# Integration along Paths

• Jun 2nd 2007, 06:56 PM
ThePerfectHacker
Integration along Paths
What is the formal definition of "orientation" of a curve? Or is this a topology issue?

Let $\gamma: [a,b] \to \mathbb{C}$ be a smooth function. And $f(z)$ is continous on $\gamma^*$ where $\gamma^* = \{\gamma(t)|t\in [a,b]\}$, i.e. the image. And $\eta:[c,d]\to \mathbb{C}$ is another such function. With the property that $\eta^* = \gamma^*$ and have the same orientation. Then show that $\int_{\gamma}f(z) dz = \int_{\eta}f(z) dz$.

Basically, what I am trying to show that if we want to integrate a complex function along some path it makes no difference how we parametrize it (as long the orientation is the same).
• Jun 3rd 2007, 02:16 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
What is the formal definition of "orientation" of a curve? Or is this a topology issue?

Let $\gamma: [a,b] \to \mathbb{C}$ be a smooth function. And $f(z)$ is continous on $\gamma^*$ where $\gamma^* = \{\gamma(t)|t\in [a,b]\}$, i.e. the image. And $\eta:[c,d]\to \mathbb{C}$ is another such function. With the property that $\eta^* = \gamma^*$ and have the same orientation. Then show that $\int_{\gamma}f(z) dz = \int_{\eta}f(z) dz$.

Basically, what I am trying to show that if we want to integrate a complex function along some path it makes no difference how we parametrize it (as long the orientation is the same).

There is really nothing to show (given that the functions $\gamma(t)$ and $\eta(t)$ are
sufficiently well behaved) as both the integrals are the integral of the same
function around the same curve in the same sense. All you have done is
change how you parameterise it.

RonL
• Jun 3rd 2007, 03:47 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlank
There is really nothing to show (given that the functions $\gamma(t)$ and $\eta(t)$ are
sufficiently well behaved) as both the integrals are the integral of the same
function around the same curve in the same sense. All you have done is
change how you parameterise it.

RonL

I would agree that it should not be so easy to show, however I do not know how to show it. Furthermore, my book assumes that it is okay without even stating that it is valid.

As I mentioned this is not completely true. Take, for example, the semi-unit upper circle. We can express it as $\gamma(t) = e^{it} \mbox{ for }0\leq t\leq \pi$ and $\eta(t) = \sin t + i \cos t \mbox{ for }\pi \leq t\leq 2\pi$.
And consider the function $f(z)=z$.
Then, (note $\gamma^* = \eta^*$)
$\int_{\gamma} f(z) dz \not = \int_{\eta} f(z) dz$
What is wrong? The answer is the orientation. One is clockwise and one is counterclockwise.
The problem is I do not know how to formally define this notion.
• Jun 6th 2007, 04:35 PM
Rebesques
\eta here does not parametrize the same portion of the circle as \gamma, that's why the integrals are different :p

Because actually, the only change of orientation on a curve that can occur, is following it from the endpoint to the startpoint. And that just gives you a change of sign in the integral.