$d(I\omega) + I(d\omega) = \omega$ for differential forms

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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• I asked this a while ago here. If the accepted answer works for you, this can be closed as a duplicate. – Pedro Tamaroff Jan 11 at 11:23
• @PedroTamaroff your link seem to take me to my post :) – ChikChak Jan 11 at 11:48
• Link should work now. – Pedro Tamaroff Jan 11 at 12:20
• @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? – ChikChak Jan 11 at 13:17
• 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 $\mathbb{R}^2$. – 0x539 Jan 15 at 17:33