The following is a weaker predecessor of Cauchy's famous integral theorem.

Proposition 1.If $f$ is holomorphic on a (closed) rectangle $R$, then

Proof.Since $f$ is holomorphic on $R$, it is defined and differentiable in some region $\Omega\supset R$. Given a rectangle $S\subset\Omega$, we will use the notation $\eta(S)$ to denote the integral of $f$ around the boundary of that rectangle, counterclockwise: By bisecting the sides of $R$, we may divide it into $4$ rectangles $R^{(1)},R^{(2)}, R^{(3)}, R^{(4)}$, so that we have Now, by the triangle inequality, we have that at least one of the $\eta(R^{(i)})$ must have magnitude at least $|\eta(R)|/4$. Let us set $R_0=R$, and set $R_1$ equal to this new rectangle. We may now repeat the process by bisecting the sides of $R_1$ and choosing $R_2$ as one of the $4$ rectangles formed in this way, and so on, so that we have a decreasing chain of nested rectangles such that $|\eta(R_n)|\leq 4|\eta(R_{n+1})|$ for each $n\in\mathbb N$. Now, because $f$ is differentiable, for any $z_0,z\in R$, we have that as $z\to z_0$. In fact, since $R$ is compact, for any $\epsilon > 0$, there exists a universal constant $\delta>0$ independent of $z,z_0$ such that whenever $|z-z_0|<\delta$. This means that where $E_\epsilon(z)$ is an error term whose magnitude is less than $\epsilon |z-z_0|$ everywhere, and $\gamma$ is a piecewise differentiable closed curve which lies within the $\delta$-ball surrounding some $z_0$. We know, however, that the first integral on the RHS vanishes because it is possible to find an antiderivative for the integrand (which is a polynomial), hence we have and therefore this integral is less than or equal to $\epsilon$ times the length of $\gamma$.Let us define $\rho$ to be the unique point lying in the intersection of all of the rectangles $R_n$. If we fix some $\epsilon > 0$ and let $\delta$ be the corresponding value described above, it is always possible to choose $n$ sufficiently large such that $R_n$ lies within a $\delta$-ball of $\rho$. Then, if $L$ is the length of the perimeter of the original rectangle, we have that the perimeter of $R_n$ is $L/2^n$, and therefore Inside of $R_n$, no two points are separated by a distance greater than $1/2$ the perimeter of the rectangle, hence we have that Finally, since $|\eta(R)|\leq 4^n|\eta(R_n)|$, we have that but since $\epsilon>0$ was arbitrary, we must have that $\eta(R)=0$, as claimed. $\blacksquare$

In fact, we can weaken the hypothesis of this theorem by requiring only that $f(z)$ be holomorphic except at finitely many points $\zeta_i$, so long as for each of those points. By subdividing the rectangle into smaller rectangles each of which contains at most one of the $\zeta_i$, it suffices to consider only the case in which there is a single "trouble point" $\zeta$. For a rectangle containing one point $\zeta$ at which $f$ fails to be holomorphic but does satisfy we may pick an arbitrary $\epsilon > 0$ and then subdivide this rectangle further until $\zeta$ is contained within a rectangle that is sufficiently small for to hold for all $z$ inside the rectangle. If we choose the rectangle to be a square, we will then be able to show that implying that the integral can be bounded in magnitude from above by arbitrary positive constants, and therefore it must vanish. Hence, we have the following generalization:

Proposition 2.If $f(z)$ is holomorphic on the rectangle $R$ for all but finitely many points $\zeta_i$, and if for each of these points $\zeta_i$, then we have