# Contour integration

Here we follow standard texts, such as Spiegel (1964) or Levinson and Redheffer (1970). 

For a complex-valued function $f(z)=u(x,y)+iv(x,y)$ of a complex variable $z=x+iy$ that is analytic in some region ${\mathcal {R}},$ we may define the process of integration along a curve $C$ via the Riemann summation formalism

$\int _{C}f(z)dz=\int _{a}^{b}f(z)dz=\lim _{\Delta z_{n}\rightarrow 0}\lim _{N\rightarrow \infty }\sum _{n=0}^{N}f(\zeta _{n})\Delta z_{n}$ $\int _{C}f(z)dz=\lim _{\Delta z_{n}\rightarrow 0}\lim _{N\rightarrow \infty }\sum _{n=0}^{N}f(\zeta _{n})\Delta z_{n}$ where $\Delta z_{n}=z_{n+1}-z_{n}$ represents an increment along the curve

$C=\left\{a=z_{0}<\zeta _{0} See Figure 1. Such a curve is called a contour, though this term does not refer to a level curve,

or a curve of equal elevation as it would in cartography, but to any curve in the complex plane that does not cross itself.

Formally we may write

$\int _{C}f(z)\;dz=\int _{C}(u(x,y)+iv(x,y))\;(dx+idy)=\left(\int _{C}u(x,y)\;dx-\int _{C}v(x,y)\;dy\right)+i\left(\int _{C}v(x,y)\;dx+\int _{C}u(x,y)\;dy\right).$ Several important results allow us to make sense of, and to make use of the properties of complex integration.

## Cauchy's Theorem

If $C$ is a closed contour, and the complex valued function $f(z)$ is analytic inside the region bounded by, and on $C$ then

$\int _{C}f(z)\;dz=0.$ This is followed by a complementary theorem by Morerra

## Morera's Theorem

If for every closed contour $C$ within a region ${\mathcal {R}}$ of the complex and

$\int _{C}f(z)\;dz=0$ then $f(z)$ is analytic everywhere in ${\mathcal {R}}$ .

## Cauchy integral theorem

If $f(z)$ is analytic on, and inside, a region bounded by a closed contour $C$ then for a point $a$ inside C,

$f(a)={\frac {1}{2\pi i}}\int _{C}{\frac {f(z)}{(z-a)}}\;dz.$ ## Cauchy integral formulas

If $f(z)$ is analytic on, and inside, a region bounded by a closed contour $C$ then for a point $a$ inside C, and for every integer $k>0$ $f^{(k)}(a)={\frac {k!}{2\pi i}}\int _{C}{\frac {f(z)}{(z-a)^{k+1}}}$ where $f^{(k)}(a)$ is the $k$ -th derivative with respect to $a$ of $f(a)$ .

## Residue Theorem

### Simple pole

If $f(z)$ is analytic on, and inside, a region bounded by a closed contour $C$ then for a point except at a point $a$ inside C, such that $a$ is a simple pole, that is $f(z)={\frac {g(z)}{(z-a)}}$ for $g(z)$ analytic inside and on $C$ then

$\int _{C}f(z)\;dz=2\pi i\;Res(f(z);a)$ where, $Res(f(z);a)$ is called the residue of $f(z)$ at $a.$ If $f(z)$ has $M$ finite simple poles inside $C$ $\int _{C}f(z)\;dz=2\pi i\;Res(f(z);a)=2\pi i\sum _{k=0}^{M}\;Res(f(z);a_{k})$ ### Multiple pole

If $f(z)={\frac {g(z)}{(z-a)^{N+1}}}$ for $N>0$ an integer where $g(z)$ is analytic inside and on $C,$ then $f(z)$ is said to have an $N$ -th order pole at $a$ and

$\int _{C}f(z)\;dz=2\pi i\;Res(f(z);a).$ Here, the $Res(f(z);a)=\lim _{z\rightarrow a}N!(z-a)^{N+1}f^{(N)}(z),$ where $f^{(N)}$ is the $N$ -th derivative of $f(z).$ 