Question
Let $\alpha=\sum_{i=1}^m\alpha_i(x)dx^i\in\Omega^1(\mathbb{R}^m)$ be a closed 1-form on $\mathbb{R}^m$. Show that $f(x)=\sum_{i=1}^mx^i\int_0^1\alpha^i(ux)du$ defines a smooth function satisfying $df=\alpha$.
Attempt
Since $\alpha$ is smooth (by the problem’s assumption), the integrals in the definition of $f$ involve smooth functions $\alpha^i$. Since the composition of smooth functions is smooth, and the integral of a smooth function wrt $u$ from $0$ to $1$ is also smooth, then $f$ is smooth.
Now, we want to show that $df=\alpha$. We have $df=\sum_{i=1}^m\frac{\partial f}{\partial x^i}dx^i$.
Now, for each $j=1,…,m$, we have
$\frac{\partial f}{\partial x^j}=\sum_{i=1}^m\frac{\partial }{\partial x^j}(x^i\int_0^1\alpha^i(ux)du)$.
When $i=j$,
$\frac{\partial }{\partial x^j}(x^j\int_0^1\alpha^j(ux)du)=\int_0^1\alpha^j(ux)du+x^j\int_0^1u\frac{\partial \alpha^j}{\partial x^j}du$.
Since $\alpha$ is closed, then $d\alpha=0$. So, the second integral vanishes.
When $i\neq j$,
$\frac{\partial }{\partial x^j}(x^i\int_0^1\alpha^i(ux)du)=x^i\int_0^1u\frac{\partial \alpha^i}{\partial x^j}du$.
Since $\alpha$ is closed, then $d\alpha=0$. So, this integral vanishes as well.
Now, we evaluate the integral we have left this way:
$\int_0^1\alpha^i(ux)du=\alpha^i(x)-\alpha^i(0)$ (I need $\alpha^i(0)$ to vanish)
Therefore,
$df=\sum_{i=1}^m\frac{\partial f}{\partial x^i}dx^i=\sum_{i=1}^m(\int_0^1\alpha^j(ux)du)dx^i=\sum_{i=1}^m\alpha_i(x)dx^i=\alpha$.
Note First, I don’t seem to have a good justification to make $\alpha^i(0)$ vanish. Second, I don’t know if there is any part of this solution I should improve. Your help would be appreciated.