Contours, primitives and the Cauchy–Goursat theorem

1. An integral records both a path and its orientation

A piecewise $C^1$ curve is a continuous map $\gamma:[a,b]\to\mathbb C$ that is continuously differentiable on finitely many subintervals, with one-sided derivatives at the joins. It is closed when $\gamma(a)=\gamma(b)$. Its length is $\ell(\gamma)=\int_a^b|\gamma'(t)|\,dt$. A contour integral of continuous $f$ along it is

$$\int_\gamma f(z)\,dz=\int_a^b f(\gamma(t))\gamma'(t)\,dt,$$

with real and imaginary parts integrated separately. The differential $dz$ includes the direction of motion; replacing it by $|dz|$ would define a different integral.

A piecewise $C^1$ increasing change of parameter preserving endpoints leaves the integral unchanged by substitution. Reversal changes its sign. Concatenating two compatible curves adds their integrals. These statements also allow zero-speed pieces. They do not require the curve to be injective.

Estimation lemma. If $|f|\le M$ on the curve, then

$$\left|\int_\gamma f(z)\,dz\right|\le M\ell(\gamma).$$

Indeed, the modulus of a real-parameter integral is at most the integral of the modulus, and $|f(\gamma)\gamma'|\le M|\gamma'|$. This simple inequality is the main tool for showing that auxiliary arcs contribute nothing in a limit.

Worked example. On the positively oriented circle $z=a+re^{it}$, $0\le t\le2\pi$, with $r>0$,

$$\int_\gamma(z-a)^n\,dz=\begin{cases}2\pi i,&n=-1,\\0,&n\in\mathbb Z,\ n\ne-1.\end{cases}$$

For $n=-1$ the integrand times $dz$ is $i\,dt$. Otherwise it is $ir^{n+1}e^{i(n+1)t}\,dt$, whose integral vanishes. The answer for $n=-1$ is independent of radius, but changes sign if orientation is reversed.

2. Primitives turn path integrals into endpoint differences

A primitive of $f$ on an open set $U$ is a holomorphic function $F$ there with $F'=f$. The real chain rule applied on each smooth piece gives

$$\int_\gamma f(z)\,dz=F(\gamma(b))-F(\gamma(a))$$

for every contour in $U$. Hence a function with a primitive has zero integral around every closed contour in $U$.

Conversely, on a domain $U$, suppose continuous $f$ has zero integral around every closed contour. Open connected subsets of the plane are polygonally connected. Fix $a\in U$ and define $F(z)$ as the integral from $a$ to $z$ along any polygonal path in $U$. Closed-path cancellation makes it path independent. For sufficiently small $h$, the final straight segment from $z$ to $z+h$ stays in $U$, and

$$\frac{F(z+h)-F(z)}h=\int_0^1 f(z+th)\,dt\longrightarrow f(z)$$

by continuity. Thus $F'=f$. The same construction works if polygonal closed paths alone have zero integrals; the resulting primitive then handles all contours.

Warning. Holomorphic does not automatically mean globally possessing a primitive on any domain. The function $1/z$ is holomorphic on $\mathbb C\setminus\{0\}$ but its integral around the unit circle is $2\pi i$. On the slit plane of C01, however, $\operatorname{Log}z$ is a primitive of $1/z$. The domain is mathematical data.

3. Goursat's triangle argument: no continuity of the derivative assumed

Theorem. If $f$ is holomorphic on an open set containing a closed triangle $T$, then its integral around the positively oriented boundary of $T$ is zero.

Proof. Subdivide $T$ by joining side midpoints into four triangles. Interior edge integrals cancel, so the outer integral $I$ is the sum of four child integrals. At least one child has integral modulus at least $|I|/4$. Repeat inside such a child, obtaining nested closed triangles $T_n$, diameter $D/2^n$, perimeter $P/2^n$, and

$$|I|\le4^n\left|\int_{\partial T_n}f\,dz\right|.$$

Compactness and shrinking diameters give one common point $z_0$. Differentiability at $z_0$ says

$$f(z)=f(z_0)+f'(z_0)(z-z_0)+(z-z_0)\varepsilon(z),\qquad\varepsilon(z)\to0.$$

The constant and linear terms have primitives, so contribute zero on the triangle boundary. For any $\eta>0$, eventually $|\varepsilon|\le\eta$ throughout $T_n$. Since $z_0\in T_n$, the estimation lemma gives

$$\left|\int_{\partial T_n}f\,dz\right|\le\eta(D/2^n)(P/2^n).$$

Thus $|I|\le\eta DP$. As $\eta$ is arbitrary, $I=0$. No continuity of $f'$ was used. ∎

Corollary: a primitive on each disk. If $f$ is holomorphic on a disk $D$, fix its centre $a$ and set $F(z)=\int_{[a,z]}f$. For $z,z+h\in D$ sufficiently close, the triangle with vertices $a,z,z+h$ lies in $D$. The triangle theorem gives $F(z+h)-F(z)=\int_{[z,z+h]}f$. The limit calculation of §2 yields $F'=f$. Consequently all closed-contour integrals in the disk vanish. The same argument works on any open convex set.

Polygonal boundary version. Suppose a compact polygonal region, possibly with finitely many polygonal holes, and a neighbourhood of it lie in the holomorphy set of $f$. Triangulate the region into finitely many triangles. Summing the triangle theorem cancels every interior edge. The outer boundary is counterclockwise and hole boundaries clockwise, so their oriented total integral is zero. A region containing a hole is not the same as the region with that hole filled in.

4. Homotopy and what “no hole” actually means

A fixed-endpoint homotopy between paths $\gamma_0,\gamma_1$ is a continuous map $H:[0,1]^2\to U$ with $H(s,0)=\gamma_0(s)$, $H(s,1)=\gamma_1(s)$, and $H(0,t),H(1,t)$ constant in $t$. It deforms one path into the other without leaving $U$. A domain is simply connected if every closed path can be contracted to its basepoint by such a homotopy.

Homotopy invariance. If $f$ is holomorphic on $U$, and two piecewise $C^1$ paths in $U$ are fixed-endpoint homotopic there, their integrals agree.

Proof using only local primitives. The compact image of $H$ is covered by finitely many disks on which $f$ has primitives. By the Lebesgue-number property and uniform continuity, subdivide the parameter square into a fine rectangular grid so that the image of each closed cell lies in one such disk. Join the images of adjacent grid vertices by straight segments. For each cell these segments lie in its disk because disks are convex. Its polygonal boundary integral is zero by its primitive. Sum over cells: internal segments cancel with their reversals. On the bottom and top edges, each original path subarc and its replacing straight segment lie in the same disk and have the same endpoints, hence the same integral by the local primitive. The side segments vanish because endpoints were fixed. What remains is the bottom integral minus the top integral. This proves the claim even if the homotopy's intermediate paths are not differentiable. ∎

Cauchy theorem on a simply connected domain. Every holomorphic function on such a domain has zero integral on every closed contour and hence has a primitive. Contract the contour and apply homotopy invariance; then use §2. Simple connectivity is sufficient, not necessary for a particular function to have a primitive. The function $1$ has primitive $z$ even on a punctured domain.

The often-used phrase “inside the curve” requires care for self-intersecting curves or domains with holes. For this chapter, the safe usable versions are a disk, a convex domain, a simply connected domain with all curves contained in it, or an explicitly oriented region whose closure has a holomorphic neighbourhood.

5. Practical contour choices

Before using Cauchy theorem, list singular or undefined points and identify the actual domain. Then decide whether an endpoint primitive is available, whether a deformation stays in the domain, or whether a direct parameterization is shorter.

Worked example: two routes, one forbidden deformation. The upper and lower semicircles from $1$ to $-1$, both traversed in that direction, give

$$\int_{\text{upper}}\frac{dz}{z}=i\pi,\qquad \int_{\text{lower}}\frac{dz}{z}=-i\pi.$$

Parameterize with angles $0\to\pi$ and $0\to-\pi$. Their difference is $2\pi i$. Therefore the two paths cannot be fixed-endpoint homotopic in the punctured plane. A drawing that slides one through zero is not a permitted homotopy.

For any contour from $a$ to $b$, $\int 2z\,dz=b^2-a^2$ because $z^2$ is a global primitive. For $\int\bar z\,dz$, by contrast, Cauchy theorem is unavailable: $\bar z$ is nowhere holomorphic. On the unit circle $\bar z=1/z$, so its integral is $2\pi i$, a nonzero result obtained without contradiction.

6. Graded exercises

P01 — Foundation. Integrate $z^2$ along the line segment from $0$ to $1+i$, directly and using a primitive.

P02 — Foundation. Evaluate $\int_{|z-2|=3}(z-2)^{-1}\,dz$ for clockwise traversal.

P03 — Core. Evaluate $\int_\gamma\bar z\,dz$ for the straight segment from zero to $1+i$, and for the broken path $0\to1\to1+i$.

P04 — Core. Show that $1/z$ has a primitive on the right half-plane, but not on the punctured plane.

P05 — Core. If $|f(z)|\le5$ on a circular arc of radius $2$ subtending angle $\pi/3$, bound the modulus of its integral.

P06 — Proof. If continuous functions $f_n$ converge uniformly to $f$ on a contour of finite length, show that their integrals converge to the integral of $f$.

P07 — Diagnosis. A student says a holomorphic function integrates to zero around any closed contour because the curve has a compact interior. Explain the missing assumption using a concrete example.

P08 — Extension. Let $U$ be a domain and continuous $f$ have a path-independent contour integral. Prove that any two primitives differ by a constant, and explain what changes if $U$ is disconnected.

7. Full solutions

P01. Parameterize $z=(1+i)t$, $0\le t\le1$. The integral is $(1+i)^3\int_0^1t^2dt=(-2+2i)/3$. The primitive $z^3/3$ gives the same endpoint difference.

P02. Counterclockwise traversal gives $2\pi i$ by §1. Reversal gives $-2\pi i$. The radius and location do not change this result.

P03. On the segment $z=(1+i)t$, $\bar z\,dz=(1-i)t(1+i)dt=2t\,dt$, giving $1$. On $0\to1$, the integral is $1/2$. On $1\to1+i$, put $z=1+it$, so $\bar z\,dz=(1-it)i\,dt=(t+i)dt$, giving $1/2+i$. Total: $1+i$. Path dependence proves there is no primitive on any domain containing these paths and their comparison loop.

P04. The principal logarithm is holomorphic on the right half-plane and has derivative $1/z$. If a primitive existed on the punctured plane, its integral on the unit circle would be zero, whereas direct integration gives $2\pi i$.

P05. The arc length is $2\pi/3$. Thus the modulus is at most $10\pi/3$. This is a bound, not an assertion that equality holds.

P06. Linearity and the estimation lemma give $|\int_\gamma(f_n-f)dz|\le\ell(\gamma)\sup_\gamma|f_n-f|\to0$. Pointwise convergence alone does not give this estimate with a vanishing supremum.

P07. Holomorphy must hold on a neighbourhood of the region filled by the contour, or another valid topological version must apply. For $1/z$ on the punctured plane, the unit circle is a closed contour but zero is missing from its interior holomorphy set. The integral is $2\pi i$, despite the closed disk being compact. Compactness is not holomorphy.

P08. The construction in §2 gives a primitive. If $F,G$ are primitives, $H=F-G$ has complex derivative zero, hence real derivative zero. Along a polygonal path in $U$, the real chain rule gives a constant value on every segment, so $H$ has the same value at any two points. On a disconnected open set, the difference is constant on each connected component, with no requirement that component constants agree.

8. Summary

Contours are oriented parameterized objects. Primitives imply endpoint evaluation; zero closed integrals imply primitives for continuous functions on domains. Goursat's subdivision establishes local Cauchy theory directly from differentiability. Local primitives then yield homotopy invariance and the simply connected version of Cauchy theorem. All domain and orientation hypotheses must survive a contour deformation. The next chapter uses these results to recover values and derivatives from boundary integrals.