Editing Cauchy's integral formula (section)

===Smooth functions===
A version of Cauchy's integral formula is the Cauchy–[[Dimitrie Pompeiu|Pompeiu]] formula,<ref>{{harvnb|Pompeiu|1905}}</ref> and holds for [[smooth function]]s as well, as it is based on [[Stokes' theorem]].  Let {{math|''D''}} be a disc in {{math|'''C'''}} and suppose that {{math|''f''}} is a complex-valued {{math|[[continuously differentiable function|''C''{{isup|1}}]]}} function on the [[closure (topology)|closure]] of {{math|''D''}}.  Then<ref>{{cite web | url = https://people.math.carleton.ca/~ckfong/S32.pdf | title = §2. Complex 2-Forms: Cauchy-Pompeiu's Formula}}</ref><ref>{{harvnb|Hörmander|1966|loc=Theorem 1.2.1}}</ref><ref>{{cite web | url = https://www.jirka.org/scv/scv.pdf#thm.4.1.1 | title = Theorem 4.1.1 (Cauchy–Pompeiu).}}</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 {{math|''D''}}.  Indeed, if {{math|''φ''}} is a function in {{math|''D''}}, then a particular solution {{math|''f''}} of the equation is a holomorphic function outside the support of {{math|''μ''}}.  Moreover, if in an open set {{math|''D''}},
<math display="block">d\mu = \frac{1}{2\pi i}\varphi \, dz\wedge d\bar{z}</math>
for some {{math|''φ'' ∈ ''C''{{isup|''k''}}(''D'')}} (where {{math|''k''&nbsp;≥&nbsp;1}}), then {{math|''f''(''ζ'', {{overline|''ζ''}})}} is also in {{math|''C''{{isup|''k''}}(''D'')}} 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]] {{math|''μ'' ∗ ''k''(''z'')}} 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 {{math|''μ''}}.  Here {{math|p.v.}} denotes the [[Cauchy principal value|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 {{math|''f''}} of compact support on {{math|'''C'''}} 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 (mathematics)|distribution]], {{math|(π''z'')<sup>−1</sup>}} is a [[fundamental solution]] of the [[Cauchy–Riemann operator]] {{math|{{sfrac|∂|∂''z̄''}}}}.<ref>{{harvnb|Hörmander|1983|pp=63, 81}}</ref>

The generalized Cauchy integral formula can be deduced for any bounded open region {{math|''X''}} with {{math|''C''{{isup|1}}}} boundary {{math|∂''X''}} from this result and the formula for the [[distributional derivative]] of the [[indicator function|characteristic function]] {{math|''χ''<sub>''X''</sub>}} of {{math|''X''}}:
<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 {{math|∂''X''}}.<ref>{{harvnb|Hörmander|1983|pp=62–63}}</ref>
{{Math proof|For <math>\varphi \in \mathcal{D}(X)</math> calculate:
:<math>
\begin{aligned}
\left\langle\frac{\partial}{\partial \bar{z}}\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:
{{Math proof|Since <math>u=\frac{\chi_X}{\pi\left(z-z_0\right)} \in \mathrm{L}_{\text{loc}}^1(X)</math> and since <math>z_0 \in X</math> this distribution is locally in <math>X</math> of the form "distribution times {{math|C<sup>&infin;</sup>}} function", so we may apply the [[Product rule|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 {{math|(π''z'')<sup>−1</sup>}} is a [[fundamental solution]] of the [[Cauchy–Riemann operator]] {{math|{{sfrac|∂|∂''z̄''}}}}, 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.
}}