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I'm new to Complex Analysis so forgive me please. An exercise in Sarason's Complex Function Theory says to find, explicitely, the "Cauchy integral" of the constant function $1$ over $[0, 1]$ on the real line. A Cauchy integral is one of the form $$\int_{\gamma} \frac{\phi(w)}{w-z} dw =: f(z),$$ where $\phi$ is continuous and $\mathbb{C}$-valued, $\gamma$ is some piecewise-$C^{1}$ curve, and $z \in \mathbb{C}\setminus \gamma^{*}$. It turns out that $f(z)$ is holomorphic, and that $$f^{(n)}(z) = n!\int_{\gamma} \frac{\phi(w)}{(w-z)^{n+1}}dw.$$ This doesn't seem to help though. However, I do have a hunch that the only branch of logarithm on $\mathbb{C}\setminus [0, 1]$ will be Log$(z) + i\pi$, where Log$(z)$ is the principal branch. This is because the only ray starting at the origin that I can cut out of $\mathbb{C}\setminus[0, 1]$ is $(0, \infty)$. And in this case, my integral would become (Log$(1) + i\pi) - (\text{Log}(0) + i\pi)$, which doesn't make sense...

rash
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Freddie
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1 Answers1

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A detailed answer (7.1) can be found here https://wolfweb.unr.edu/homepage/alex/410/8s.pdf (see the snippet below)

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dnqxt
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