="block">\nabla G\left(\mathbf r, \mathbf r'\right) = \delta\left(\mathbf r- \mathbf r'\right).</math>

It is this useful property that can be used, in conjunction with the generalized Stokes theorem: <math display="block">\oint_{\partial V} d\mathbf S \; f(\mathbf r) = \int_V d\mathbf V \; \nabla f(\mathbf r)</math> where, for an Template:Math-dimensional vector space, Template:Math is an Template:Math-vector and Template:Math is an Template:Math-vector. The function Template:Math can, in principle, be composed of any combination of multivectors. The proof of Cauchy's integral theorem for higher dimensional spaces relies on the using the generalized Stokes theorem on the quantity Template:Math and use of the product rule: <math display="block">\oint_{\partial V'} G\left(\mathbf r, \mathbf r'\right)\; d\mathbf S' \; f\left(\mathbf r'\right) = \int_V \left(\left[\nabla' G\left(\mathbf r, \mathbf r'\right)\right] f\left(\mathbf r'\right) + G\left(\mathbf r, \mathbf r'\right) \nabla' f\left(\mathbf r'\right)\right) \; d\mathbf V</math>

When Template:Math, Template:Math is called a monogenic function, the generalization of holomorphic functions to higher-dimensional spaces — indeed, it can be shown that the Cauchy–Riemann condition is just the two-dimensional expression of the monogenic condition. When that condition is met, the second term in the right-hand integral vanishes, leaving only <math display="block">\oint_{\partial V'} G\left(\mathbf r, \mathbf r'\right)\; d\mathbf S' \; f\left(\mathbf r'\right) = \int_V \left[\nabla' G\left(\mathbf r, \mathbf r'\right)\right] f\left(\mathbf r'\right) = -\int_V \delta\left(\mathbf r - \mathbf r'\right) f\left(\mathbf r'\right) \; d\mathbf V =- i_n f(\mathbf r)</math> where Template:Math is that algebra's unit Template:Math-vector, the pseudoscalar. The result is <math display="block">f(\mathbf r) =- \frac{1}{i_n} \oint_{\partial V} G\left(\mathbf r, \mathbf r'\right)\; d\mathbf S \; f\left(\mathbf r'\right) = -\frac{1}{i_n} \oint_{\partial V} \frac{\mathbf r - \mathbf r'}{S_n \left|\mathbf r - \mathbf r'\right|^n} \; d\mathbf S \; f\left(\mathbf r'\right)</math>

Thus, as in the two-dimensional (complex analysis) case, the value of an analytic (monogenic) function at a point can be found by an integral over the surface surrounding the point, and this is valid not only for scalar functions but vector and general multivector functions as well.