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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.

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I don't understand the title. –  Qiaochu Yuan Feb 22 '13 at 1:31

2 Answers 2

up vote 4 down vote accepted

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 $[a,b]\subset\Bbb C$.

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It should be added that, on a simply-connected domain, every holomorphic function has an antiderivative. –  Robert Israel Feb 21 '13 at 20:34
    
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? –  bob.sacamento Feb 21 '13 at 22:29
    
@bob.sacamento: Yes. –  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.)

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