Cauchy's integral theorem

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Template:Complex analysis sidebar In mathematics, the Cauchy integral theorem (also known as the Cauchy–Goursat theorem) in complex analysis, named after Augustin-Louis Cauchy (and Édouard Goursat), is an important statement about line integrals for holomorphic functions in the complex plane. Essentially, it says that if <math>f(z)</math> is holomorphic in a simply connected domain Ω, then for any simply closed contour <math>C</math> in Ω, that contour integral is zero.

<math display="block">\int_C f(z)\,dz = 0. </math>

StatementEdit

Fundamental theorem for complex line integralsEdit

If Template:Math is a holomorphic function on an open region Template:Mvar, and <math>\gamma</math> is a curve in Template:Mvar from <math>z_0</math> to <math>z_1</math> then, <math display="block">\int_{\gamma}f'(z) \, dz = f(z_1)-f(z_0).</math>

Also, when Template:Math has a single-valued antiderivative in an open region Template:Mvar, then the path integral <math display="inline">\int_{\gamma}f(z) \, dz</math> is path independent for all paths in Template:Mvar.

Formulation on simply connected regionsEdit

Let <math>U \subseteq \Complex</math> be a simply connected open set, and let <math>f: U \to \Complex</math> be a holomorphic function. Let <math>\gamma: [a,b] \to U</math> be a smooth closed curve. Then: <math display="block">\int_\gamma f(z)\,dz = 0. </math> (The condition that <math>U</math> be simply connected means that <math>U</math> has no "holes", or in other words, that the fundamental group of <math>U</math> is trivial.)

General formulationEdit

Let <math>U \subseteq \Complex</math> be an open set, and let <math>f: U \to \Complex</math> be a holomorphic function. Let <math>\gamma: [a,b] \to U</math> be a smooth closed curve. If <math>\gamma</math> is homotopic to a constant curve, then: <math display="block">\int_\gamma f(z)\,dz = 0. </math>where z є U

(Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy (within <math>U</math>) from the curve to the constant curve. Intuitively, this means that one can shrink the curve into a point without exiting the space.) The first version is a special case of this because on a simply connected set, every closed curve is homotopic to a constant curve.

Main exampleEdit

In both cases, it is important to remember that the curve <math>\gamma</math> does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve: <math display="block">\gamma(t) = e^{it} \quad t \in \left[0, 2\pi\right] ,</math> which traces out the unit circle. Here the following integral: <math display="block">\int_{\gamma} \frac{1}{z}\,dz = 2\pi i \neq 0 , </math> is nonzero. The Cauchy integral theorem does not apply here since <math>f(z) = 1/z</math> is not defined at <math>z = 0</math>. Intuitively, <math>\gamma</math> surrounds a "hole" in the domain of <math>f</math>, so <math>\gamma</math> cannot be shrunk to a point without exiting the space. Thus, the theorem does not apply.

DiscussionEdit

As Édouard Goursat showed, Cauchy's integral theorem can be proven assuming only that the complex derivative <math>f'(z)</math> exists everywhere in <math>U</math>. This is significant because one can then prove Cauchy's integral formula for these functions, and from that deduce these functions are infinitely differentiable.

The condition that <math>U</math> be simply connected means that <math>U</math> has no "holes" or, in homotopy terms, that the fundamental group of <math>U</math> is trivial; for instance, every open disk <math>U_{z_0} = \{ z : \left|z-z_{0}\right| < r\}</math>, for <math>z_0 \in \Complex</math>, qualifies. The condition is crucial; consider <math display="block">\gamma(t) = e^{it} \quad t \in \left[0, 2\pi\right]</math> which traces out the unit circle, and then the path integral <math display="block">\oint_\gamma \frac{1}{z}\,dz = \int_0^{2\pi} \frac{1}{e^{it}}(ie^{it} \,dt) = \int_0^{2\pi}i\,dt = 2\pi i </math> is nonzero; the Cauchy integral theorem does not apply here since <math>f(z) = 1/z</math> is not defined (and is certainly not holomorphic) at <math>z = 0</math>.

One important consequence of the theorem is that path integrals of holomorphic functions on simply connected domains can be computed in a manner familiar from the fundamental theorem of calculus: let <math>U</math> be a simply connected open subset of <math>\Complex</math>, let <math>f: U \to \Complex</math> be a holomorphic function, and let <math>\gamma</math> be a piecewise continuously differentiable path in <math>U</math> with start point <math>a</math> and end point <math>b</math>. If <math>F</math> is a complex antiderivative of <math>f</math>, then <math display="block">\int_\gamma f(z)\,dz=F(b)-F(a).</math>

The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given <math>U</math>, a simply connected open subset of <math>\Complex</math>, we can weaken the assumptions to <math>f</math> being holomorphic on <math>U</math> and continuous on <math display="inline">\overline{U}</math> and <math>\gamma</math> a rectifiable simple loop in <math display="inline">\overline{U}</math>.<ref>Template:Cite journal</ref>

The Cauchy integral theorem leads to Cauchy's integral formula and the residue theorem.

ProofEdit

If one assumes that the partial derivatives of a holomorphic function are continuous, the Cauchy integral theorem can be proven as a direct consequence of Green's theorem and the fact that the real and imaginary parts of <math>f=u+iv</math> must satisfy the Cauchy–Riemann equations in the region bounded by Template:Nowrap and moreover in the open neighborhood Template:Mvar of this region. Cauchy provided this proof, but it was later proven by Goursat without requiring techniques from vector calculus, or the continuity of partial derivatives.

We can break the integrand Template:Nowrap as well as the differential <math>dz</math> into their real and imaginary components:

<math display="block"> f=u+iv </math> <math display="block"> dz=dx+i\,dy </math>

In this case we have <math display="block">\oint_\gamma f(z)\,dz = \oint_\gamma (u+iv)(dx+i\,dy) = \oint_\gamma (u\,dx-v\,dy) +i\oint_\gamma (v\,dx+u\,dy)</math>

By Green's theorem, we may then replace the integrals around the closed contour <math>\gamma</math> with an area integral throughout the domain <math>D</math> that is enclosed by <math>\gamma</math> as follows:

<math display="block">\oint_\gamma (u\,dx-v\,dy) = \iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right) \,dx\,dy </math> <math display="block">\oint_\gamma (v\,dx+u\,dy) = \iint_D \left( \frac{\partial u}{\partial x} -\frac{\partial v}{\partial y} \right) \,dx\,dy </math>

But as the real and imaginary parts of a function holomorphic in the domain Template:Nowrap <math>u</math> and <math>v</math> must satisfy the Cauchy–Riemann equations there: <math display="block">\frac{ \partial u }{ \partial x } = \frac{ \partial v }{ \partial y } </math> <math display="block">\frac{ \partial u }{ \partial y } = -\frac{ \partial v }{ \partial x } </math>

We therefore find that both integrands (and hence their integrals) are zero

<math display="block">\iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right )\,dx\,dy = \iint_D \left( \frac{\partial u}{\partial y} - \frac{\partial u}{\partial y} \right ) \, dx \, dy =0</math> <math display="block">\iint_D \left( \frac{\partial u}{\partial x}-\frac{\partial v}{\partial y} \right )\,dx\,dy = \iint_D \left( \frac{\partial u}{\partial x} - \frac{\partial u}{\partial x} \right ) \, dx \, dy = 0</math>

This gives the desired result <math display="block">\oint_\gamma f(z)\,dz = 0</math>

See alsoEdit

ReferencesEdit

Template:Reflist

External linksEdit

|_exclude=urlname, _debug, id |url = https://mathworld.wolfram.com/{{#if:CauchyIntegralTheorem%7CCauchyIntegralTheorem.html}} |title = Cauchy Integral Theorem |author = Weisstein, Eric W. |website = MathWorld |access-date = |ref = Template:SfnRef }}