Multi-dimensional theorem of calculus as a multidimensional integral

Question

As indicated in the title, I am looking for a multi-dimensional version of the theorem of calculus that I think to already have seen before but I cannot remember where. What I remember is something looking like $$f(x) - f(y) = \int_{\mathbb{R}^d} \frac{(x-y)·∇f(z)}{|x-z|^d} \,\mathrm{d}\,z.$$ Is this formula true? Or something in the same spirit? What would be the proof?

Thank you.

Answer 1

So, I was not far, the formula I was looking for was $$ f(x) = \frac{1}{|\mathbb{S}^{d-1}|} \int_{\mathbb{R}^d} \frac{\left(x-y\right)·\nabla f(y)}{|x-y|^d}\,\mathrm{d}y. $$

This follows from $$ f = -\Delta(-\Delta)^{-1}f = -\nabla\cdot\left(\frac{1}{(2-d)\,|\mathbb{S}^{d-1}|\,|x|^{d-2}} * ∇f\right) = \frac{1}{|\mathbb{S}^{d-1}|}\frac{x}{|x|^{d}}*∇f $$

Answer 2

The given formula is incorrect. I will prove by giving a counter example.

Let $f(z)=z_1$ then the gradient is $\nabla f \equiv e_1$. Let $y=0$.

Insert into the given formula \begin{align} f(x)-f(y)=x_1 \overset{?}{=} \int_{\mathbb{R}^n} \frac{(x-y)\cdot \nabla f (z)}{|x-z|^n} \, dz = x_1 \int_{\mathbb{R}^n} \frac{1}{|x-z|^n} \, dz = x_1 \int_{\mathbb{R}^n} \frac{1}{|z|^n} \, dz \end{align}

The last integral is not unity for standard 3-dimensional space and also not for other dimensions. This is obvious using multidimensional spherical coordinates.