# Path Integral
## Introduction

This page on the topic "Path Integral" can be displayed as Wiki2Reveal Slides. Individual sections are considered as slides, and changes to the slides immediately affect the content of the slides. The following subtopics are treated in detail:
(1) Paths as continuous mappings from an interval $[a,b]$ into the complex numbers $\mathbb {C}$ over which integration is performed,
(2) Derivatives of curves/paths as a prerequisite for the definition of path integrals,
(3) Definition of path integrals

## Learning requirements

The learning resource on the topic "Path Integral" has the following learning prerequisites, which are helpful or necessary for understanding the subsequent explanations:
* Concept ofPaths in a topological Space,
* Differentiability in real analysis,
* Integration in real analysis.

## Basic Geometric Idea of the Path Integral

The following curve $\gamma$ loops around a point $z_{0}\in \mathbb {C}$ twice.

## Integral over an Interval

Let $G\subseteq \mathbb {C}$ be a domain and $g\colon [a,b]\to \mathbb {C}$ a complex-valued function. The function $g$ is called integrable if
       $\operatorname {Re} (g):G\to \mathbb {R}$ and $\operatorname {Im} (g):G\to \mathbb {R}$ with $g=\operatorname {Re} (g)+i\cdot \operatorname {Im} (g)$ are integrable functions.
It is defined as
    $\int \limits _{a}^{b}g(x)\mathrm {d} x:=\int \limits _{a}^{b}\operatorname {Re} (g)(x)\mathrm {d} x+\mathrm {i} \int \limits _{a}^{b}\operatorname {Im} (g)(x)\mathrm {d} x$.
Thus, the integral is $\mathbb {C}$-linear. If $g$ is continuous and $G$ is an antiderivative of $g$, then as in the real case,
    $\int \limits _{a}^{b}g(x)\mathrm {d} x=G(b)-G(a)$.

## Extension of the Integral Concept

The integral concept is extended through the definition of an integration path in the complex plane as follows: If $f\colon G\to \mathbb {C}$ is a complex-valued function on a domain $G\subseteq \mathbb {C}$, and $\gamma \colon [a,b]\to G$ is a piecewise continuously differentiable path in $G$, then the path integral of $f$ along the path $\gamma$ is defined as
    $\int \limits _{\gamma }f:=\int \limits _{\gamma }f(z),\mathrm {d} z:=\int \limits _{a}^{b}f(\gamma (t))\cdot \gamma '(t),\mathrm {d} t.$
Here, the multiplication sign refers to complex multiplication.

## Cauchy's Integral Theorem

The central statement about path integrals of complex functions is the Cauchy Integral Theorem: For a holomorphic function $f$, the path integral depends only on the homotopy class of $\gamma$. If $U$ is simply connected, then the integral depends not on $\gamma$, but only on the starting and ending points.
Analogous to the real case, the length of the path $\gamma :[a,b]\rightarrow \mathbb {C}$ is defined as
    ${\mathcal {L}}(\gamma ):=\int \limits _{a}^{b}\left|\gamma '(t)\right|\mathrm {d} t$.
For theoretical purposes, the following inequality, called the standard estimate, is of particular interest:
    $\left|\int _{\gamma }f(z),\mathrm {d} z\right|\leq {\mathcal {L}}(\gamma )\cdot C$, if $\left|f(z)\right|\leq C$ for all $z\in \gamma ([0,1])$.
As in the real case, the path integral is independent of the parametrization of the path $\gamma$, i.e., it is not strictly necessary to choose $[0,1]$ as the parameter domain, as can be shown by substitution. This allows the definition of complex curve integrals by replacing the above formulas with a curve ${\mathcal {C}}$ in $\mathbb {C}$.

## Exercises

* Be $\gamma \colon [a,b]\to G$ with $t\mapsto \gamma (t)=\sin(t)+i\cdot t^{2}$. Determine $\gamma '(t)$!
* Compute the path integral $\int \limits _{\gamma }{\frac {1}{z}},\mathrm {d} z$ for the path $\gamma \colon [0,2\pi ]\to \mathbb {C}$ with $t\mapsto \gamma (t)=r\cdot e^{i\cdot t}$.
* Calculate the length of the path ${\mathcal {}\gamma )$ with $t\mapsto \gamma (t)=r\cdot e^{i\cdot t}$.