“doubt regarding theorem on pg 135 of Churchill-Brown reference book:i think i understand from the theorem that if there is a function f continuous on a domain(any domain; no restrictions about simple connectedness were given) having an antiderivative on the same domain(the antiderivative just has to exist) would give 0 as the result of an integral along a contour lying entirely in the domain. they even proceed to show the application for f(z)=1/(z^2) on |z|>0. the slides seem to state sth different: requiring the function to be holomorphic on a simply connected domain at least, not to mention additional constraints on the contour of integration. please elaborate on the same. p.s.:thank you”
- doubtful existence
01/09/2020, 21:03
TL;DR - If you happen to know that \(f\) has an antiderivative, then you need not care about the domain.
However, given just \(f\), the existence of antiderivative in not guaranteed. In the event that the domain is simply-connected, we have hope and can conclude that integral around closed loops is zero.
Full answer -
Hey, thanks for an actual math question. Here’s the point:
If \(f:\Omega \to \mathbb{C}\) has an antiderivative \(F:\Omega \to \mathbb{C}\) (i.e., \(F' = f\)), then it is true that
for any closed path \(\gamma\) in \(F\). (Here, you do need the constraint that \(\gamma\) is closed. Otherwise, even the example of \(z^{-2}\) will give a counterexample. For example, take the path \(\gamma:[0, 1] \to \mathbb{C}\setminus\{0\}\) given as \(\gamma(t) = e^{\iota\pi t/2}\).)
Moreover, we restrict ourselves to the case of \(\gamma\) being (piecewise) smooth so that we can define the integral as
\[\int_{a}^b f(\gamma(t))\gamma'(t){\mathrm{d}}t.\]Note that even Churchill has the following assumption in the definition of contour:
A contour, or piecewise smooth arc, is an arc consisting of a finite number of smooth arcs joined end to end.
With just these constraints on \(\gamma\), it is true that
\[\int_\gamma f = 0.\]However, note that this is under the assumption that \(f\) has an antiderivative \(F\) on \(\Omega\).
The issue arises precisely because we don’t know that a holomorphic function has an antiderivative on the same domain. For example, consider the holomorphic function
\[f:\mathbb{C}\setminus\{0\} \to \mathbb{C}\]defined as
\[f(z) = \dfrac{1}{z}.\]This is indeed holomorphic but
\[\int_\gamma f = 2\pi\iota,\]where \(\gamma\) is the unit circle centered at \(0\) oriented counterclockwise.
So, what went wrong? Well, it’s precisely that \(f\) has no antiderivative on the given domain. (The above computation is a proof of that! This is similar to how integrating in MA 105 let us conclude that a certain vector field is not the grad of anything, even though its curl was zero.)
And indeed, our domain was not simply-connected!
If you do want the (holomorphic) function to have integral zero about every closed loop, then you need to make further assumptions about the “shape” (read: topology) of the domain, which is why the slides had the further assumptions.
By the way, what you say as “additional constraints on the contour of integration”, I guess you mean the stuff after the piece-wise smoothness. Note that that was actually weakening the conditions on \(\gamma\), requiring it to be just rectifiable and so forth. So, to summarise - the slides did not state anything weaker.