Contour Integration

By [L1], $x$ and $y$ have bounded variation. The four real functions $u\circ \gamma$ and $v\circ \gamma$ are continuous, so [L3] gives all four Stieltjes integrals in the complex definition.

The function $\mathrm{\mid }f\circ \gamma \mathrm{\mid }$ is continuous, and [L2] makes $s_{\gamma }$ a bounded-variation integrator, so [L3] gives the absolute integral.

Thus both definitions are well-defined. On a singleton or constant path the relevant integrators are constant and every integral is $0$.

On a nondegenerate smooth piece $[t_j,t_{j+1}]$ the integrands $u(\gamma (t))$ and $v(\gamma (t))$ are continuous, being composites of the continuous $f$ with the continuous $\gamma$, so [L5] makes each of the four component integrands Riemann integrable; the integrators $x_j,y_j$ are $C^1$ on the piece, hence continuous with continuously extending derivative. The hypotheses of [L1] therefore hold, and applying [L1] to the four component Stieltjes integrals and recombining gives $\int f(\gamma (t))(x_j^{\mathrm{\prime }}(t)+i{y}_j^{\mathrm{\prime }}(t))dt=\int f(\gamma (t)){\gamma }_j^{\mathrm{\prime }}(t)dt$.

On the same piece the arc-length integrator is $s_{{\gamma }_j}$, which by [L6] is differentiable with $s_{{\gamma }_j}^{\mathrm{\prime }}(t)=\mathrm{\mid }{\gamma }_j^{\mathrm{\prime }}(t)\mathrm{\mid }$, continuous because ${\gamma }_j$ is $C^1$; and $\mathrm{\mid }f(\gamma (t))\mathrm{\mid }$ is continuous, hence Riemann integrable by [L5]. So [L1] applies with $\alpha =s_{{\gamma }_j}$ and yields $\int \mathrm{\mid }f(\gamma )\mathrm{\mid }d{s}_{{\gamma }_j}=\int \mathrm{\mid }f(\gamma (t))\mathrm{\mid }\mathrm{\mid }{\gamma }_j^{\mathrm{\prime }}(t)\mathrm{\mid }dt$; summing over pieces and using [L3] to identify the total arc length gives the absolute-integral formula, which is the scalar line integral in [L2].

The real and imaginary parts in step 1.1 are exactly the vector line integrals of $(u,-v)$ and $(v,u)$ from [L2]. This uses the published real construction in a numbered step, with its piecewise-$C^1$ hypothesis unchanged.

Summing the identities over the pieces proves both displays. No equality of one-sided derivatives at corners is needed, and zero-speed pieces contribute $0$.

Expand the real and imaginary parts of $\alpha f+\beta g$ and apply [L2] to each component integral in [L1].

Recombining the real and imaginary identities gives the displayed complex linearity. Zero scalars and singleton paths are included.

By [L1] the reversal is ${\gamma }^{-}(t)=\gamma (a+b-t)$. The map $t\mapsto a+b-t$ sends a partition $a=t_0<\cdots <t_n=b$ to the partition with points $a+b-t_n<\cdots <a+b-t_0$ and sends tags to tags, so it is a mesh-preserving bijection between tagged partitions of $[a,b]$ and tagged partitions of $[a,b]$. Under it each coordinate increment for ${\gamma }^{-}$ is the negative of the matching increment for $\gamma$, since the two subinterval endpoints are exchanged; each arc-length increment is instead unchanged, because by [L1] length is unaffected by monotone reparametrization. Hence every Riemann–Stieltjes sum in the coordinate integrators of [L4] for ${\gamma }^{-}$ is the negative of the corresponding sum for $\gamma$, and every sum in the arc-length integrator is equal to it. Passing to the limit over refinements yields the two reversal identities.

By [L1] the concatenation $\alpha \ast \beta$ is given on $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$ by the two standard affine pieces, each a strictly increasing continuous bijection onto its factor's parameter interval. Apply [L5] to each component Stieltjes integral in [L4] to transport it to the factor's own interval. The integrators are coordinates and arc length of a rectifiable path, hence of bounded variation, and the integrand is continuous, so the join $\frac{1}{2}$ satisfies the additivity hypotheses of [L2]; splitting there and recombining gives both additive identities.

On piecewise-$C^1$ contours these conclusions agree exactly with [L3], whose hypotheses and orientation distinction are preserved. Constant pieces contribute $0$.

Assume first that both intervals are nondegenerate. Apply [L1] to each of the four component integrals in [L4]; their recombination is unchanged.

If both intervals are singletons, both complex and absolute integrals are $0$ by definition.

For the absolute integral in [L4], [L2] identifies the reparametrized arc-length integrator, and [L1] gives the same Stieltjes integral.

The cases exhaust the Statement and prove both identities. On piecewise-$C^1$ contours this is exactly the increasing half of [L3]; decreasing reparametrization is excluded and instead changes the complex integral's sign.

For a tagged partition $P$, the complex polygonal sum $S_P=\sum f(\gamma ({\xi }_j))(\gamma (t_{j+1})-\gamma (t_j))$ satisfies $\mathrm{\mid }{S}_P\mathrm{\mid }\le \sum \mathrm{\mid }f(\gamma ({\xi }_j))\mathrm{\mid }\mathrm{\mid }\gamma (t_{j+1})-\gamma (t_j)\mathrm{\mid }$ by [L2].

By [L3], each chord in step 1.1 is at most $s_{\gamma }(t_{j+1})-s_{\gamma }(t_j)$. The resulting tagged arc-length sums converge to the existing absolute line integral from [L1].

Letting the mesh tend to $0$, [L1] identifies the limit of $S_P$ with the complex line integral and step 2.1 identifies the majorant limit with the absolute integral, proving the inequality with sharp constant $1$. The same argument includes a constant contour: every chord and arc-length increment is $0$, so both sides vanish.

Apply [L3] to $\mathrm{\mid }f\circ \gamma \mathrm{\mid }$ and the nondecreasing $s_{\gamma }$; by [L2], ${\int }_{\gamma }\mathrm{\mid }f\mathrm{\mid }\mathrm{\mid }dz\mathrm{\mid }\le ML(\gamma )$.

Combine step 1.1 with [L1].

This agrees with the published piecewise-$C^1$ estimate [L4] on its exact domain and extends it to rectifiable contours. The cases $M=0$ and $L(\gamma )=0$ give zero directly.

With $f\equiv 1$, [L1] becomes ${\int }_a^{b}1d{s}_{\gamma }=s_{\gamma }(b)-s_{\gamma }(a)$.

Substitute [L2] to obtain $L(\gamma )$. Singleton and constant paths have both sides $0$.

Fix $\epsilon >0$. For a point $p$ of the trace let $S_p$ be the set of $r\in (0,1]$ such that $B(p,r)$ lies in the open domain and $\mathrm{\mid }f(u)-f(p)\mathrm{\mid }<\epsilon \mathrm{/}2$ for every $u\in B(p,r)$. Continuity of $f$ at $p$ and openness of the domain make $S_p$ nonempty, and $S_p$ is downward closed in $(0,1]$, so $r_p:=\frac{1}{2}\sup S_p$ is a positive real belonging to $S_p$; the assignment $p\mapsto r_p$ is defined outright, not selected, so no choice principle is used. The balls $B(p,r_p)$ cover the compact trace by [L6], so [L7] supplies $\delta >0$ such that any two trace points at distance below $\delta$ lie in a single $B(p,r_p)$. That ball is convex, so the segment joining them stays inside it, and $\mathrm{\mid }f(u)-f(z)\mathrm{\mid }<\epsilon$ all along that segment.

For trace points $z,w$ as in step 1.1, apply [L4] componentwise to the straight segment and then [L8] to $f-f(z)$. This gives $F(w)-F(z)=f(z)(w-z)+r(z,w)$ with $\mathrm{\mid }r(z,w)\mathrm{\mid }\le \epsilon \mathrm{\mid }w-z\mathrm{\mid }$.

By [L6] and [L10], choose $\eta >0$ so that every partition with mesh below $\eta$ has consecutive trace points within $\delta$. For every such partition, sum the identity of step 2.1: the left side telescopes to $F(\gamma (b))-F(\gamma (a))$, and by [L2] and repeated use of [L9] the total remainder is at most $\epsilon L(\gamma )$. As the mesh tends to $0$, [L5] identifies the limit of the main sums with the complex integral, so $\mathrm{\mid }F(\gamma (b))-F(\gamma (a))-{\int }_{\gamma }fdz\mathrm{\mid }\le \epsilon L(\gamma )$.

Letting $\epsilon \downarrow 0$ proves the formula for every rectifiable contour. The argument also covers constant paths, for which [L2] gives length $0$ and both sides vanish.

Apply [L1] and then [L2]: the endpoint increment is $F(\gamma (a))-F(\gamma (a))=0$.

Thus the integral vanishes, including for a constant closed contour.

By [L1], $F(z)=cz$ is a primitive of the constant function $c$.

The constant function $c$ is continuous on all of $\mathbb{C}$, and $F^{\mathrm{\prime }}=c$ is that same constant, so the hypotheses of [L2] hold on any open set containing the trace. Apply [L2] and simplify $F(\gamma (b))-F(\gamma (a))=c\gamma (b)-c\gamma (a)$. The cases $c=0$ and a constant path are included.

If $f$ has a primitive $F$ on $U$, then $F^{\mathrm{\prime }}=f$ is continuous by the Given, so [L5] applies to every rectifiable contour in $U$ and gives endpoint independence; endpoint independence makes every closed-contour integral zero because the constant contour with the same endpoint has integral $0$.

Assume every closed-contour integral is zero. By [L1] and [L2], fix a basepoint $z_0\in U$; for each $z\in U$ at least one polygonal path in $U$ runs from $z_0$ to $z$. Any two such paths carry the same integral: concatenating one with the reversal of the other is a closed contour, whose integral is by [L3] the difference of the two, and the closed-loop hypothesis makes that difference $0$.

So for each $z\in U$ there is a unique complex number shared by the integrals of $f$ along all polygonal paths in $U$ from $z_0$ to $z$; define $F(z)$ to be that number. This specifies $F$ uniquely from the data of step 1.2, with no path selected and no choice principle used.

For sufficiently small $h$, the segment from $z$ to $z+h$ lies in $U$. By [L3] and [L4], $F(z+h)-F(z)={\int }_0^{1}f(z+th)hdt$, so division by $h\mathrm{\ne }0$ gives an average tending to $f(z)$ by continuity. Thus $F^{\mathrm{\prime }}(z)=f(z)$.

The construction proves that zero closed integrals imply a primitive, completing both directions of the equivalence. On piecewise-$C^1$ contours the componentwise statement agrees with the real vector-field equivalence [L6], whose open and path-connected hypotheses hold by [L1] and [L2].

If $L(\gamma )=0$, [L2] gives ${\int }_{\gamma }(f_n-f)dz=0$ for every $n$.

If $L(\gamma )>0$, given $\epsilon >0$ choose $N$ such that $\mathrm{\mid }{f}_n-f\mathrm{\mid }<\epsilon \mathrm{/}L(\gamma )$ on the trace for $n\ge N$.

By [L1] all integrals exist, and [L2] applied to $f_n-f$ gives $\mathrm{\mid }{\int }_{\gamma }{f}_{n}dz-{\int }_{\gamma }fdz\mathrm{\mid }<\epsilon$ for $n\ge N$.

The two length cases are exhaustive and prove convergence without ever dividing by zero.

Since $r>0$, $\gamma (t)-a\mathrm{\ne }0$, so all integer powers in [L2] are defined. By [L1] and [L3], the integrand becomes $i{r}^{m+1}\exp (i(m+1)t)$.

If $m=-1$, the expression in step 1.1 is the constant $i$, whose integral from $0$ to $2\pi$ is $2\pi i$.

If $m\mathrm{\ne }-1$, an antiderivative is $r^{m+1}\exp (i(m+1)t)\mathrm{/}(m+1)$ by [L3]. Apply the real theorem [L5] to its two components using [L6]; the complex integral is the endpoint difference $r^{m+1}(\exp (2\pi i(m+1))-1)\mathrm{/}(m+1)$. Write $k:=m+1$, a nonzero integer. For $k>0$ the addition law in [L3] gives $\exp (2\pi ik)=\exp (i\pi {)}^{2k}$, and $\exp (i\pi )=-1$ by [L4], so $\exp (2\pi ik)=(-1{)}^{2k}=1$; for $k<0$ the addition law gives $\exp (2\pi ik)\exp (-2\pi ik)=\exp (0)=1$ with $\exp (-2\pi ik)=1$ by the previous case, so again $\exp (2\pi ik)=1$. The endpoint difference is therefore $0$.

The integer cases $m=-1$ and $m\mathrm{\ne }-1$ are exhaustive, proving the formula.

By [L3], $2\pi \mathrm{\ne }0$ in the embedded real field; the unique complex-coordinate form in [L2] then gives $2\pi i\mathrm{\ne }0$, so division is licensed.

Substitute [L1] and divide to obtain $1$. The value is independent of the positive radius, and no winding-number or Cauchy theorem is used.

Apply [L6] to $P=1$ and $Q(z)=z$: it makes $U$ open and $1\mathrm{/}z$ holomorphic there, hence continuous by [L7]. The point $1$ lies in $U$, and [L2] makes $U$ connected, so it is a domain by [L1].

Suppose, contrary to the desired refutation, that $1\mathrm{/}z$ had a primitive on $U$. Then [L4] would make its integral around the unit circle zero.

But [L3] gives that integral as $2\pi i\mathrm{\ne }0$, a contradiction. Hence the false claim fails, and [L5] gives the corrected zero-closed-contour criterion.