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I have been re-reading through my complex analysis text and wanted to try something different. Cauchy's Integral Theorem is typically proved using an application of Green's Theorem and then by virtue of the Cauchy-Riemann Equations the integral vanishes. I have been trying to do the same to Cauchy's Integral Formula. That is, starting with $\int_{\gamma}\frac{f(z)}{z-a}dz$, with $a$ inside the region defined by the curve $\gamma$ I want to get back $f(a) 2 \pi i$. However, whenever I try Green's theorem on the integral, I get that it vanishes. Here is my work so far (mostly based on the proof of Cauchy's Integral Theorem):

$$\begin{align} & \int_\gamma \frac{f(z)}{z-a}dz \\[10pt] & = \int_\gamma \frac{u(x,y)+iv(x,y)}{z-a}(dx+i\,dy) \end{align} $$

Now let $l(x,y)=\dfrac1{x+iy+a}$

So we have $$ \begin{align} & = \int_\gamma \frac{u(x,y)+iv(x,y)}{z-a}(dx+i\,dy) \\ & = \int_\gamma ul\;dx -vl\; dy + i \int_{\gamma}vl\; dx+ ul\; dy \\ & = \int\int_D -\frac{\partial vl }{\partial x} -\frac{\partial ul }{\partial y} dx\,dy + i \int\int_D \frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \end{align} $$

$\dfrac{\partial ul }{\partial x}=\dfrac{\partial l }{\partial x}u+\dfrac{\partial u }{\partial x}l$ and we know $$\frac{\partial l }{\partial x}=\frac{\partial }{\partial x} \frac1{x+iy+a}=\frac{-1}{(x+iy+a)^2} $$

and similarly

$$\frac{\partial }{\partial x}\frac1{x+iy+a}=\frac{-i}{(x+iy+a)^2} $$

Thus we have $$ \begin{align} & {} \quad \int\int_{D}-\frac{\partial vl }{\partial x} -\frac{\partial ul }{\partial y} dx\,dy + i \int\int_{D}\frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \\[8pt] & =-\int\int_{D}\frac{\partial vl }{\partial x} + \frac{\partial ul }{\partial y} dx\,dy + i \int\int_{D}\frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \\[8pt] & =-\int\int_{D}\frac{\partial l }{\partial x}v+\frac{\partial v }{\partial x}l + \frac{\partial l }{\partial y}u+\frac{\partial u }{\partial y}l \, dx\,dy + i \int\int_{D}\frac{\partial l }{\partial x}u+\frac{\partial u }{\partial x}l-\frac{\partial l }{\partial y}v-\frac{\partial v }{\partial y}l \, dx\,dy \\[8pt] & =-\int\int_{D}\frac{-1}{(x+iy+a)^2}v+\frac{\partial v }{\partial x}l + \frac{-i}{(x+iy+a)^2}u+\frac{\partial u }{\partial y}l\; dx\,dy + i \int\int_{D}\frac{-1}{(x+iy+a)^2}u+\frac{\partial u }{\partial x}l-\frac{-i}{(x+iy+a)^2}v-\frac{\partial v }{\partial y}l\;dx\,dy \end{align} $$

edit: Taking out a small area around the singularity gives the correct answer. Thank you to froggie for pointing this out!

This also gives:

If $g(z)$ is a holomorphic function that is bounded for $| z| < 1$ then for all $|a|< 1$ we have $\displaystyle\pi g(a)=\int\int_{|z|<1} \dfrac{g(z)}{(1-a\bar z)^2}\,dx\,dy$ where $z=x+iy$.

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    I don't have time to look at this right now, but I wanted to say that I'm very impressed by your proactiveness so far. Keep up the good work.2012-05-08
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    I'll have to set aside time to look at this, but it seems like you have to be careful in the sense that $\gamma$ encloses a singularity of the integrand. There is a version of Stokes that deals with this, but maybe you can use a limit argument?2012-05-08
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    This is a great exercise! It's important to remember that Green's theorem doesn't work as usual when the integrand has a singularity in $D$. Here there is a singularity at $a$. I recommend removing a small disk $D(a,\varepsilon)$ around $a$ from the region $D$ before applying Green's theorem. Then take the limit as $\varepsilon\to 0$.2012-05-08
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    So I need to subtract the integral of the line curve around this small disk (with opposite orientation)?2012-05-08
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    Yes, that should work.2012-05-08
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    But what I've done so far makes it seem as though the contributing part will just be the line integral of that tiny curve. And that cannot be right, can it? How could someone get that last formula that I just added to the post? I am wondering if I somehow not taking care of my parameterization using $z=x+iy$ or something. Or perhaps should I not worry about vanishing since the C-R equations do not even hold at the point $\bar z=1/a$. Then again, for that last formula, we are told $|a|<1$ so $|1/\bar a|>1$ and thus that integral has no singularities.2012-05-08
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    It doesn't seem like you've made any mistakes. After applying Green's theorem to $D - D(a,\varepsilon)$ you should end up seeing that the integral around $\gamma$ is the same as the integral around the boundary of $D(a,\varepsilon)$. What happens to this integral as $\varepsilon\to 0$?2012-05-09
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    Parameterizing using little circles we get that the tiny integral goes to $2 \pi f(a)$. Neat! Thank you. How do you think someone got that final result?2012-05-09
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    @froggie I just figured it out, it's the same technique2012-05-09

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