Chapter 26Cauchy's Integral Formula

26.1 A Concrete Example of Green's Theorem — Verification on the Upper Half-Disk

Let us first recall Green's theorem, which we stated without derivation in Section 25.3. For a region $D$ in the plane and its boundary $C=\partial D$ (oriented counterclockwise, that is, in the direction in which $D$ is always on our left as we move along the boundary), if $f(x,y)$ has continuous partial derivatives on a slightly larger set containing $D$ and its boundary, then

holds; this is (part of) Green's theorem. The claim that a “one-dimensional integral over the boundary” equals a “two-dimensional integral over the interior” seems mysterious at first. In this section we choose one specific $f$ and $D$, compute both sides head-on, and see with our own eyes that they really agree. This is exactly the computation on p. 035 of the source notes.

26.1.1 Setting up the region and its boundary

Let $r\gt0$ be a constant, and set

That is, $D$ is the upper half of the disk of radius $r$ (the upper half-disk), and $C$ is its boundary—the arc part and the diameter part on the $x$-axis—joined together into one curve. The function we use for the check is

This is the square of the distance from the origin, and it is simple and easy to compute with on $D$. Incidentally, the source notes draw this region as a perspective view with three-dimensional depth, in which the $xy$-plane is seen obliquely from above. However, the problem itself concerns the plane (two dimensions), so in this book we redraw it, as in Figure 26.1, as a two-dimensional picture seen from directly above.

26.1.2 Computing both sides

We split the boundary $C$ into the arc part ($C_{\text{arc}}$) and the diameter part ($C_{\text{diam}}$), and compute $\int_C f\,dx=\int_{C_{\text{arc}}}f\,dx+\int_{C_{\text{diam}}}f\,dx$. First we parametrize the arc part by polar coordinates:

As $\theta$ increases from $0$ to $\pi$, the point $(x,y)$ traces exactly the arc in Figure 26.1 counterclockwise. Then

(Differentiating $x=r\cos\theta$ with respect to $\theta$ gives $-r\sin\theta$; since $r$ is a constant, it can be taken outside.) Also, on the arc we have $x^2+y^2=r^2$ (because the points lie on the circle), so $f(x,y)=r^2$ is constant on the arc. Hence

(We used $\int_0^\pi\sin\theta\,d\theta=[-\cos\theta]_0^\pi=-\cos\pi-(-\cos0)=-(-1)+1=2$.)

Next, the diameter part. Here we stay on the line $y=0$, and $x$ moves from $-r$ to $r$. Since $f(x,0)=x^2+0^2=x^2$,

Adding these together, the line integral over the boundary (the left-hand side) is

Next we compute the right-hand side, that is, the double integral over the interior. We have $\pdiff{f}{y}=\pdiff{}{y}(x^2+y^2)=2y$ (we regarded $x$ as a constant and differentiated with respect to $y$ alone). Expressing the upper half-disk in polar coordinates $x=r'\cos\theta,\,y=r'\sin\theta$ (with radial coordinate $r'$, $0\le r'\le r$, $0\le\theta\le\pi$), the area element is $dx\,dy=r'\,dr'\,d\theta$ (the Jacobian, from Chapter 7, Section 7.4), so

(We substituted $y=r'\sin\theta$.) The integrals in $\theta$ and $r'$ are independent of each other and can be separated (one factor for each variable), so

Therefore the right-hand side is $-\dfrac43r^3$, which agrees exactly with the value $-\dfrac43r^3$ of the left-hand side. Thus, for a specific function and region, we have seen with our own eyes that Green's theorem

really does hold.

26.1.3 Example: What happens on the lower half-disk?

26.2 Path Integrals of $\log z$ in a Multiply Connected Domain and Branch Points

In Section 25.2, we saw that computing $\displaystyle\oint_{|z|=r}\log z\,dz$ along the unit circle (the circle centered at the origin) does not give $0$. The reason was that $\log z$ is not holomorphic at $z=0$ (or rather, cannot even be defined at $z=0$). So what happens if we avoid the origin itself and integrate $\log z$ once around a closed path in a region from which only the neighborhood around the origin has been removed? In this section we follow this investigation on p. 036 of the source notes—the computation using a keyhole-shaped contour—through to the end. In fact, this computation hides a small “trap” concerning the choice of branch, and the source notes end with the remark “It doesn't come out to 0…!?” still written at the end. The goal of this section is to avoid this “trap,” keep the branch consistent from beginning to end, and arrive at the correct answer (the result is $0$).

26.2.1 The multivalued function $\log z$ and branches

Write a complex number $z\ne0$ in polar form $z=|z|(\cos\theta+i\sin\theta)=|z|e^{i\theta}$ (Chapter 21, Section 21.2). Then $\theta=\arg z$ (the argument of $z$) is ambiguous up to integer multiples of $2\pi$: both $\theta$ and $\theta+2\pi$ represent the same point $z$. The complex logarithm $\log z$ is defined as follows, so that the property $e^{\log z}=z$ of the real logarithm continues to hold, unchanged, for complex numbers as well:

This is the definition given in Chapter 23. Here $\log|z|$ is the ordinary natural logarithm of a real number. The problem is that there are infinitely many candidates $\theta_0,\theta_0+2\pi,\theta_0-2\pi,\dots$ for $\theta$, so $\log z$ has no single determined value for a single $z$. Such a function is called a multivalued function.

26.2.2 Setting up the keyhole contour

Fix positive real numbers $\varepsilon,R$ ($0\lt\varepsilon\lt R$) and consider the closed curve $C=C_1+C_2+C_3+C_4$ consisting of the following four curves (exactly the setup of the source notes).

Connecting these four curves in the order $C_1\to C_2\to C_3\to C_4$ gives a closed curve shaped just like a keyhole (Figure 26.2). The domain $D_{\varepsilon,R}=\{z\mid \varepsilon\le|z|\le R,\ \operatorname{Im}z\ge0\}\setminus(\text{neighborhood of the origin})$ enclosed by this closed curve does not contain the origin $z=0$ (the branch point of $\log z$). Therefore, if we use the branch that fixes the range of $\theta$ in this domain to $0\le\theta\le\pi$—that is, the branch with $\theta=0$ on the positive real axis ($C_4$) and $\theta=\pi$ on the negative real axis ($C_2$)—then $\log z=\log|z|+i\theta$ is continuous and holomorphic on the whole domain. Particular care is needed with the point that “on $C_2$ we use $\theta=\pi$” (we confirm this in detail in Section 26.2.4).

26.2.3 The integral along each of the four paths

At last, we compute $\displaystyle\int_C\log z\,dz=\int_{C_1}+\int_{C_2}+\int_{C_3}+\int_{C_4}$ carefully, one path at a time. On every path we use the branch fixed earlier (it starts from $\theta=0$ on $C_4$, passes through the upper half-plane, and reaches $\theta=\pi$ on $C_2$).

26.2.4 Computing $C_2$

In computing the last remaining piece, $C_2$ (the part on the negative real axis), we must take particular care in choosing the argument. What we must watch is the argument at a point $z=x$ on $C_2$ (where $x$ is a negative real number, $-R\le x\le-\varepsilon$). The branch fixed in Section 26.2.2 was this: “set $\theta=0$ on the positive real axis, and increase $\theta$ continuously as we move counterclockwise through the upper half-plane along $C_1$.” If we stay on this branch and arrive at $C_2$ (the negative real axis), then $\theta$ has increased to exactly $\pi$. That is, the expression

must be used. If we carelessly reuse the same expression $\log z=\log|x|$ (the branch with $\theta=0$) as on $C_4$, we drop this contribution $+i\pi$, so be careful.

With this in mind, let us compute the integral over $C_2$. We have $z=x$ (with $x:-R\to-\varepsilon$), $\log z=\log|x|+i\pi=\log(-x)+i\pi$ (since $x\lt0$, we have $|x|=-x$), and $dz=dx$.

The second term is easy: $\displaystyle i\pi\int_{-R}^{-\varepsilon}dx=i\pi\big((-\varepsilon)-(-R)\big)=i\pi(R-\varepsilon)$. For the first term, $\displaystyle\int_{-R}^{-\varepsilon}\log(-x)\,dx$, we must be careful in constructing the antiderivative. Integrating by parts with $u=\log(-x),\ dv=dx$, we have $v=x$ (an antiderivative of $dv=dx$ is $x$, not $-x$), and since $\dfrac{d}{dx}\log(-x)=\dfrac{1}{-x}\cdot(-1)=\dfrac1x$ (by the chain rule: differentiating the outer function $\log(\cdot)$ gives $\frac{1}{-x}$, differentiating the inner function $-x$ gives $-1$, and multiplying these gives $\frac{1}{-x}\cdot(-1)=\frac1x$), we get $du=\dfrac1x\,dx$. Therefore, we obtain

(note that an antiderivative of $dv=dx$ is $v=x$, not $v=-x$—if we carelessly took $v=-x$, the signs in what follows would be off). Using this antiderivative $x\log(-x)-x$, we get

(We expanded term by term, keeping in mind that $-(-x)=x$. When $x=-\varepsilon$, we have $-x=\varepsilon$, so $\log(-x)=\log\varepsilon$; when $x=-R$, we have $-x=R$, so $\log(-x)=\log R$.)

From the above, the integral over $C_2$ is

This is the result.

26.2.5 The sum of the four integrals — it comes out to a clean $0$

We list all four integrals found in Sections 26.2.3 and 26.2.4.

Collecting the $R\log R$ terms gives $-2R\log R+R\log R+R\log R=0$. Collecting the $\varepsilon\log\varepsilon$ terms gives $-\varepsilon\log\varepsilon+2\varepsilon\log\varepsilon-\varepsilon\log\varepsilon=0$. Collecting the $R$ terms (those containing no logarithm) gives $2R-R+R=2R$. “Hmm, this does not vanish…” you might think, but since $C_2$ has a $-R$ and $C_4$ has a $-R$, the total is $2R-R-R=0$ (compare the four expressions carefully and be sure to pick up every like term without missing any). Similarly, the $\varepsilon$ terms give $\varepsilon-2\varepsilon+\varepsilon=0$. Finally, the terms containing $i\pi$ give $-i\pi R+i\pi R-i\pi\varepsilon+i\pi\varepsilon=0$. All the terms cancel out neatly, giving

This is an equality that holds exactly for any $0\lt\varepsilon\lt R$ (even without taking the limits $\varepsilon\to0$ or $R\to\infty$). The function $\log z$ is holomorphic in the keyhole domain $D_{\varepsilon,R}$ (once the branch is fixed), and $D_{\varepsilon,R}$ can be regarded as a simply connected domain (Section 25.4: if we regard $C_2,C_4$ as two walls with opposite orientations along the “slit” of the keyhole, the whole becomes a single simply connected domain). Therefore, by Cauchy's integral theorem (Section 25.3), we should have $\oint_C\log z\,dz=0$, and Eq. \eqref{eq:26-logz-total} agrees perfectly with this expectation. We have now confirmed that, as long as the choice of branch and the sign of the antiderivative are handled correctly, there is no problem at all with $\log z$ itself or with Cauchy's integral theorem.

26.3 Cauchy's Integral Formula — Statement and Proof

Up to the previous section, we have become familiar with four tools—Green's theorem, Cauchy's integral theorem, the contour integral formula, and the principle of path deformation—by working through concrete examples. Now we put all of them to work to state and prove, in complete form, the main theorem of this chapter—the theorem that is written as “Theorem” on p. 039 of the source notes, where its proof breaks off at “Proof…”.

26.3.1 Statement of the theorem

26.3.2 Proof strategy

The idea of the proof consists of the following three steps.

1. First, using the principle of path deformation (Section 25.4), we replace the integral along the boundary $\partial D$ by an integral along a small circle $C_\rho:\ |\zeta-z|=\rho$ centered at the point $z$.
2. We split the numerator $f(\zeta)$ of the integrand into two parts: the constant $f(z)$ and the difference $f(\zeta)-f(z)$. By the contour integral formula (Section 25.5), the constant part gives exactly $2\pi i\,f(z)$.
3. For the remaining difference part, we use the continuity of $f$ and the $ML$ inequality (Section 25.1) to show that it can be made as small as we like by making $\rho$ small enough. Moreover, the value of this difference part itself must be a constant independent of $\rho$ (by the principle of path deformation), so by the reasoning that “the only constant that can be made arbitrarily small is $0$,” we conclude that the difference part is exactly $0$.

26.3.3 Proof

26.3.4 Examples: Applying the formula directly

26.4 Putting Cauchy's Integral Formula to Work — Examples and Corollaries

In this section, we put Cauchy's integral formula (Theorem 26.1) to further use and look at (1) how to handle the case of several singularities and (2) two important corollaries of the formula (the mean value property and the infinite differentiability of holomorphic functions). The source notes contain nothing beyond this point (the notes end at p. 040), but these results follow naturally as applications of Theorem 26.1 and are also a gateway to later developments in complex analysis (Taylor expansion and the residue theorem), so this book supplies them on its own.

26.4.1 The case of several singularities

When the denominator of the integrand is an expression of degree two or higher that becomes $0$ at several points, it suffices to use partial fraction decomposition (the same technique used for rational expressions in high school mathematics, extended to the complex numbers) and apply Cauchy's integral formula to each singularity one by one.

26.4.2 Corollary: A holomorphic function can be written as an average over a circle

In Theorem 26.1, let us take the boundary $\partial D$ to be the circle $C_\rho:\ \zeta=z+\rho e^{i\theta}$ ($0\le\theta\le2\pi$) itself, with center $z$ and radius $\rho$ (this is nothing other than Theorem 26.1 for the case where $D$ itself is the disk $|\,\zeta-z\,|\le\rho$). Substituting $d\zeta=i\rho e^{i\theta}d\theta$, we get

(The $\rho e^{i\theta}$ in the numerator and denominator cancel, and only $i$ remains.) Since $i/i=1$, we obtain

This means that “the value of a holomorphic function at the center $z$ equals the average of its values on any circle centered at that point,” and it is called the mean value property. For example, for $f(z)=e^z$, $z=1+i$, $\rho=0.7$, averaging the values $f(z+\rho e^{i\theta})$ on the circle from $\theta=0$ to $2\pi$ returns exactly $f(1+i)=e^{1+i}$, which can also be confirmed numerically (verify/univmath/ch26_cauchy_examples.py).

26.4.3 Corollary: A holomorphic function can be differentiated any number of times

We introduce another surprising consequence of Theorem 26.1, without going into the details of the proof. Let us formally differentiate the right-hand side of Cauchy's integral formula

with respect to $z$. Differentiating $\dfrac{1}{\zeta-z}$ inside the integral sign with respect to $z$ gives $\dfrac{1}{(\zeta-z)^2}$ ($\dfrac{d}{dz}(\zeta-z)^{-1}=-(\zeta-z)^{-2}\cdot(-1)=(\zeta-z)^{-2}$). It can be shown that the order of integration and differentiation may be interchanged (we omit the proof in this book, but it follows from the fact that the integrand is continuous in $\zeta$ and smooth in $z$), so we have

Repeating the same operation, we find that in general the following formula holds for the $n$th derivative as well:

($n!$ is the factorial of $n$; for $n=0$ we have $0!=1$, and we return to Eq. \eqref{eq:26-cif}.) For example, for $f(z)=z^4$, $z=1$, $\rho=0.5$, $n=2$, computing the right-hand side of Eq. \eqref{eq:26-higher-deriv} by numerical integration gives $12$, which agrees with the value $f''(1)=12$ computed directly from $f''(z)=12z^2$ (verify/univmath/ch26_cauchy_examples.py).

26.5 Summary and Exercises

26.5.1 Summary

• We checked Green's theorem $\displaystyle\int_C f\,dx=-\iint_D\pdiff{f}{y}\,dx\,dy$ for $f(x,y)=x^2+y^2$ on the upper half-disk and on the lower half-disk, and confirmed that reversing the orientation of the boundary (the positive direction is the one in which the region lies on your left) reverses the sign.
• A multivalued function such as $\log z$ must be considered in a region that avoids the branch point ($z=0$ for $\log z$), and one branch, with the range of $\theta$ (the argument) fixed to a single choice, must be used consistently along the entire path. If the correct branch is used, the integral of $\log z$ along the keyhole contour $C_1+C_2+C_3+C_4$ is exactly $0$ regardless of the values of $\varepsilon,R$ (consistent with Cauchy's integral theorem).
• Cauchy's integral formula $\displaystyle f(z)=\frac{1}{2\pi i}\oint_{\partial D}\frac{f(\zeta)}{\zeta-z}\,d\zeta$ (Eq. \eqref{eq:26-cif})—the value at an interior point $z$ of a function $f$ holomorphic on $D$ is determined by the values on the boundary $\partial D$ alone.
• Outline of the proof: ① deform the path of integration to a small circle $C_\rho$ by the principle of path deformation; ② split the numerator into $f(z)$ (a constant) and $f(\zeta)-f(z)$, where the former gives $2\pi i\,f(z)$ by the contour integral formula; ③ for the latter, continuity and the $ML$ inequality show that, although it is a constant independent of $\rho$, it can be made arbitrarily small, and hence we conclude that it is $0$.
• A tip for computation: $\displaystyle\oint\frac{f(z)}{z-\alpha}\,dz=2\pi i\,f(\alpha)$ (when $\alpha$ lies in the interior and $f$ is holomorphic near it). When there are several singularities, decompose into partial fractions and apply the formula at each singularity.
• Corollaries: the mean value property $\displaystyle f(z)=\frac{1}{2\pi}\int_0^{2\pi}f(z+\rho e^{i\theta})\,d\theta$ (Eq. \eqref{eq:26-mean-value}), and the infinite differentiability of holomorphic functions $\displaystyle f^{(n)}(z)=\frac{n!}{2\pi i}\oint_{\partial D}\frac{f(\zeta)}{(\zeta-z)^{n+1}}\,d\zeta$ (Eq. \eqref{eq:26-higher-deriv}).

26.5.2 Exercises

26.5.3 References