A recent question from Juan Saloman reminded me of something that has nagged me for years, and I have never understood and never heard explained. (or maybe I just don't remember, but anyway ...) In the complex plane, can an integral of the general form $\int dz f(z)$ over the complex plane be treated as a straightforward 1-D integral if no path in the complex plane is specified? I have been in classes where the professors seemed to do just that. But, I mean, $z$ has two independent parts, right? In other words, it's the complex plane, not the complex line. I have seen these kinds of integrals before and sort learned to deal with them in a monkey-see-monkey-do fashion, but never really understood what was going on to my own satisfaction. Thanks.

  • $\begingroup$ I don't understand the title. $\endgroup$ – Qiaochu Yuan Feb 22 '13 at 1:31

In the complex plane $\Bbb C$ certain functions $z\mapsto f(z)$ are distinguished as being holomorphic. Any such function comes with a domain $\Omega\subset\Bbb C$, which by definition is an open and connected set. Given such an $f$ and a curve $$\gamma:\ [a,b]\to\Omega, \quad t\mapsto z(t)\tag{1}$$ one can consider the integral $$\int_\gamma f(z)\ dz:=\int_a^b f\bigl(z(t)\bigr)\ z'(t)\ dt\ ,$$ whose value depends on $f$ as well as on $\gamma$.

When $f$ happens to be the derivative of some other analytic function $F:\ \Omega\to\Bbb C$, as $\ \cos\ $ is the derivative of $\ \sin$, and if $(1)$ is an arbitrary curve in $\Omega$ connecting the point $\gamma(a)=z_1\in\Omega$ with the point $\gamma(b)=z_2\in\Omega$ then $$\int_\gamma f(z)\ dz= F(z_2)-F(z_1)\ .\tag{2}$$ In order to prove $(2)$ consider the auxiliary function $$\phi(t):=F\bigl(z(t)\bigr)\qquad(a\leq t\leq b)$$ and rewrite the formula $\int_a^b \phi'(t)\ dt=\phi(b)-\phi(a)$ in terms of $f$ and $F$.

In this light the formula $\int_a^b f(x)\ dx=F(b)-F(a)$ from ordinary calculus can be considered as the special case where $\gamma$ is the directed segment $[a,b]\subset\Bbb C$.

  • $\begingroup$ It should be added that, on a simply-connected domain, every holomorphic function has an antiderivative. $\endgroup$ – Robert Israel Feb 21 '13 at 20:34
  • $\begingroup$ Much obliged. Just to make sure I understand, any two paths give the same integral as long as the two paths together don't encircle any singularities. Right? $\endgroup$ – bob.sacamento Feb 21 '13 at 22:29
  • $\begingroup$ @bob.sacamento: Yes. $\endgroup$ – Christian Blatter Feb 22 '13 at 9:16

You need to integrate over a curve, but if $f(z)$ has an antiderivative, it won't matter which curve you integrate over, as long as the curve is contained in an open set on which $f$ has an antiderivative. (Only the endpoints of the curve matter.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.