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  "body": "\n## Proof\n\nWe compute the partial derivatives of $g$ by differentiating under the integral sign.\n\n1. Let $U \\subseteq \\R^n$ be [[open|01ZQQDCERP]] and [[convex|0K06X9QXDX]], and let $\\omega = \\sum_{i=1}^n f_i \\, dx_i$ be a [[one-form|0MTENFBW3Q]] on $U$. Each $f_i \\colon U \\to \\R$ has continuous [[partial derivatives|05AXG3QRKR]] $\\partial_k f_i = \\partial f_i / \\partial x_k$. Assume that $\\omega$ is [[closed|0F1AVGX26N]].\n2. By [[continuous partial derivatives imply differentiability|0KXSK8TEEB]], each $f_i$ is [[differentiable|0135XZZQ8M]] at every point of $U$.\n3. For $a \\in U$ and $v \\in \\R^n$, we have $Df_i(a)(v) = \\sum_{k=1}^n \\partial_k f_i(a) \\, v_k$. Indeed, taking $h = s e_k$ in the definition of $Df_i(a)$ gives $Df_i(a)(e_k) = \\partial_k f_i(a)$, where $e_k$ is the $k$-th standard basis vector. Linearity of $Df_i(a)$ then gives the formula.\n4. We show that $\\partial_j f_i = \\partial_i f_j$ on $U$ for all $i, j$. By the definition of the [[exterior derivative|004FRCXH6M]], $d\\omega = \\sum_{i=1}^n \\sum_{k=1}^n \\partial_k f_i \\, dx_k \\wedge dx_i$.\n5. By the rules of the [[wedge product|0S166D363M]], $dx_i \\wedge dx_i = 0$ and $dx_i \\wedge dx_k = -dx_k \\wedge dx_i$. Grouping the terms with indices $\\set{k, i}$ gives $d\\omega = \\sum_{k \u003c i} (\\partial_k f_i - \\partial_i f_k) \\, dx_k \\wedge dx_i$.\n6. The two-forms $dx_k \\wedge dx_i$ with $k \u003c i$ are linearly independent. Since $d\\omega = 0$, every coefficient vanishes, so $\\partial_k f_i = \\partial_i f_k$ for $k \u003c i$. The case $k = i$ is trivial, and the case $k \u003e i$ follows by swapping the names of the indices.\n7. Fix $p \\in U$. For $x \\in U$ and $t \\in [0, 1]$, the point $p + t(x - p) = (1 - t) p + t x$ lies in $U$, because $U$ is convex. So the integrand in the definition of $g(x)$ is defined. It is continuous in $t$, as a composition and product of continuous functions. By [[continuous functions are Riemann integrable|06SESV7GQY]], $g(x)$ is well defined.\n8. Fix $x \\in U$ and an index $j$. Since $U$ is open, choose $r \u003e 0$ such that the closed ball of radius $2r$ about $x$ lies in $U$.\n9. For $t \\in [0, 1]$ and $\\abs{s} \u003c 2r$, put $q(t, s) = p + t(x + s e_j - p)$. The point $x + s e_j$ lies in $U$, so $q(t, s) \\in U$ by convexity.\n10. Define $H(t, s) = \\sum_{i=1}^n f_i(q(t, s)) \\, (x_i + s \\delta_{ij} - p_i)$, where $\\delta_{ij}$ is $1$ if $i = j$ and $0$ otherwise. Then $g(x + s e_j) = \\int_0^1 H(t, s) \\, dt$. The function $H$ is continuous on $[0, 1] \\times [-r, r]$.\n11. Fix $t \\in [0, 1]$. The curve $s \\mapsto q(t, s)$ on $(-2r, 2r)$ has derivative $t e_j$. By the [[chain rule for a curve|0A8TJ0VHC7]] and Step 3, $\\frac{\\partial}{\\partial s} f_i(q(t, s)) = Df_i(q(t, s))(t e_j) = t \\, \\partial_j f_i(q(t, s))$.\n12. The factor $x_i + s \\delta_{ij} - p_i$ has derivative $\\delta_{ij}$ in $s$. By the [[product rule|0V9A43H7WV]] and Step 11,\n$$\\frac{\\partial H}{\\partial s}(t, s) = \\sum_{i=1}^n t \\, \\partial_j f_i(q(t, s)) \\, (x_i + s \\delta_{ij} - p_i) + f_j(q(t, s)).$$\n13. The right side of Step 12 is continuous on $[0, 1] \\times [-r, r]$, because each $\\partial_j f_i$ and each $f_j$ is continuous and $q$ is continuous.\n14. By [[differentiation under the integral sign|0AR06W1E1B]], applied to $H$ on $[0, 1] \\times [-r, r]$, the function $s \\mapsto g(x + s e_j)$ is differentiable at $s = 0$. Its derivative there is $\\partial_j g(x)$. So, with $\\gamma(t) = p + t(x - p) = q(t, 0)$,\n$$\\partial_j g(x) = \\int_0^1 \\Big( t \\sum_{i=1}^n \\partial_j f_i(\\gamma(t)) \\, (x_i - p_i) + f_j(\\gamma(t)) \\Big) \\, dt.$$\n15. By Step 6, $\\partial_j f_i = \\partial_i f_j$. So the integrand in Step 14 equals $\\psi(t) = t \\sum_{i=1}^n \\partial_i f_j(\\gamma(t)) \\, (x_i - p_i) + f_j(\\gamma(t))$. This function $\\psi$ is continuous on $[0, 1]$.\n16. Put $\\varphi(t) = t \\, f_j(\\gamma(t))$ for $t \\in [0, 1]$. It is continuous on $[0, 1]$. The curve $\\gamma$ has derivative $x - p$.\n17. For $t \\in (0, 1)$, apply the chain rule for a curve to $\\gamma$ restricted to $(0, 1)$. With Step 3, it gives $\\frac{d}{dt} f_j(\\gamma(t)) = \\sum_{i=1}^n \\partial_i f_j(\\gamma(t)) \\, (x_i - p_i)$. By the product rule, $\\varphi'(t) = f_j(\\gamma(t)) + t \\sum_{i=1}^n \\partial_i f_j(\\gamma(t)) \\, (x_i - p_i) = \\psi(t)$.\n18. So $\\varphi'$ extends to the continuous function $\\psi$ on $[0, 1]$, which is Riemann integrable. By the [[second fundamental theorem of calculus|0XYAFCN0PT]], $\\int_0^1 \\psi(t) \\, dt = \\varphi(1) - \\varphi(0) = f_j(x) - 0 = f_j(x)$.\n19. By Steps 14, 15 and 18, $\\partial_j g(x) = f_j(x)$ for every $x \\in U$ and every $j$.\n20. The partial derivatives $\\partial_j g = f_j$ are continuous, so $g$ is [[continuously differentiable|06SKCHDQT5]]. By the definition of the exterior derivative of a function, $dg = \\sum_{j=1}^n \\partial_j g \\, dx_j = \\sum_{j=1}^n f_j \\, dx_j = \\omega$.\n21. Hence $\\omega$ is exact, and the given function $g$ satisfies $dg = \\omega$ on $U$.\n\n## Idea\n\nThe function $g(x)$ is the integral of $\\omega$ along the straight segment from $p$ to $x$, which lies in $U$ by convexity. Differentiating under the integral sign gives an integrand with the terms $\\partial_j f_i$. The closedness condition $\\partial_j f_i = \\partial_i f_j$ turns this integrand into the exact derivative $\\frac{d}{dt}\\big(t f_j(p + t(x - p))\\big)$. The fundamental theorem of calculus then gives $f_j(x)$.\n",
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