]> 21.2 The Fundamental Theorem for Integration in on a Path in the Complex Plane

## 21.2 The Fundamental Theorem for Integration in on a Path in the Complex Plane

The generalization of area used to define integration along a path in the complex plane obeys a fundamental theorem exactly analogous to that for ordinary real integrals.

We can break up the path into tiny pieces and the value of the integral on each piece will closely approximate $f ( z + d z ) − f ( z )$ for sufficiently small $d z$ and $f$ continuous at $z$ . We will not repeat the arguments given above but note the consequences

$∫ P , z = a z = b d f d z d z = f ( b ) − f ( a )$

Again, this means that all the methods discussed in Chapter 19 for finding anti-derivatives work exactly the same way in the complex plane.

Of particular interest here is what happens when $b = a$ so that the path closes on itself;

We would expect then that the integral would be 0. This is the case when $f$ is a well defined function.

It is not necessarily so when $f$ has logarithmic factors, because these are not true functions, and can take on different values at the same point.

In general, the integral of $f$ on a closed path will be 0 if $f$ has no singularities inside the path, but in general it is not 0 when there are singularities inside it, which contribute logarithmic factors.

Functions like polynomials, $exp ⁡ ( x )$ , polynomials in sines and cosines exponentials and polynomials all are always well defined and their integrals always obey the equation above unambiguously. Functions with logarithmic factors obey it in a sense, but the values of $f ( b )$ and $f ( a )$ in it may depend on the path.

The simplest example of the peculiarity associated with logarithms here occurs with integrand $z − 1$ .

Suppose we integrate this function on a circular path with constant $r$ around the origin. We then have $z = r e i θ$ so that with fixed $r$ we have $d z = i r e i θ d θ$ . $d z z$ therefore becomes $i d θ$ . And its integral around a circle of constant $r$ is $2 π i$ .

This is equivalent to saying that in integrating from $a$ to $b$ with this integrand we get a real part plus any integer multiple of $2 π i$ , depending on how many times the path $P$ winds around the origin.

If the function $f$ is not singular in the area of the plane surrounded by path $P$ , then the integral of $f$ around $P$ will be 0.

If $f$ can be expanded in a power series with both positive and negative terms about some argument $z 0$ , then all powers but the minus first power are derivatives of other powers and their integrals are path independent.

The minus first power gives a problem as described above. The coefficient of the minus first power in a power series expansion around that point is called the residue of the function at the point. The integral of the function around that point is $2 π i$ times its residue there.

To summarize, the fundamental theorem of calculus applies perfectly well in the context of complex integration, but nevertheless some functions have integrals with the same endpoints but have different values.

This is because the integrals evaluate to entities that are really not true functions but are multivalued.

An interesting question then becomes, can one detect this strange pheonomenon happening from the integrand?

The answer is that you can; if the integrand is singular between the two paths the values of the integrals will differ, and will differ by $2 π i$ times the sum of the residues between the paths.

This fact gives you a valuable tool for evaluating integrals beyond all those available from anti-differentiating.

The second form of the fundamental theorem, that the derivative of a complex integral with respect to its upper endpoint is the integrand holds exactly as for ordinary real integrals. It is rarely used in this context, in my experience.

Exercises:

21.1 Evaluate the integral of $z − 1 d z$ in the complex plane over a circular path around the origin.

21.2 Integrate it over a path that goes around the origin twice in counterclockwise direction.

21.3 Integrate it over a path that forms a circle of radius 1 around the point $z = 2$ .

21.4 Integrate $1 sin ⁡ z$ around the origin on a circle of radius 1. Over a circle of radius 2 about the point $z = 1.5$ .