Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Stein's Complex Analysis, the integral of $f$ along $\gamma$ is defined by $$\int_{\gamma}f(z)\text{d}z=\int_a^bf(z(t))z'(t)\text{d}t$$

where $z:[a,b]\rightarrow \Bbb{C}$ is a parametrization of $\gamma$

the right hand of the equation is a definite integral, and it is defined by: $$\int_a^bf(z(t))z'(t)\text{d}t =\lim_{||\Delta t||\rightarrow 0} \sum_{k=1}^{n}f(z(\xi_k))z'(\xi_k)(t_{k}-t_{k-1})$$

where $a=t_0<t_1<\cdots<t_{n-1}<t_n=b$ is a segmentation of $[a,b]$ and $t_{k-1}<\xi_k<t_k$ and $||\Delta t||=\max\{|t_{1}-t_{0}|,|t_{2}-t_{1}|,\cdots,|t_{n}-t_{n-1}|\}$

If I want to define the integral by
$$\lim_{||\Delta z||\rightarrow 0} \sum_{k=1}^{n}f(\zeta_k)(z_k-z_{k-1})$$ where $z_k$ is points on $\gamma$ which divided $\gamma$ into some segments

assume $f$ is continuous and $z'(t)$ is continuous too. then they are all exists and equal.

My Question is: why author define $\int_{\gamma}f(z)\text{d}z=\int_a^bf(z(t))z'(t)\text{d}t$,but not directly use Riemann sum to define:$\int_{\gamma}f(z)\text{d}z=\lim_{||\Delta z||\rightarrow 0} \sum_{k=1}^{n}f(\zeta_k)(z_k-z_{k-1})$?

thanks very much

share|improve this question

2 Answers 2

Well, only the author can tell you why, but a common reason for doing it that way is expediency. You just rely on the standard definition and properties of the integral, and can move on to topics of more interest in complex analysis. The definition using Riemann sums (really some sort of Riemann–Stieltjes sums), on the other hand, is more fundamental and at the same time more general, as it does only requires the integration path to have bounded variation. But it takes more time to develop the necessary properties rigorously, and besides one is rarely interested in integrating along non-smooth paths anyhow.

share|improve this answer

Because the second form is not quite a Riemann sum. You have terms like $z_k - z_{k-1}$ there, which are now complex numbers, i.e. two-dimensional real vectors. It's impossible to take subdivision of these (how do you subdivide a vector?), so you'd need to build whole Riemann integration theory for complex numbers from scratch. This seems like a waste of time for most applications.

On the other hand, the first form reduces to the usual real Riemann integration, so you can directly apply those results in this situation. The only price paid in doing so is that the definition now depends on the parametrization $z(t)$ and one needs to prove that the integral is actually reparametrization invariant.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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