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Let $U$ be a convex connected and open set in $\mathbb{R}^n$, such that $0\in U$. For every $k$-differential form $\omega$, $$\omega =\sum_{i_1\lt\dots\lt i_k} c_{i_1,\dots,i_k}(x)dx{_{i_1}}\wedge\dots\wedge dx{_{i_k}},$$ we define a new $k-1$ differential form, $I\omega$ as follows -

$$ \begin{split} (I\omega) = \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^kx_{i_j} &\left( \int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx)ds\right)\\ & dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k} \end{split} $$

Prove that $d(I\omega) + I(d\omega) = \omega$.

So I know that -

$d\omega = \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^n\frac{\partial c_{i_1,\dots,i_k}(x)}{\partial x_j}dx_j\wedge dx_{i_1}\wedge\dots\wedge dx_{i_k}$

and therefore -

$I(d\omega) = \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^k x_{i_j}(\int_0^1\frac{\partial c_{i_1,\dots,i_k}(sx)}{\partial x_j}ds)dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k}$

All I need now is to calculate $d(I\omega)$, and put it into $d(I\omega) + I(d\omega)$ and show that it is equal to $\omega$.

However, I'm not sure - How can I calculate $d(I\omega)$?

What is $d(I\omega)=\sum_{i_1\lt\dots\lt i_k}x_{i_j}\sum_{j=1}^n\frac{\partial(\int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx)ds)}{\partial d_{x_j}}dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k}$?

Or perhaps, is there any other way(some tricks) so I won't need to calculate it explicitly?

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This question has an open bounty worth +50 reputation from ChikChak ending in 10 hours.

The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.

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    $\begingroup$ I asked this a while ago here. If the accepted answer works for you, this can be closed as a duplicate. $\endgroup$ – Pedro Tamaroff Jan 11 at 11:23
  • $\begingroup$ @PedroTamaroff your link seem to take me to my post :) $\endgroup$ – ChikChak Jan 11 at 11:48
  • $\begingroup$ Link should work now. $\endgroup$ – Pedro Tamaroff Jan 11 at 12:20
  • $\begingroup$ @PedroTamaroff In the answer to your question, he assumes that the differential forms are on $[0,1]\times \mathbb{R}^n$. Why can we assume that? $\endgroup$ – ChikChak Jan 11 at 13:17
  • $\begingroup$ I think sure your formula for $I(\omega)$ is missing a factor of $(-1)^{j-1}$ inside the sum. Try setting $\omega = dx \wedge dy$ in $\mathbb{R}^2$. $\endgroup$ – 0x539 Jan 15 at 17:33

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