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I have a list of the components of the gradients, $\partial f/\partial x_i$, of a function $f(x_1,x_2,\cdots)$. Is there some neat way to reconstruct the function $f$?

One approach to doing this would be treat this as a system of PDEs and use DSolve. However, Mathematica is unable to solve PDEs with more than 3 variables – see, for instance here.

Another approach is to integrate the gradients $\int (\partial f/\partial x_i) dx_i $ and then take the Union of the terms from all the integrals. This isn't quite a robust way of doing things as it fails if the expressions for the integrals are not simple enough (ExpandAll does not help). A code for doing this is the following : Table[act[m]=ExpandAll[Integrate[gradient[m],Subscript[x, m]],{m,1,NN}]; f=Fold[Union,act[1],Table[act[m],{m,2,NN}]];

Any better ideas?

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Problems with code often require the code (or "All unhappy codes are unhappy in their own way"), but here's a somewhat complicated example that works:

vars = {w, x, y, z};
vf = Grad[(Log[w]^2 Sqrt[1 + x^3 y] + y^5 ArcTan[z^7])/Log[1 + x z], vars];
Fold[
 #1 + Integrate[First[#2] - D[#1, Last[#2]], Last[#2],
   Assumptions -> vars \[Element] Reals] &, 
 0, Transpose@{vf, vars}]
(*  (y^5 ArcTan[z^7])/Log[1 + x z] + (Sqrt[1 + x^3 y] Log[w]^2)/Log[1 + x z]  *)

Note: Fold[#1 + Integrate @@ #2 - #1 &, 0, Transpose@{F, vars}] is shorter, but the integrals are more complicated.

Of course I've seen Integrate fail on very complicated expressions, but I don't know what one could expect to work better than it, other than perhaps DSolve which sometimes takes a different route in edge cases.

Alternate method of integration:

Here's a way to use DSolve, which ends with a constant of integration C[5] that is omitted in the Integrate method:

iter[F_, {dF_, vars_, c_}] := 
  F /. First@DSolve[D[F, First@vars] == dF, c @@ vars, vars];
Fold[
 iter,
 C[1] @@ vars,
 Transpose@{
   vf,
   NestList[Rest, vars, Length@vars - 1],
   Array[C, Length@vars]}
 ]

Use FoldList instead of Fold and you see the process mentioned in a comment below:

{C[1][w, x, y, z],
 (Sqrt[1 + x^3 y] Log[w]^2)/
   Log[1 + x z] + C[2][x, y, z],
 (y^5 ArcTan[z^7])/Log[1 + x z] + (Sqrt[1 + x^3 y] Log[w]^2)/
   Log[1 + x z] + C[3][y, z],
 (y^5 ArcTan[z^7])/Log[1 + x z] + (Sqrt[1 + x^3 y] Log[w]^2)/
   Log[1 + x z] + C[4][z],
 (y^5 ArcTan[z^7])/Log[1 + x z] + (Sqrt[1 + x^3 y] Log[w]^2)/
   Log[1 + x z] + C[5]}

Update: Error check

iter::nxact = "The vector field is not conservative: the derivative of `` with respect to `` minus `` depends on ``.";
iter[F_, {dF_, v_, c_}] := F /. First@ DSolve[
     If[Internal`DependsOnQ[#, Complement[vars, v]],
         Message[iter::nxact, F, First@v, dF, Complement[vars, v]];
         Throw[$Failed],
         #
         ] &@ Simplify[D[F, First@v] - dF] == 0,
     c @@ v, v];

Catch@ Fold[
  iter,
  C[1] @@ vars,
  Transpose@{
    vf,
    NestList[Rest, vars, Length@vars - 1],
    Array[C, Length@vars]}
  ]
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  • $\begingroup$ Thanks a lot! This works like a charm for the case I have, in which the gradients are multivariate polynomials in the vars. May I ask what is the logic you are using here: it is iterated integrals and subtracting out something? $\endgroup$ – TheTwistedSector Jul 30 '19 at 19:07
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    $\begingroup$ @TheTwistedSector It's how I teach it in Multivariable/Diff Eq. Basically, to find an $F$ such that $\nabla F$ equals a given $\nabla f$, alternate integrating and differentiating w.r.t the variables. First integrate the $w$ component $F=\int f_w\,dw+g(x,y,z)=F^{(1)}+g$; then differentiate w.r.t. $x$: $F^{(1)}_x+g_x=f_x$. Then integrate $g=\int(f_x-F^{(1)}_x)\,dx+h(y,z)=F^{(2)}+h$. And so forth. One should check the integration constants $g,h,\dots$ are free of any dependency on $(w),\,(w,x),\dots$, respectively. The code above does not do that. The answer is $F=F^{(1)}+F^{(2)}+\cdots$. $\endgroup$ – Michael E2 Jul 30 '19 at 19:44
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    $\begingroup$ One advantage is that the integrations get easier as you go along because of cancelation. You also can tell when the vector field is not conservative/closed/exact. $\endgroup$ – Michael E2 Jul 30 '19 at 19:44
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It is worth pointing out that as of V12.2, DSolve can solve the example in my other answer:

vars = {w, x, y, z};
vf = Grad[(Log[w]^2 Sqrt[1 + x^3 y] + y^5 ArcTan[z^7])/Log[1 + x z], 
   vars];

DSolve[
 Grad[f @@ vars, vars] == vf,
 f, vars]
(*
{{f -> Function[{w, x, y, z}, 
    C[1] + (y^5 ArcTan[z^7])/Log[1 + x z] + (
     Sqrt[1 + x^3 y] Log[w]^2)/Log[1 + x z]]}}
*)
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  • $\begingroup$ Looks like some progress is happening on this front! Does DSolve also work with the examples here? I would have tested this out by myself, but do not have v12.2 yet. $\endgroup$ – TheTwistedSector Feb 14 at 12:34
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    $\begingroup$ @TheTwistedSector Yes, it works. I added an answer. $\endgroup$ – Michael E2 Feb 14 at 15:16

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