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I found a nice paper about inverse vector operators here. I have successfully implemented a Mathematica function for most of them, however I can't figure out how to do inverse gradient (page 7 in the paper). According the paper, the inverse gradient can be computed as a path independent line integral, but in my attempts I got improper results. The paper doesn't show up an example for it and I'm not really familiar with line integrals.
If anyone could help me I would be really thankful.
Perform the line integral over the straight-line path from the origin to $(x,y)$ via the parametrization $u(t)=(tx,ty)$ for $0\le t\le1$. Then $\mathrm du=(x,y)\,\mathrm dt$, so the integral can be computed as
Integrate[b[t x, t y].{x, y}, {t, 0, 1}]
Note that if $b = \nabla a$, from $b$ one can only recover $a$ up to a constant. In this case, since we are integrating from the origin, what we obtain is $a(x,y)-a(0,0)$. This may be a problem if $a$ has a singularity at the origin, in which case you'll need to integrate from a different point.
