As we know, according to the Cauchy integral theorem we can easily evaluate an integral of an analytic complex function along the curve connecting two points in a complex plane. Thus for a curve $\gamma:[a,b]\subset\mathbb{R} \to \mathbb{C}$ which can be parameterized by $\gamma (t)=u(t)+iv(t)$ (where $u(t), v(t)$ are some real-valued functions) the contour integral of a function $\;f(z)$ of a complex variable $z$ can be computed easily using the rules for integration from standard calculus as follows $$ \int_{\gamma}f(z)dz=\int_a^bf(\gamma(t)) \dot\gamma(t)dt \tag{1} $$

We have an even more powerful tool for evaluating complex integrals without having to parameterize the curve. For an analytic function we have something magical called The Fundamental Theorem of Integration: $$ \int_a^bf(z)dz=F(b)-F(a), \tag{2} $$

where $F(z)$ is an antiderivative or an indefinite integral of a function $f(z)$. Using this theorem we can even easier compute the integral of an analytic function along the contour without having to parameterize the curve. Both $(1)$ and $(2)$ yield the same result when applied to the same function along the same contour.

My question is simple: it seems to me, that we ain't don't ever need $(1)$, as $(2)$ allows us to do the same computation easier. Is there something I am missing or $(2)$ is just a more powerful tool?

  • 1
    $\begingroup$ The first integral you use does not depend upon $f$ being analytic. That is the definition when $\gamma$ is a differentiable curve, at least as I've seen it in texts. Thus, (1) is more general than (2), so whenever $f(z)$ is not analytic you cannot use (2) but can use (1). $\endgroup$
    – Hayden
    Commented Jul 14, 2014 at 16:02
  • $\begingroup$ @Hayden I guess you are right. $(1)$ is more general, than $(2)$ as it can be applied to any function, however the function must be analytic in order to get the use of $(2)$. That's the point $\endgroup$ Commented Jul 14, 2014 at 16:07
  • $\begingroup$ Precisely, so the answer would be: "Using (2) is easier when $f$ is analytic, but using (1) is the only available method when $f$ isn't analytic (and thus would be more useful in that case)". $\endgroup$
    – Hayden
    Commented Jul 14, 2014 at 16:09
  • $\begingroup$ @Hayden that makes much sense, thanks $\endgroup$ Commented Jul 14, 2014 at 16:10

1 Answer 1


(Comment turned answer)

The equation $$\int_\gamma{f(z)dz}=\int_a^b{f(\gamma(t))\gamma'(t)dt} \tag{1}$$ when $\gamma:[a,b]\rightarrow \mathbb{C}$ is a differentiable curve is the definition of contour integration, and does not depend upon $f$ at all (except that the resulting integral exists), and in particular does not require $f$ to be analytic. (It should be noted that $\int_\gamma{f(z)dz}$ can be defined more generally when $\gamma$ is rectifiable.)

The second does depend upon $f$ being analytic, as having an antiderivative $F$ implies that $F$ is complex-differentiable. Complex-differentiability implies that analyticity, and thus every derivative of $F$ is analytic, including $f$. (The integral depending upon the end-points only is also equivalent to the antiderivative existing) For this reason, the equation $$\int_\gamma{f(z)dz}=F(b)-F(a) \tag{2}$$ cannot be used in the more general case where $f$ is not analytic, even if the integral exists in every case.

For these reasons it follows that (1) is more useful in the more general case because there is no guarantee that (2) is even well-defined. However, when $f$ is analytic, then (2) is much more easily applied if the antiderivative can be found easily.


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