What is the inverse operation of a gradient? 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?).
 A: 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 $\gamma$.
A: Considering the usual notation abuse $\nabla=\left(\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},\cdots,\frac{\partial}{\partial x_n}\right)$, you might try something similar with the integral where $I=\left(\int dx_1,\int dx_2,\ldots,\int dx_n\right)$, so that $I\cdot\nabla=n$. Then you can divide the integral by $n$ to get the original function like so:
$$\frac{1}{3}I\cdot\nabla f=\frac{1}{3}\left(\int dx,\int dy,\int dz\right)\cdot\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right)=\frac{1}{3}\left(f+f+f\right)=\frac{3f}{3}=f$$
