# Cauchy's Integral Formula - Related

We know from Cauchy's Integral formula that if $f:D \rightarrow \mathbb C$ is holomorphic and $\gamma$ is some closed simple curve in the disc $D$, that $$f(z) = \frac{1}{2\pi i}\int_\gamma \! \frac{f(\zeta)}{\zeta-z} \, d\zeta$$ for all $z \in D$ with $z$ inside of the image of $\gamma$. If instead of being holomorphic $f$ is merely continuous on the image of $\gamma$, we still get a holomorphic function $$F(z) = \int_\gamma \! \frac{f(\zeta)}{\zeta-z} \, d\zeta.$$ My question is: what do we know about how $f$ and $F$ will relate?

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The first sentence, "We know $\dots$", isn't quite right. TIf $\gamma$ doesn't cross itself, then the formula is correct only when $z$ is inside the curve $\gamma$. More generally, the formula becomes correct (whether or not $\gamma$ crosses itself) if you multiply the left side by the winding number of $\gamma$ around $z$. –  Andreas Blass Jan 8 '13 at 20:05

The transform $$F(z)=\frac{1}{2\pi i}\int_{\mathbb{T}}\frac{f(\zeta)}{\zeta-z}d\zeta,$$ where $\mathbb{T}$ denotes the unit circle, is strongly related to the so called Cauchy transform.
In order to say something nice about the transform on general curves $\gamma$ you would certainly need condition on the parametrization of the curve.
Also note that for $|z|<1$ we may consider the geometric series expansion $$\frac{1}{\zeta-z}=\frac{1}{\zeta}\frac{1}{1-\bar{\zeta}z}=\frac{1}{\zeta}\sum_{k=0}^\infty \bar{\zeta}^kz^k$$ (here we assume $\zeta\in\mathbb{T}$ so that $\bar{\zeta}\zeta=|\zeta|^2 =1$ and hence $\bar{\zeta}=1/\zeta$) which converges uniformly in $\zeta$. Using that in the above integral formula leads to $$\frac{1}{2\pi i}\int_\mathbb{T} \frac{f(\zeta)}{\zeta-z}d\zeta = \frac{1}{2\pi i}\int_\mathbb{T} \frac{1}{\zeta}\sum_{k=0}^\infty f(\zeta)\bar{\zeta}^kz^k d\zeta$$ Then, if we set $\zeta=e^{it}$ so that $d\zeta=ie^{it}dt =i\zeta dt$, we arrive at $$F(z)=\frac{1}{2\pi}\int_{-\pi}^\pi \sum_{k=0}^\infty f(e^{it})e^{-ikt}z^k dt$$ switching the order of summation and integration leads to $$F(z)=\sum_{k=0}^\infty\frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})e^{-ikt}z^k dt =\sum_{k=0}^\infty\hat{f}(k)z^k$$ That is $F$ is the analytic projection of the Fourier expansion of $f$.