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  "body": "\n## Proof\n\nWe integrate $f$ along a horizontal and then a vertical segment, and check the Cauchy–Riemann equations.\n\n1. Since $f$ is [[holomorphic|0DG0Z72CQD]], it is [[complex differentiable|0D1C9Y1EXA]] at each point, hence continuous: $f(z_1 + h) - f(z_1) = h \\cdot \\frac{f(z_1 + h) - f(z_1)}{h} \\to 0$.\n2. Write $c = x_c + i y_c$. For $z = x + iy \\in D$, let $Q_z$ be the set of points $s + it$ with $s$ between $x_c$ and $x$ and $t$ between $y_c$ and $y$.\n3. For $s + it \\in Q_z$, $\\abs{s - x_c} \\le \\abs{x - x_c}$ and $\\abs{t - y_c} \\le \\abs{y - y_c}$. So $\\abs{(s + it) - c} \\le \\abs{z - c} \u003c R$, and $Q_z \\subseteq D$.\n4. For $z = x + iy \\in D$ define two numbers, with oriented integrals $\\int_\\alpha^\\beta = -\\int_\\beta^\\alpha$ taken on real and imaginary parts:\n   $$F(z) = \\int_{x_c}^{x} f(s + i y_c) \\, ds + i \\int_{y_c}^{y} f(x + it) \\, dt, \\qquad G(z) = i \\int_{y_c}^{y} f(x_c + it) \\, dt + \\int_{x_c}^{x} f(s + iy) \\, ds.$$\n   The segments used lie in $Q_z \\subseteq D$ by step 3, and the integrands are continuous by step 1.\n5. For real $\\alpha_1, \\beta_1, \\alpha_2, \\beta_2$ such that the rectangle with these side coordinates lies in $D$, put\n   $$E(\\alpha_1, \\beta_1, \\alpha_2, \\beta_2) = \\int_{\\alpha_1}^{\\beta_1} f(s + i \\alpha_2) \\, ds + i \\int_{\\alpha_2}^{\\beta_2} f(\\beta_1 + it) \\, dt - i \\int_{\\alpha_2}^{\\beta_2} f(\\alpha_1 + it) \\, dt - \\int_{\\alpha_1}^{\\beta_1} f(s + i \\beta_2) \\, ds.$$\n   Then $F(z) - G(z) = E(x_c, x, y_c, y)$. If $x = x_c$, both $s$-integrals vanish and the two $t$-integrals agree, so $E = 0$. In the same way, $E = 0$ if $y = y_c$.\n6. Swapping $\\alpha_1$ and $\\beta_1$ changes the sign of $E$. Swapping $\\alpha_2$ and $\\beta_2$ also changes the sign of $E$. This follows from $\\int_\\alpha^\\beta = -\\int_\\beta^\\alpha$ and a regrouping of the four terms.\n7. Now let $x \\ne x_c$ and $y \\ne y_c$. Then $Q_z$ is a closed rectangle $\\set{s + it : p_1 \\le s \\le q_1, \\ p_2 \\le t \\le q_2}$ in $D$, with $p_1 \u003c q_1$ and $p_2 \u003c q_2$.\n8. Parametrize the sides of $\\partial Q_z$ counterclockwise, using [[additivity over intervals|05RCDKHZM6]] and the [[substitution|0D37QFPEZE]] $s \\mapsto p_1 + q_1 - s$ on the backward sides. The [[contour integral|00FWNP9TXY]] is then\n   $$\\oint_{\\partial Q_z} f = \\int_{p_1}^{q_1} f(s + i p_2) \\, ds + i \\int_{p_2}^{q_2} f(q_1 + it) \\, dt - \\int_{p_1}^{q_1} f(s + i q_2) \\, ds - i \\int_{p_2}^{q_2} f(p_1 + it) \\, dt.$$\n9. This is $E(p_1, q_1, p_2, q_2)$. By step 6, $E(x_c, x, y_c, y) = \\pm \\oint_{\\partial Q_z} f$. By [[Goursat's theorem|0BRG85WXYC]] the integral is $0$. With step 5, $F = G$ on $D$.\n10. Fix $z = x + iy \\in D$. The set $J = \\set{t \\in \\R : x + it \\in D}$ is an open interval, since $D$ is open and convex. It contains $y_c$ and $y$, because $x + i y_c$ and $z$ lie in $Q_z$.\n11. For $t \\in J$, $F(x + it) = \\int_{x_c}^{x} f(s + i y_c) \\, ds + i \\int_{y_c}^{t} f(x + i \\tau) \\, d\\tau$. The first term does not depend on $t$.\n12. By [[the integral of a continuous function is an antiderivative|0NC2A3HWGT]], applied to real and imaginary parts, the [[partial derivative|05AXG3QRKR]] in $y$ is $\\partial F / \\partial y (z) = i f(z)$.\n13. In the same way, using $F = G$ and the open interval $\\set{s \\in \\R : s + iy \\in D}$, which contains $x_c$ and $x$, we get $\\partial F / \\partial x (z) = f(z)$.\n14. Write $F = U + iV$ and $f = a + ib$ with real $U, V, a, b$. By steps 12 and 13, $U_x = a$, $V_x = b$, $U_y = -b$ and $V_y = a$. These are continuous on $D$ by step 1.\n15. Since [[continuous partial derivatives imply differentiability|0KXSK8TEEB]], $U$ and $V$ are differentiable on $D$. Also $U_x = V_y$ and $U_y = -V_x$.\n16. By the [[Cauchy–Riemann equations|0T71SSYZ39]], $F$ is complex differentiable at each $z \\in D$, with $F'(z) = U_x + i V_x = a + ib = f(z)$. So $F$ is holomorphic on $D$ and $F' = f$.\n\n## Idea\n\nGoursat's theorem says that going around a rectangle gives $0$. So integrating first horizontally and then vertically gives the same value as the other order. One order makes the $y$-derivative easy to compute, the other the $x$-derivative. Together they give the Cauchy–Riemann equations for the primitive.\n",
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