# 2.3 Reverse paths and contours

We now introduce the concept of the *reverse path* (some texts use the name *opposite path*) of a smooth path . This is simply the path we obtain by traversing the original path in the opposite direction, starting from the final point of the original path and finishing at the initial point of the original path. In order to define the reverse path formally, we use the fact that as increases from to , so decreases from to .

## Definition

Let be a smooth path. Then the **reverse path** of , denoted by , is the path with parametrisation , where

Note that the initial point of is the final point of , and the final point of is the initial point of (see Figure 20). The path is smooth because is smooth. Also note that, as *sets*, and are the same.

## Exercise 6

Write down the reverse path of the path with parametrisation

### Answer

Since and , the reverse path is (), where

We can also define a *reverse contour*. This is done in the natural way – namely by reversing each of the constituent smooth paths of a contour and reversing the order in which they are traversed.

## Definition

Let be a contour. The **reverse contour** of is

A contour and its reverse contour are illustrated in Figure 21.

As an example, if is the contour from 0 to in Figure 22(a), with smooth parametrisations

then is the contour from to 0 in Figure 22(b), with smooth parametrisations

## Example 6

Evaluate

where is the reverse path of the line segment from to .

### Solution

We use the standard parametrisation

of . For the reverse path , the corresponding parametrisation is

Then , so we substitute

to give

In Example 3 we saw that

which is the negative of the value that we obtained in Example 6. This illustrates the general result that if we integrate a function along a reverse contour , then the answer is the negative of the integral of the function along .

## Theorem 6 Reverse Contour Theorem

Let be a contour, and let be a function that is continuous on . Then the integral of along the reverse contour of satisfies

## Proof

The proof is in two parts. We first prove the result in the case when is a smooth path, and then extend the proof to contours.a.Let be a smooth path. Then the parametrisation of is

It follows that , by the Chain Rule, so

where, in the second-to-last line, we have made the real substitution

b.To extend the proof to a general contour , we argue as follows.

Let , for smooth paths . Then

and we can apply part (a) to see that

In Example 4 we saw that

where is the unit circle . The next exercise asks you to check Theorem 6 for this contour integral.

## Exercise 7

Verify that

where is the unit circle.

### Answer

In Example 4 we used the parametrisation

For the reverse path we use the parametrisation

Since , we have

and . Hence

(Therefore, by Example 4,