This question is in relation to exercise 4.3 as found in Sjamaar's book "Manifolds and Differential Form", which asks the reader to solve following:
Let $\alpha = \sum_{i=1}^n f_i \hspace{0.2cm}dx_i$ be a closed 1-form defined on all of $\mathbb{R}^n$. Verify that the function:
$$g(\hat{x}) = \sum_{i=1}^n x_i \int_0^1 f_i(t\hat{x}).dt$$
Satisfies $dg = \alpha$, where $\hat{x} = [x_1, ... x_n] \in \mathbb{R}^n$.
My tried approach to this has been to explicitly use the formula that for any function ($0$-form) $\psi:\mathbb{R}^n \to \mathbb{R}$, we can find it's $1$-form via:
$$d\psi = \sum_{i=1}^n \frac{\partial \psi}{\partial x_i} \hspace{0.2cm}dx_i$$
This gives:
$$dg = \sum_{j=1}^n \frac{\partial}{\partial x_j} \left(\sum_{i=1}^n x_i \int_0^1 f_i(t\hat{x}).dt\right) \hspace{0.2cm}dx_j$$
$$=\sum_{j=1}^n \left[ \sum_{i=1}^n \frac{\partial}{\partial x_j} \left( x_i \int_0^1 f_i(t\hat{x}).dt\right) \right] dx_j$$
$$= \sum_{j=1}^n \left[ \sum_{i=1}^n \left\{ \frac{\partial x_i}{\partial x_j} \left( \int_0^1 f_i(t\hat{x}).dt\right)\right\} + \left\{ x_i \left( \int_0^1 \frac{\partial}{\partial x_j}f_i(t\hat{x}).dt\right)\right\}\right] dx_j$$
$$= \sum_{j=1}^n \left(\int_0^1 f_j(t\hat{x}).dt\right) + \left[ \sum_{i=1}^n x_i \left( \int_0^1 \frac{\partial}{\partial x_j}f_i(t\hat{x}).dt\right)\right] dx_j$$
Where the third line is by application of the product rule, and the fourth line is due to the fact that $\frac{\partial x_i}{\partial x_j}$ is $0$ unless $i=j$.
I do not know where to go next and am unsure of what I am missing. Whether there is some properties from calculus which I need to invoke or some fact orienting around multivariate functions.
Any help would be appreciated.
So far you have shown $$ \frac{\partial g}{\partial x_j} = \int_0^1 f_j(t\hat x)\,dt + \sum_{i=1}^n x_i \int_0^1 \frac{\partial}{\partial x_j} f_i(t\hat x)\,dt $$ By the chain rule, $$ \frac{\partial}{\partial x_j} f_i(t\hat x) = \frac{\partial f_i}{\partial x_j}(t\hat x)t $$ Since $\alpha$ is closed, $$\dfrac{\partial f_i}{\partial x_j} = \dfrac{\partial f_j}{\partial x_i}$$ So $$\begin{aligned} \frac{\partial g}{\partial x_j} &= \int_{0}^1 \left(f_j(t\hat x) + \sum_{i=1}^n x_i \frac{\partial f_j}{\partial x_i} (t\hat x)t\right)\,dt \\&= \int_{0}^1 \left(f_j(t\hat x) + t \sum_{i=1}^n \frac{\partial f_j}{\partial x_i} (t\hat x)x_i\right)\,dt \\&= \int_{0}^1 \frac{d}{dt}\left(t f_j(t\hat x)\right)\,dt \\&= \left.tf_j(t\hat x)\right|^{t=1}_{t=0} = f_j(\hat x) \end{aligned}$$