# 2.1 Integration along a smooth path

Motivated by the discussion of the preceding section, we make the following definition of the integral of a complex function.

## Definition

Let be a smooth path in , and let be a function that is continuous on . Then the **integral of** **along the path** , denoted by , is

The integral is evaluated by splitting into its real and imaginary parts and , and evaluating the resulting pair of real integrals,

## Remarks

Since is continuous on and is a smooth parametrisation, the functions and are both continuous on , so the function is continuous on . It follows that the real functions and are continuous on , and hence

exist, so also exists.

An important special case is when , so is the real line segment from to . Since , we see that equals

where and . This equation is a formula for the integral of a

*complex*function over a real interval.An alternative notation for is .

If the path of integration has a standard parametrisation , then, unless otherwise stated, we use in the evaluation of the integral of along .

To help to remember the formula used to define , notice that it can be obtained by ‘substituting’

We consider to be a shorthand for .

The following examples demonstrate how to evaluate integrals along paths. In each case, we follow the convention of Remark 3 and use the standard parametrisation of the path.

## Example 2

Evaluate

where is the line segment from to .

### Solution

Here , and we use the standard parametrisation

of , which satisfies .

Then , so

You need not include every line of working of Example 2 if you do not need to. Here is another example.

## Example 3

Evaluate

where is the line segment from 0 to .

### Solution

Here , and again we use the standard parametrisation

of , which satisfies . Then

so

We set out our solution to the next example using the observation and notation of Remark 4.

## Example 4

Evaluate

where is the unit circle .

### Solution

Here , and we use the standard parametrisation

of . Then , and . Hence

Sometimes when evaluating integrals we will use the alternative notation of Example 4 instead of the notation of Example 2 and Example 3; both notations are commonly used in complex analysis.

In the examples above, we used the standard parametrisation in each case. The following exercise suggests that the value of the integral is not affected by the choice of parametrisation.

## Exercise 3

a.Verify that the result of Example 3 is unchanged if we use the smooth parametrisation

b.Verify that the result of Example 4 is unchanged if we use the smooth parametrisation

### Answer

a.Here . Let . Then

and, since , we obtain

in accordance with Example 3.

b.We set out this solution in a similar style to Example 4.

Here . Then

Hence

in accordance with Example 4.

The reason why we have obtained the same values in Exercise 3 as those in Example 3 and Example 4 is because of the following theorem.

## Theorem 4

Let and be two smooth parametrisations of paths with the same image set , and let be a function that is continuous on . Then

does not depend on which parametrisation or is used.

The proof of Theorem 4 uses the Inverse Function rule and the Chain rule for the derivatives of complex functions, which are not covered within this course. So we shall omit the details of this proof.

In practical terms, this theorem allows you to choose any convenient smooth parametrisation when evaluating a complex integral along a given path. We will see how this can be helpful in the next subsection.

For further practice in integration, try the following exercise.

## Exercise 4

Evaluate the following integrals.

a., where is the line segment from 0 to .

b., where is the circle with centre and radius .

### Answer

a.The standard parametrisation of is

Then

Hence

b.The standard parametrisation of is

Then

Hence