# Line Integrals & Primitives

## By Sébastien Boisgérault, Mines ParisTech, under CC BY-NC-SA 4.0

### December 8, 2016

# Contents

# Exercises

## Primitives of Power Functions

### Question

Determine the primitives of the power \(z \mapsto z^n\) – defined on \(\mathbb{C}\) if \(n\) nonnegative and on \(\mathbb{C}^*\) otherwise – or prove that no such function exist.### Answer

If \(n \neq -1\), the function \(z \mapsto z^{n+1}/(n+1)\) is a primitive of \(z \mapsto z^n\). As \(\mathbb{C}\) and \(\mathbb{C}^*\) are path-connected, the other primitives differ from this one by a constant.If \(n=-1\), no primitive exist: the function \(\gamma: t \in [0,1] \to e^{i2\pi t}\) is a closed rectifiable path of \(\mathbb{C}^*\) and \[ \int_{\gamma} \frac{dz}{z} = \int_0^1 \frac{e^{i2\pi t} i2\pi}{e^{i2 \pi t}} \, dt = i2\pi, \] which is nonzero.

## Primitive of a Rational Function

### Question

Let \(\Omega = \mathbb{C} \setminus \{0,1\}\) and let \(f: \Omega \to \mathbb{C}\) be defined by \[ f(z) = \frac{1}{z(z-1)}. \] Show that \(f\) has no primitive on \(\Omega\), but that it has a primitive on \(\mathbb{C} \setminus [0,1]\) and determine its expression.### Answer

We have \[ f(z) = -\frac{1}{z} + \frac{1}{z-1}. \] The function \(z\mapsto -1/z\) has no primitive on \(D(0,1)\setminus \{0\}\): indeed if \(\gamma(t) = 1/2 \times e^{i2\pi t}\), we have \[ \int_{\gamma} \frac{dz}{z} = i2\pi \neq 0. \] On the other hand, on the same set, \(z \mapsto \log (z-1)\) is a primitive of \(z \mapsto 1/(z-1)\). Hence \(f(z)\) has no primitive.The function \[ g(z) = \log \frac{z-1}{z} = \log \left(1 - \frac{1}{z} \right) \] is defined on \(\mathbb{C} \setminus [0,1]\) and is a primitive of \(f\). Indeed \(g(z)\) is defined as long as neither of the conditions \(z = 0\) and \(1 - 1/z \in \mathbb{R}_-\) are met; they are equivalent to the condition \(z \in [0,1]\), which is excluded. Moreover, \(g\) satisfies \[ g'(z) = \frac{1/z^2}{1 - 1/z} = \frac{1}{z(z-1)} \] hence it is a primitive of \(f\).

## Reparametrization of Paths

### Questions

Let \(\alpha:[0,1] \to \mathbb{C}\) be a continuously differentiable path. Let \(\phi:[0,1] \to [0,1]\) be a continuously differentiable function such that \(\phi(0) = 0\), \(\phi(1) = 1\) and \(\phi'(t)>0\) for any \(t \in [0,1]\).Show that \(\beta = \alpha \circ \phi\) is a rectifiable path which has the same initial point, terminal point and image as \(\alpha\).

Prove that for any continuous function \(f: \alpha([0,1]) \to \mathbb{C}\), \[ \int_{\alpha} f(z) \, dz = \int_{\beta} f(z) \, dz. \]

Prove that the paths \(\alpha\) and \(\beta\) have the same length.

### Answers

The statement about the initial and terminal points is obvious. The one relative to the image holds because, under the assumptions that were made, the function \(\phi\) is a bijection from \([0,1]\) on itself (and its inverse is also continuously differentiable).

We have \[ \int_{\beta} f(z) \, dz = \int_{0}^1 (f \circ \beta)(t) \beta'(t)\,dt = \int_{0}^1 (f \circ \alpha)(\phi(t)) \alpha'(\phi(t))\,(\phi'(t) dt). \] The change of variable \(s = \phi(t)\) leads to \[ \int_{\beta} f(z) \, dz = \int_{0}^1 (f \circ \alpha)(s) \alpha'(s)\,ds = \int_{\alpha} f(z) \, dz. \]

We have \[ \int_0^1 |\beta'(t)|\, dt = \int_{0}^1 |\alpha' (\phi(t)) \phi'(t)| \, dt = \int_{0}^1 |\alpha' (\phi(t))| \, \phi'(t) dt \] The change of variable \(s = \phi(t)\) leads to \[ \int_0^1 |\beta'(t)|\, dt = \int_0^1 |\alpha'(s)| \, ds, \] hence the lengths of \(\alpha\) and \(\beta\) are equal.

## The Logarithm: Alternate Choices

### Question

Show that for any \(\alpha \in \mathbb{R}\), the function \(z \in \mathbb{C}_{\alpha} \mapsto 1/z\) defined on \[ \mathbb{C}_{\alpha} = \mathbb{C} \setminus \{r e^{i\alpha} \;|\; r \geq 0 \}. \] has a primitive; describe the set of all its primitives.### Answer

Let \(\gamma\) be a closed rectifiable path of \(\mathbb{C}_{\alpha}\). The path \(\mu: [0,1] \mapsto e^{i(\pi-\alpha)} \gamma(t)\) is closed, rectifiable and its image is included in \(\mathbb{C} \setminus \mathbb{R}_-\). Additionally \[ \int_{\gamma} \frac{dz}{z} = \int_{\gamma} \frac{d(e^{i(\pi-\alpha)}z)}{e^{i(\pi-\alpha)}z} = \int_{\mu} \frac{dz}{z}. \] Since the principal value of the logarithm is a primitive if \(z \mapsto 1/z\) on \(\mathbb{C} \setminus \mathbb{R}_-\), the integral of \(z\mapsto 1/z\) on \(\mu\) is equal to zero. Therefore, there are primitives of \(z\mapsto 1/z\) on \(\mathbb{C}_{\alpha}\); since \(\mathbb{C}_{\alpha}\) is connected, they all differ from an arbitrary constant.Alternatively, we can build explicitly such a primitive: the function \[ f: z \mapsto \log (z e^{i(\pi - \alpha)}); \] it is defined and holomorphic on \(\mathbb{C}_{\alpha}\) and for any \(z \in \mathbb{C}_{\alpha}\), \[ f'(z) = \frac{1}{z e^{i(\pi - \alpha)}} \times e^{i(\pi - \alpha)}= \frac{1}{z}. \]