# Zero Staircase Integral Condition for Primitive

## Theorem

Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.

Let $z_0 \in D$.

Suppose that $\displaystyle \oint_C f \left({z}\right) \rd z = 0$ for all closed staircase contours $C$ in $D$.

Then $f$ has a primitive $F: D \to \C$ defined by:

$\displaystyle F \left({w}\right) = \int_{C_w} f \left({z}\right) \rd z$

where $C_w$ is any staircase contour in $D$ with start point $z_0$ and end point $w$.

## Proof

From Connected Domain is Connected by Staircase Contours, it follows that there exists a staircase contour $C_w$ in $D$ with start point $z_0$ and end point $w$.

If $C_w'$ is another staircase contour with the same endpoints as $C_w$, then $C_w' \cup \left({- C_w}\right)$ is a closed staircase contour.

Then the definition of $F$ is independent of the choice of contour, as:

 $\displaystyle \int_{C_w} f \left({z}\right) \rd z$ $=$ $\displaystyle \int_{C_w} f \left({z}\right) \rd z + \int_{C_w' \cup \left({- C_w}\right) } f \left({z}\right) \rd z$ by assumption $\displaystyle$ $=$ $\displaystyle \int_{C_w} f \left({z}\right) \rd z + \int_{C_w'} f \left({z}\right) \rd z + \int_{-C_w} f \left({z}\right) \rd z$ Contour Integral of Concatenation of Contours $\displaystyle$ $=$ $\displaystyle \int_{C_w} f \left({z}\right) \rd z + \int_{C_w'} f \left({z}\right) \rd z - \int_{C_w} f \left({z}\right) \rd z$ Contour Integral along Reversed Contour $\displaystyle$ $=$ $\displaystyle \int_{C_w'} f \left({z}\right) \rd z$

We now show that $F$ is the primitive of $f$.

Let $\epsilon \in \R_{>0}$.

By definition of continuity, there exists $r \in \R_{>0}$ such that the open ball $B_r \left({w}\right) \subseteq D$, and for all $z \in B_r \left({w}\right)$:

$\left\vert{f \left({z}\right) - f \left({w}\right) }\right\vert < \dfrac \epsilon 2$

Let $h = x+iy \in \C \setminus \left\{ {0}\right\}$ with $x, y \in \R$ such that $\left\vert{h}\right\vert < r$.

Let $\mathcal L$ be the staircase contour that goes in a horizontal line from $w$ to $w + x$, and continues in a vertical line from $w + x$ to $w + h$.

As $w + x, w + h \in B_r \left({w}\right)$, it follows from Open Ball is Convex Set that $\mathcal L$ is a contour in $B_r \left({w}\right)$.

Then $C_w \cup \mathcal L$ is a staircase contour from $z_0$ to $w + h$, so:

 $\displaystyle F \left({w + h}\right) - F \left({w}\right)$ $=$ $\displaystyle \int_{C_w \cup \mathcal L} f \left({z}\right) \rd z - \int_{C_w} f \left({z}\right) \rd z$ $\displaystyle$ $=$ $\displaystyle \int_{\mathcal L} f \left({z}\right) \rd z$ Contour Integral of Concatenation of Contours

From Derivative of Complex Polynomial, it follows that $\dfrac \rd {\rd z} f \left({w}\right) z = f \left({w}\right)$, so:

 $\displaystyle \int_{\mathcal L} f \left({w}\right) \rd z$ $=$ $\displaystyle f \left({w}\right) \left({w + h}\right) - f \left({w}\right) w$ Fundamental Theorem of Calculus for Contour Integrals $\displaystyle$ $=$ $\displaystyle h f \left({w}\right)$

We can now show that $F' \left({w}\right) = f \left({w}\right)$, as:

 $\displaystyle \left\vert{\dfrac{F \left({w + h}\right) - F \left({w}\right)} h - f \left({w}\right) }\right\vert$ $=$ $\displaystyle \left\vert{\dfrac 1 h \int_{\mathcal L} f \left({z}\right) \rd z - \dfrac 1 h h f \left({w}\right) }\right\vert$ by the above calculations $\displaystyle$ $=$ $\displaystyle \left\vert{\dfrac 1 h }\right\vert \left\vert{\int_{\mathcal L} \left({f \left({z}\right) - f \left({w}\right) }\right) \rd z }\right\vert$ Linear Combination of Contour Integrals $\displaystyle$ $<$ $\displaystyle \left\vert{\dfrac 1 h }\right\vert \dfrac \epsilon 2 L \left({\mathcal L}\right)$ Estimation Lemma, as $z \in B_r \left({w}\right)$ $\displaystyle$ $=$ $\displaystyle \dfrac{\left\vert{x}\right\vert + \left\vert{y}\right\vert }{\left\vert{h}\right\vert } \dfrac \epsilon 2$ the lengths of the line segments are $\left\vert{x}\right\vert$ and $\left\vert{y}\right\vert$ $\displaystyle$ $\le$ $\displaystyle \epsilon$ Modulus Larger than Real Part and Imaginary Part

When $h$ tends to $0$, we have $F' \left({w}\right) = f \left({w}\right)$ by definition of differentiability.

$\blacksquare$