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# contour integral

Let $f$ be a complex-valued function defined on the image of a curve $\alpha$: $[a,b]\rightarrow\mathbb{C}$, let $P=\{a_{{0}},...,a_{{n}}\}$ be a partition of $[a,b]$. We will restrict our attention to *contours*, i.e. curves for which the parametric equations consist of a finite number of continuously differentiable arcs.
If the sum

$\sum_{{i=1}}^{{n}}f(z_{{i}})(\alpha(a_{{i}})-\alpha(a_{{i-1}})),$ |

where $z_{{i}}$ is some point $\alpha(t_{{i}})$ such that $a_{{i-1}}\leqslant t_{{i}}\leqslant a_{{i}}$, converges as $n$ tends to infinity and the greatest of the numbers $a_{{i}}-a_{{i-1}}$ tends to zero, then we define the *contour integral* of $f$ along $\alpha$ to be the integral

$\int_{{\alpha}}f(z)dz:=\int_{a}^{b}f(\alpha(t))d\alpha(t)$ |

# Notes

(i) If $\operatorname{Im}(\alpha)$ is a segment of the real axis, then this definition reduces to that of the Riemann integral of $f(x)$ between $\alpha(a)$ and $\alpha(b)$.

(ii) An alternative definition, making use of the Riemann-Stieltjes integral, is based on the fact that the definition of this can be extended without any other changes in the wording to cover the cases where $f$ and $\alpha$ are complex-valued functions.

Now let $\alpha$ be any curve $[a,b]\rightarrow\mathbb{R}^{{2}}$. Then $\alpha$ can be expressed in terms of the components $(\alpha_{{1}},\alpha_{{2}})$ and can be associated with the complex-valued function

$z(t)=\alpha_{{1}}(t)+i\alpha_{{2}}(t).$ |

Given any complex-valued function of a complex variable, $f$ say, defined on $\operatorname{Im}(\alpha)$ we define the contour integral of $f$ along $\alpha$, denoted by

$\int_{{\alpha}}f(z)dz$ |

by

$\int_{{\alpha}}f(z)dz=\int_{{a}}^{{b}}f(z(t))dz(t)$ |

whenever the complex Riemann-Stieltjes integral on the right exists.

(iii) Reversing the direction of the curve changes the sign of the integral.

(iv) The contour integral always exists if $\alpha$ is rectifiable and $f$ is continuous.

(v) If $\alpha$ is piecewise smooth and the contour integral of $f$ along $\alpha$ exists, then

$\int_{{\alpha}}fdz=\int_{{a}}^{{b}}f(z(t))z^{{\prime}}(t)dt.$ |

## Mathematics Subject Classification

30A99*no label found*30E20

*no label found*

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