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C. Results from Analytic Function Theory

C.7 Integrals Revisited

Theorem C.7 (Cauchy Integral Theorem)   If $f(z)$ is analytic in some simply connected domain $D$, then $ \int f(z) dz$ is independent of path in $D$ and


\begin{displaymath}\oint_C f(z) dz =0 \end{displaymath} (C.7.1)

where $C$ is a simple closed path in $D$.

Proof

This follows from the Cauchy-Riemann conditions together with Theorem C.2.

$\Box \Box \Box $

We are also interested in the value of integrals in various limiting situations. The following examples cover relevant cases.

We note that if $L_C$ is the length of a simple curve $C$, then


\begin{displaymath}\left\vert\int_C f(z)dz \right\vert\leq \max_{z\in C}(\vert f(z)\vert)L_C \end{displaymath} (C.7.2)

Example C.5  Assume that $C$ is a semicircle centered at the origin and having radius $R$. The path length is then $L_C=\pi R$. Hence,

  • if $f(z)$ varies as $z^{-2}$, then $\vert f(z)\vert$ on $C$ must vary as $R^{-2}$ - hence, the integral on $C$ vanishes for $R\rightarrow \infty$.

  • if $f(z)$ varies as $z^{-1}$, then $\vert f(z)\vert$ on $C$ must vary as $R^{-1}$ - then, the integral on $C$ becomes a constant as $R\rightarrow \infty$.

Example C.6   Consider the function $f(z)=\ln(z)$ and an arc of a circle, $C$, described by $z=\epsilon e^{j\gamma}$ for $\gamma\in[-\gamma_1,\gamma_1]$. Then

\begin{displaymath}I_\epsilon\stackrel{\rm\triangle}{=}\lim_{\epsilon\rightarrow 0}\int_C f(z) dz=0 \end{displaymath} (C.7.3)

This is proven as follows. On $C$, we have that $f(z)=\ln(\epsilon)$. Then


\begin{displaymath}I_\epsilon=\lim_{\epsilon\rightarrow 0} \left[(\gamma_2-\gamma_1)\epsilon\ln (\epsilon)\right ] \end{displaymath} (C.7.4)

We then use the fact that $\lim_{\vert x\vert \rightarrow 0}(x\ln x)=0$, and the result follows.

Example C.7  Consider the function


\begin{displaymath}f(z)=\ln \left(1+\frac{a}{z^n} \right ) \hspace{10mm}n\geq 1 \end{displaymath} (C.7.5)

and a semicircle, $C$, defined by $z=Re^{j\gamma}$ for $\gamma\in \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]$. Then, if $C$ is followed clockwise,


\begin{displaymath}I_R \stackrel{\rm\triangle}{=}\lim_{R\rightarrow \infty}\int_... ...0 &\mbox{ for $n>1$ }\\ -j\pi a &\mbox{ for $n=1$ } \end{cases}\end{displaymath} (C.7.6)

This is proven as follows.

On $C$, we have that $z=Re^{j\gamma}$; then


\begin{displaymath}I_R= \lim_{R\rightarrow \infty}j\int_{\frac{\pi}{2}}^{-\frac{... ...left(1+\frac{a}{R^n}e^{-jn\gamma} \right )Re^{j\gamma} d\gamma \end{displaymath} (C.7.7)

We also know that


\begin{displaymath}\lim_{\vert x\vert\rightarrow 0}\ln (1+x)=x \end{displaymath} (C.7.8)

Then


\begin{displaymath}I_R=\lim_{R\rightarrow \infty}\frac{a}{R^{n-1}}j\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} e^{-j(n-1)\gamma}d\gamma \end{displaymath} (C.7.9)

From this, by evaluation for $n=1$ and for $n>1$, the result follows.

$\Box \Box \Box $

Example C.8   Consider the function


\begin{displaymath}f(z)=\ln \left(1+e^{-z\tau}\frac{a}{z^n} \right ) \hspace{10mm}n\geq 1;\hspace{3mm}\tau>0 \end{displaymath} (C.7.10)

and a semicircle, $C$, defined by $z=Re^{j\gamma}$ for $\gamma\in \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]$. Then, for clockwise $C$,


\begin{displaymath}I_R \stackrel{\rm\triangle}{=}\lim_{R\rightarrow \infty}\int_C f(z) dz=0 \end{displaymath} (C.7.11)

This is proven as follows.

On $C$, we have that $z=Re^{j\gamma}$; then


\begin{displaymath}I_R= \lim_{R\rightarrow \infty}j\int_{\frac{\pi}{2}}^{-\frac{... ...\frac{z}{e^{z\tau}} \right )z \right] _{z=Re^{j\gamma}}d\gamma \end{displaymath} (C.7.12)

We recall that, if $\tau$ is a positive real number and $\Re \{z\}>0$, then


\begin{displaymath}\lim_{\vert z\vert\rightarrow \infty}\frac{z}{e^{z\tau}}=0 \end{displaymath} (C.7.13)

Moreover, for very large $R$, we have that


\begin{displaymath}\left. \ln\left(1+\frac{a}{z^{n+1}}\frac{z}{e^{z\tau}} \right... ...\frac{1}{z^n} \frac{z}{e^{z\tau}}\right\vert _{z=Re^{j\gamma}} \end{displaymath} (C.7.14)

Thus, in the limit, this quantity goes to zero for all positive $n$. The result then follows.

$\Box \Box \Box $

Example C.9  Consider the function


\begin{displaymath}f(z)=\ln \left(\frac{z-a}{z+a} \right ) \end{displaymath} (C.7.15)

and a semicircle, $C$, defined by $z=Re^{j\gamma}$ for $\gamma\in \left[- \frac{\pi}{2}, \frac{\pi}{2} \right]$. Then, for clockwise $C$,


\begin{displaymath}I_R \stackrel{\rm\triangle}{=}\lim_{R\rightarrow \infty}\int_C f(z) dz=j2\pi a \end{displaymath} (C.7.16)

This result is obtained by noting that


\begin{displaymath}\ln \left(\frac{z-a}{z+a} \right )=\ln \left(\frac{1-\frac{a}... ...left( 1-\frac{a}{z} \right )-\ln \left( 1+\frac{a}{z} \right ) \end{displaymath} (C.7.17)

and then applying the result in Example C.7.

$\Box \Box \Box $

Example C.10  Consider a function of the form


\begin{displaymath}f(z)=\frac{a_{-1}}{z}+\frac{a_{-2}}{z^2}+\ldots \end{displaymath} (C.7.18)

and $C$, an arc of circle $z=Re^{j\theta}$ for $\theta\in [\theta_1, \theta_2]$. Thus, $dz=jzd\theta$, and


\begin{displaymath}\int_C\frac{dz}{z}=\int _{\theta_1}^{\theta_2}jd\theta=-j(\theta_2-\theta_1) \end{displaymath} (C.7.19)

Thus, as $R\rightarrow \infty$, we have that


\begin{displaymath}\int_C f(z) dz=-ja_{-1}(\theta_2-\theta_1) \end{displaymath} (C.7.20)

$\Box \Box \Box $

Example C.11  Consider, now, $f(z)=z^n$. If the path $C$ is a full circle, centered at the origin and of radius $R$, then


$\displaystyle \oint_C z^n dz$ $\textstyle =$ $\displaystyle \int_{-\pi}^\pi \left(R^ne^{jn\theta}\right)jRe^{j\theta} d\theta$ (C.7.21)
  $\textstyle =$ $\displaystyle \begin{cases} 0 &\mbox{ for $n\neq -1$\space }\\ -2\pi j&\mbox{ for $n=-1$\space (integration clockwise)} \end{cases}$ (C.7.22)

$\Box \Box \Box $

We can now develop Cauchy's Integral Formula.

Say that $f(z)$ can be expanded as


\begin{displaymath}f(z)=\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\ldots \end{displaymath} (C.7.23)

the $a_{-1}$ is called the residue of $f(z)$ at $z_0$.


Figure C.3: Path for integration of a function having a singularity
\begin{figure} \begin{center} \leavevmode \begin{picture} (0,0)% \epsfig{file=... ...lt}{\mddefault}{\updefault}$\epsilon$ }}} \end{picture} \end{center}\end{figure}

Consider the path shown in Figure C.3. Because $f(z)$ is analytic in a region containing $C$, we have that the integral around the complete path shown in Figure C.3 is zero. The integrals along $C_1$ and $C_2$cancel. The anticlockwise circular integral around $z_0$ can be computed by following Example C.11 to yield $2\pi ja_{-1}$. Hence, the integral around the outer curve $C$ is minus the integral around the circle of radius $\epsilon$. Thus,


\begin{displaymath}\oint_Cf(z)dz= -2\pi ja_{-1} \end{displaymath} (C.7.24)

This leads to the following result.

Theorem C.8 (Cauchy's Integral Formula)  Let $g(z)$be analytic in a region. Let $q$ be a point inside the region. Then $\frac{g(z)}{z-q}$ has residue $g(q)$ at $z=q$, and the integral around any closed contour $C$ enclosing $q$ in a clockwise direction is given by


\begin{displaymath}\oint_C \frac{g(z)}{z-q}dz=-2\pi jg(q) \end{displaymath} (C.7.25)

$\Box \Box \Box $

We note that the residue of $g(z)$ at an interior point, $z=q$, of a region $D$ can be obtained by integrating $\frac{g(z)}{z-q}$ on the boundary of $D$. Hence, we can determine the value of an analytic function inside a region by its behaviour on the boundary.