I notice that the function
$$f(x,y,x;a,b,c) = ke^{-a/x-b/y-c/z}$$
has partial derivatives
$$\nabla f = \begin{bmatrix}
\partial f / \partial x
\\
\partial f / \partial y
\\
\partial f / \partial z
\end{bmatrix} = \begin{bmatrix}
a f / x^2
\\
b f / y^2
\\
c f / z^2
\end{bmatrix}$$
I want an operation, let's call it $\beta(\nabla f)$, that does this:
$$\beta(\nabla f)=ke^{-a/x-b/y-c/z}$$
I.e., takes the gradient and returns the original function. I was thinking along the lines of
$$\beta(\nabla f)=\int\int\int \nabla f\cdot \mathbf{n} \:\:dx \:dy\:dz \:\:; \: \mathbf{n} = [1,1,1]$$
But it seems pretty arbitrary to throw that $\mathbf{n}$ in there. Plus, this would basically be an indefinite volume integral, which I'm not sure exists (volume integrals always have to be definite integrals, right?).
Best Answer
There is no such operation, because (for any domain $D \subset \Bbb R^n$) the linear operator $$\nabla : C^{\infty}(D) \to \{\textrm{conservative vector fields on $D$}\}$$ is not injective: $$\ker \nabla = \{\textrm{constant functions $D \to \Bbb R$}\} \cong \Bbb R.$$ (For simplicity we take all objects to be smooth, i.e., infinitely differentiable, but this can be weakened considerably.)
On the other hand, recognizing $\nabla$ as linear tells us how to find the next-best thing to an inverse. Since $\nabla$ is surjective onto the set of conservative vector fields (this is just the definition of conservative), taking the quotient by the kernel induces an isomorphism $$C^{\infty}(D) / \{\textrm{constant functions $D \to \Bbb R$}\} \stackrel{\cong}{\to} \{\textrm{conservative vector fields on $D$}\} .$$ By construction, we can interpret its inverse, $$\tilde\beta : \{\textrm{conservative vector fields}\} \stackrel{\cong}{\to} C^{\infty}(D) / \{\textrm{constant functions $D \to \Bbb R$}\} ,$$ as the map that assigns to any conservative vector field its $1$-parameter family of potentials.
Unwinding definitions shows that computing $\tilde\beta({\bf X})$ for a conservative vector field $${\bf X} = \pmatrix{P\\Q\\R}$$ (here taking $n = 3$ for notational simplicity) just amounts to integrating: A function $f$ is a representative of $\tilde\beta({\bf X})$ iff it satisfies $f_x = P$, $f_y = Q$, $f_z = R$, so $$f(x, y, z) = \int_{\gamma} {\bf X} \cdot d{\bf s} = \int_{\gamma} P \,dx + Q \,dy + R \,dz,$$ where $\gamma$ is a path from some fixed reference point $(x_0, y_0, z_0) \in D$ to $(x, y, z)$. The conservativeness of $\bf X$ guarantees precisely that the above integrals are independent of the choice of path $\gamma$.