# 1-form is exact

Let $\omega = f_1 dx_1 +f_2dx_2 + \cdots + f_ndx_n$ be a closed $C^{\infty}$ $1-$form on $\mathbb R ^n$. Define a function $g$ by

$\displaystyle{ g(x_1, x_2,\cdots, x_n) = \int_{0}^{x_1} f_1(t,x_2 , x_3, \cdots ,x_n) +\int_{0}^{x_2} f_1(0,t, x_3, x_4, \cdots ,x_n) + \int_{0}^{x_3} f_1(0,0,t, x_4 ,x_5, \cdots ,x_n) + \cdots +\int_{0}^{x_n} f_1(0,0, \cdots ,t) }$

Show that $dg =\omega$.

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Nice question!! – WishingFish Jul 25 '13 at 4:30

$\omega$ being closed gives $\partial_i f_j = \partial_j f_i$. So we have \begin{align*} \partial_i g(x) &= \sum_{k = 1}^{i-1} \int_0^{x_k} \partial_i f_k(0,\ldots, 0, t, x_{k+1}, \ldots, x_n)\,dt + f_i(0,\ldots, 0, x_i, \ldots, x_n)\\ &= \sum_{k=1}^{i-1} \int_0^{x_k} \partial_k f_i(0,\ldots, 0, t, x_{k+1}, \ldots, x_n)\,dt + f_i(0,\ldots, 0, x_i, \ldots, x_n)\\ &= \sum_{k=1}^{i-1} \bigl(f_i(0,\ldots, x_k, \ldots, x_n) - f_i(0,\ldots, 0, x_{k+1}, \ldots, x_n)\bigr)+ f_i(0,\ldots, 0, x_i, \ldots, x_n)\\ &= f_i(x) - f_i(0, \ldots, 0, x_i,\ldots, x_n) + f_i(0,\ldots, 0, x_i,\ldots, x_n)\\ &= f_i(x) \end{align*} So $dg = \sum_i \partial_i g\,dx_i = \sum_i f_i dx_i = \omega$.