This example shows how to calculate complex
line integrals using the `'Waypoints'`

option of
the `integral`

function. In MATLAB^{®}, you use
the `'Waypoints'`

option to define a sequence of
straight line paths from the first limit of integration to the first
waypoint, from the first waypoint to the second, and so forth, and
finally from the last waypoint to the second limit of integration.

**Define the Integrand with an Anonymous Function**

Integrate

$${\oint}_{C}\frac{{e}^{z}}{z}dz},$$

where *C* is a closed contour that encloses
the simple pole of $$\frac{{e}^{z}}{z}$$ at
the origin.

Define the integrand with an anonymous function.

fun = @(z) exp(z)./z;

**Integrate Without Using Waypoints**

You can evaluate contour integrals of complex-valued functions with a parameterization. In general, a contour is specified, and then differentiated and used to parameterize the original integrand. In this case, specify the contour as the unit circle, but in all cases, the result is independent of the contour chosen.

g = @(theta) cos(theta) + 1i*sin(theta); gprime = @(theta) -sin(theta) + 1i*cos(theta); q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi)

q1 = -0.0000 + 6.2832i

This method of parameterizing, although reliable, can be difficult
and time consuming since a derivative must be calculated before the
integration is performed. Even for simple functions, you need to write
several lines of code to obtain the correct result. Since the result
is the same with any closed contour that encloses the pole (in this
case, the origin), instead you can use the `'Waypoints'`

option
of `integral`

to construct a square or triangular
path that encloses the pole.

**Integrate Along a Contour That Encloses No Poles**

If any limit of integration or element of the waypoints vector
is complex, then `integral`

performs the integration
over a sequence of straight line paths in the complex plane. The natural
direction around a contour is counterclockwise; specifying a clockwise
contour is akin to multiplying by `-1`

. Specify the
contour in such a way that it encloses a single functional singularity.
If you specify a contour that encloses no poles, then Cauchy's
integral theorem guarantees that the value of the closed-loop integral
is zero.

To see this, integrate `fun`

around a square
contour away from the origin. Use equal limits of integration to form
a closed contour.

```
C = [2+i 2+2i 1+2i];
q = integral(fun,1+i,1+i,'Waypoints',C)
```

q = 2.7756e-16 - 7.7716e-16i

The result is on the order of `eps`

and effectively
zero.

**Integrate Along a Contour with a Pole in the Interior**

Specify a square contour that completely encloses the pole at the origin, and then integrate.

```
C = [1+i -1+i -1-i 1-i];
q2 = integral(fun,1,1,'Waypoints',C)
```

q2 = 0.0000 + 6.2832i

This result agrees with the `q1`

calculated
above, but uses much simpler code.

The exact answer for this problem is 2πi.

2*pi*i

ans = 0.0000 + 6.2832i

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