柯西积分公式

This is a very remarkable statement. It demonstrates the strong inter-relation among the values of a regular function. In order to know the values of the function in the interior of $C$ we have only to specify the values on the boundary.

The representation of an analytic function by means of the Cauchy integral enables us to obtain an integral representation for its derivative. In fact we have in general:

Since $\varphi (z)$ is any continuous function whatever we cannot hope that $F(\zeta )$ will be analytic on $C$ . However, it might be supposed that $F(\zeta )$ tends to $2\pi i\varphi (z)$ as $\zeta$ approaches a point $z$ on $C$ . But this will not generally be true. For example, let $C$ be the unit circle $|z|=1$ and take $\varphi (z)=\frac{1}{z}$ . We have, for a point $\zeta$ inside the circle,

Hence the limiting value of $F(\zeta )$ on the boundary is zero.

If $C$ is a closed path in a simply connected domain of analyticity of $f(z)$ it follows by this lemma that for all points $\zeta$ in the interior of $C$ . The function (3.03) is differentiable once again and, in fact, any number of times.

We have proved an important result:

In these proofs we have avoided the assumption that the first partial derivatives of the real and imaginary parts of $f(z)=u+iv$ are continuous. The continuity of these derivatives – and even the existence of derivatives of higher order – is assured if only they exist and satisfy the Cauchy-Riemann equations.

The converse of the Cauchy Integral Theorem , the second half of Theorem 3.2.3 , is generally known as Morera’s Integral Theorem .

The hypothesis of the theorem is equivalent to the assertion that $F(z)={\int }_{z_0}^{z}f(z)dz$ takes on the same value on all possible paths joining $z_0$ to $z$ . If a function $f(z)$ has this property it is said to be integrable . We have proved ( Theorem 3.2.1 ) that an integrable function is the derivative of an analytic function. We conclude that an integrable function must in turn be analytic.

We see now that we could have defined an analytic function in terms of integrability instead of differentiability, the one property implying the other in contrast with the more complicated situation in the theory of real functions.

3.3.1 Consequences of the Cauchy Integral Formula

An immediate and interesting result is the Mean Value Theorem .

An important consequence of the formula (3.04) is the fact that the derivatives of $f(z)$ are bounded in a very definite way in terms of the bound of the original function. There is of course no analogous statement for real functions.

Let $R$ be a region of analyticity of $f(z)$ bounded by a simple closed curve $C$ . If $\zeta$ is any point in the interior of $C$ , $\rho$ the nearest distance any point of $C$ approaches $\zeta$ , then we have by (3.04) $$|f^{(n)}(\zeta )|=\frac{n!}{2\pi }\left|{\int }_C\frac{f(z)}{(z-\zeta {)}^{n+1}}dz\right|\le \frac{n!}{2\pi }\frac{ML}{{\rho }^{n+1}}.$$ Note that this bound does not depend on $\zeta$ but only on its distance from the boundary. More specially, let $C$ be the circle of radius $\rho$ about $\zeta$ . We then have where $M(\rho )$ is now the maximum value of $|f(z)|$ on $C_{\rho }$ . In particular, for $f(\zeta )$ itself we have  

This last result could have been obtained as a direct consequence of the mean value theorem. From the special result (3.12) it is easy to prove the Maximum Modulus Theorem .

Proof. Let ${\zeta }_0$ be any interior point of $R$ . We may then find a circle $C_{\rho }$ about ${\zeta }_0$ which lies completely in the interior of $R$ . Now, in (3.12) , unless $|f(z)|=M(\rho )$ everywhere on the circle, we must have the strict inequality Suppose now, that the maximum $M$ of $|f(z)|$ is attained at the interior point ${\zeta }_0$ . It will follow that $f(z)$ is constant throughout $R$ and from this the theorem follows.

First, if $|f({\zeta }_0)|=M$ then the above argument insures that $|f(z)|=M$ on every circle in $R$ about ${\zeta }_0$ . In other words, $|f(z)|=M$ in the largest circle that can be drawn about ${\zeta }_0$ in $R$ .

Now, let be any point of $R$ . We can find a path in the interior of $R$ which connects ${\zeta }_0$ to ${\zeta }_1$ . Suppose that . By the continuity of $|f(z)|$ there must be a first point on the path for which and immediately following, . But, since $|f(z)|=M$ in some circle about , this implies a contradiction. Hence $|f(z)|=M$ everywhere in $R$ . It follows from the Cauchy-Riemann equations that a function whose modulus is constant everywhere in a domain of regularity must itself be constant. For, setting $f(z)=u+iv$ , $z=x+iy$ , we have $$|f(z){|}^2=u^2+v^2=M^2.$$ Differentiating with respect to $x$ and $y$ we obtain $$u{u}_x+v{v}_x=0;u{u}_y+v{v}_y=0.$$ By the Cauchy-Riemann equations we obtain $$u{u}_x=v{u}_y;v{u}_x=-u{u}_y$$ whence $$u_x^2+u_y^2=v_y^2+v_x^2=0$$ Thus, the first partial derivatives must vanish. The only possibility is that $f(z)=u+iv=\mathit{\text{constant}}$ . ◻

Another consequence of the formula (3.11) is Liouville’s Theorem .

Liouville’s Theorem figures importantly in many ways. Consider, for example, the following proof of The Fundamental Theorem of Algebra :

3.3.2 Residues, Evaluations of Integrals

Let $f(z)$ be a regular function in a simply-connected domain $D$ with the possible exception of a single point $\zeta$ . Then if $C$ is any simple closed curve containing $\zeta$ we define the residue of $f(z)$ at $\zeta$ to be the integral This integral is certainly independent of the curve containing $\zeta$ .

More generally, if $C$ is contained in a domain of analyticity of $f(z)$ and $f(z)$ is regular everywhere in the interior of $C$ except at a finite number of points $z_1,z_2,\dots ,z_n$ , then $\frac{1}{2\pi i}{\int }_{C}f(z)dz$ is equal to the sum of the residues at these points. The proof is evident. The notion of residue is a very useful tool for the evaluation of complex integrals.

As an example, let us determine the residue of the function $f(z)=z^{-n}$ , $n$ an integer. Setting $z=r{e}^{i\theta }$ and integrating around a circle of radius $\rho$ about the origin we obtain $$\frac{1}{2\pi i}{\int }_{C_{\rho }}\frac{dz}{z^n}=\frac{1}{2\pi {\rho }^{n-1}}{\int }_0^{2\pi }{e}^{-i(n-1)\theta }d\theta$$ Since $e^{2\pi i}=e^0=1$ we conclude that the residue is zero if and if $n=1$ the residue is $1$ .

From the fact that $\frac{d\log z}{dz}=\frac{1}{z}$ , we have This characterization displays the multivalued character of the logarithms. If we join $1$ to the point $z$ by means of two non-intersecting curves which enclose the origin then the two values obtained by (3.22) will differ by $2\pi i$ . Clearly there are an infinite number of values of $\log z$ , the difference of any two of them being a multiple of $2\pi i$ .

By means of residues we can evaluate a large class of infinite integrals in the real domain:

Proof. Take so large that the $z_i$ are all contained in the circle .

Consider the integral around the semicircle in the upper half-plane: The integral on the curved part of the path can be made arbitrarily small by increasing $R$ sufficiently. For we have $$|{\int }_0^{\pi }R{e}^{i\theta }f(R{e}^{i\theta })d\theta |\le \pi \underset{|z|=R}{max}\{|zf(z)|\}$$ and by (3.23) the right hand side must go to zero. We conclude that $$\underset{R\to \mathrm{\infty }}{lim}{\int }_{-R}^{R}f(x)dx={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f(x)dx=2\pi i\sum r_{\nu }.$$ ◻

This result may easily be generalized to the case where there is a denumerable infinity of points $z_{\nu }$ for which $f(z)$ is not regular, provided only that the $z_{\nu }$ have no point of accumulation. The summation is taken in the order of increasing $\left|z_{\nu }\right|$ .

Examples

The function $$f(z)=\frac{1}{a{z}^2+bz+c}$$ with , $a$ , $b$ , $c$ real and is regular for $\mathrm{Im}(z)\ge 0$ with the exception of the point $z_1=\frac{1}{2a}(-b+i\sqrt{4ac-b^2})$ , where the sign in front of the radical is to be the same as that of $a$ . Clearly, $f(z)$ satisfies the condition (3.23) . Hence by the preceding theorem ${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{a{x}^2+bx+c}dx$ is equal to $2\pi i$ times the residue $r_1$ of $f(z)$ at $z_1$ . But, setting $z=z_1+\rho e^{i\varphi }$ , we have We conclude that $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)dx=\frac{2\pi }{\sqrt{4ac-b^2}}.$$ 

The method used in the above theorem may be adapted to other improper integrals. Take, for example, $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin z}{z}dz$$ To evaluate this integral we consider $$\int \frac{e^{iz}}{z}dz.$$ We choose a semicircular domain of integration as before but we cut out a small piece of radius $\epsilon$ about the singular point at the origin.

Denote the integral over the large semicircle $S_R$ by $-I_R$ , over the small semicircle $S_{\epsilon }$ by $I_{\epsilon }$ , and the part of the integral over the real axis exclusive of the interval by $I_H$ . Then, by Cauchy’s Integral Theorem , $$I_H=I_R-I_{\epsilon }.$$ It is easily verified that $$I_H={\int }_{-R}^{-\epsilon }+{\int }_{\epsilon }^R\frac{e^{ix}}{x}dx=2i{\int }_{\epsilon }^R\frac{\sin x}{x}dx.$$ We have only to evaluate the integrals over $S_{\epsilon }$ and $S_R$ as $\epsilon \to 0$ and $R\to \mathrm{\infty }$ . $$I_R={\int }_{S_R}\frac{e^{iz}}{z}dz=-i{\int }_0^{\pi }{e}^{-R\sin \theta }{e}^{iR\cos \theta }d\theta .$$ Hence $$\left|I_R\right|\le {\int }_0^{\pi }{e}^{-R\sin \theta }d\theta =2{\int }_0^{\pi /2}{e}^{-R\sin \theta }d\theta .$$ By choosing $R$ sufficiently large we can make $\left|I_R\right|$ as small as we please. We consider the integral in two parts $$\left|I_R\right|\le 2({\int }_0^{\delta }+{\int }_{\delta }^{\pi /2}{e}^{-R\sin \theta }d\theta ).$$ We have $\left|e^{-R\sin \theta }\right|\le 1$ for $\theta \le \delta$ and by taking $R$ sufficiently large we may be sure that in the interval $\delta \le \theta \le \frac{\pi }{2}$ . Thus we have the estimate  

On the other hand we have $$I_{\epsilon }={\int }_{S_{\epsilon }}\frac{e^{iz}}{z}dz=-i{\int }_0^{\pi }{e}^{-\epsilon \sin \theta }{e}^{i\epsilon \cos \theta }d\theta .$$ Since the integrand is uniformly continuous in $\epsilon$ and $\theta$ we may take the limit as $\epsilon \to 0$ inside the integral sign. We conclude that $$\underset{\epsilon \to 0}{lim}{I}_{\epsilon }=-\pi i.$$ Hence we obtain $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sin x}{x}dx=-i\underset{\begin{array}{c}R\to \mathrm{\infty }\\ \epsilon \to 0\end{array}}{lim}{I}_H=\pi .$$ 

As a second example we propose to evaluate the Fresnel integrals ${\int }_0^{\mathrm{\infty }}\cos x^{2}dx$ and ${\int }_0^{\mathrm{\infty }}\sin x^{2}dx$ . To this end we introduce the function $e^{-z^2}$ and integrate around a circular sector of radius $R$ and central angle $\frac{\pi }{4}$ .

Setting $z=r{e}^{i\theta }$ we have We shall now show that the second integral tends to zero as $R\to \mathrm{\infty }$ . We have Now $\frac{d^2\sin \varphi }{d{\varphi }^2}=-\sin \varphi$ , whence it follows that the sine function is convex in the interval $0\le \varphi \le \pi /2$ and therefore lies above the straight line between the endpoints; thus $$\sin \varphi \ge \frac{2}{\pi }\varphi .$$ Thus $$\left|I_2\right|\le \frac{R}{2}{\int }_0^{\pi /2}{e}^{-\frac{2}{\pi }{R}^2\varphi }d\varphi =\frac{\pi }{4R}(1-e^{-R^2})$$ and this clearly goes to zero as $R$ tends to infinity. Consequently we have $${\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx-{\int }_0^{\mathrm{\infty }}{e}^{-i{r}^2}\frac{\sqrt{2}}{2}(1+i)dr=0.$$ Now ${\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx$ is a well known integral. Its value is $\sqrt{\pi }/2$ . ² Thus we have $${\int }_0^{\mathrm{\infty }}\cos r^{2}dr-i{\int }_0^{\mathrm{\infty }}\sin r^{2}dr=\frac{\sqrt{\pi }}{2}\frac{(1-i)}{\sqrt{2}},$$ whence $${\int }_0^{\mathrm{\infty }}\cos x^{2}dx={\int }_0^{\mathrm{\infty }}\sin x^{2}dx=\frac{1}{2}\sqrt{\frac{\pi }{2}}.$$ 

As a last example we propose to evaluate the following integral which plays a role in the study of the $\mathrm{\Gamma }$ -function: To this end we consider the closed path $C$ as in the adjoining figure and the integral $${\int }_C\frac{z^{\alpha -1}}{1+z}dz=2\pi i{r}_1$$ where $r_1$ is the residue of $\frac{z^{\alpha -1}}{1+z}$ at $z=-1$ .

显然 $r_1=(-1{)}^{\alpha -1}=e^{i\pi (\alpha -1)}$ 。因此 或 但当 $R\to \mathrm{\infty }$ 时，${\int }_0^{2\pi }\frac{R^{\alpha }{e}^{i\alpha \theta }}{1+R{e}^{i\theta }}d\theta \to 0$ ，因为 $$\left|{\int }_0^{2\pi }\frac{R^{\alpha }{e}^{i\alpha \theta }}{1+R{e}^{i\theta }}d\theta \right|\le \frac{2\pi R^{\alpha }}{R-1}.$$ 类似地，当 $\rho \to 0$ 时，${\int }_0^{2\pi }\frac{{\rho }^{\alpha }{e}^{i\alpha \theta }}{1+\rho e^{i\theta }}d\theta \to 0$ ，因为我们有一个奇异不等式 $$\left|{\int }_0^{2\pi }\frac{{\rho }^{\alpha }{e}^{i\alpha \theta }}{1+\rho e^{i\theta }}d\theta \right|\le \frac{2\pi {\rho }^{\alpha }}{1-\rho }.$$ 因此

练习

练习 3.5 。通过在如下矩形上对 ${\displaystyle {\int }_C{e}^{-z^2}dz}$ 积分，计算 ${\int }_0^{\mathrm{\infty }}{e}^{-x^2}\cos 2axdx$ 。