TITLE: Intuition about the values of Cauchy integrals on the unit circle
QUESTION [13 upvotes]: Considering the Cauchy integral $$\oint \frac{f(x)}{(x-z)} dx $$ over the unit circle. The integral equals zero if $z$ is outside the circle, yet it has a value if $z$ is inside the circle. 
Why is this the case if one just looks at it computationally (without using the Fundamental Theorem of Calculus, etc.). I.e., what can be said along the lines of: when $z$ in in the interior, this and this happens to give a value; whereas when it's exterior, that and that happens to result in $0$. Thanks!

REPLY [6 votes]: This phenomenon merits attention, indeed. On one hand, we certainly want to assimilate Cauchy's formula into our intuition. On the other hand, it is an example of strange phenomena in integral representations of (holomorphic?) functions. 
That is, the "Cauchy kernel" $K(z,w)=\frac{1}{z-w}$ has the property asserted by Cauchy's formula, namely, that integrating against $f(z)$ on a circle containing $w$ produces essentially $f(w)$, while for $w$ outside the circle, we get $0$. Examination of the integral's behavior as $w$ crosses the circle is not so elementary, and is usually and reasonably not discussed in introductory situations. Indeed, even a radial limit is at best a principal value, etc. And there is a definite discontinuity as $w$ crosses the circle. I think it is fair to say that the disparity of behaviors inside and outside the circle is not obvious from the appearance of the integral.
To put this behavior in a larger context, one might consider other kernels $K(z,w)$ and integrate over circles or lines. Obviously many cases reduce to Cauchy's, or, the same thing, to "residues". 
There are examples with less discontinuity than in Cauchy's basic example, although maybe it's not trivial to produce elementary ones. That there are non-trivial possibilities was forcefully brought to my attention in some basic analytic number theory/automorphic forms business (perhaps a fancier issue than directly desired by the questioner, but there is a further, potentially useful point): part of a resolvent or fundamental solution in automorphic forms includes an integral 
$$
\int_{\Re(z)=1/2} \frac{E_{1-z}\,E_z\,dz}{(z-w)(z-1+w)}
$$ 
where $E_z$ is an Eisenstein series (with its own issues/properties...). This expression converges nicely for $\Re(w)>1/2$, and (here comes the deceitful part) the integrand is symmetric under $w\rightarrow 1-w$ and in fact is continuous as $\Re(w)\rightarrow 1/2^+$. So, yes, that integral extends to a meromorphic function on the whole plane. However, as $w$ passes through the line $\Re(w)=1/2$, its nature changes in several subtle, technical ways, among which is the violently non-intuitive point that the meromorphic continuation is not given by that integral! (What?!) Extra terms are picked up!!! The argument is by residues, that is, by Cauchy's formula... 
The details are not so much my point, but that the thought-provoking aspects of 
Cauchy's formula have considerable not-so-innocent potential!