Undergrad Is there such a thing as an antiderivative of a multivariable function?

[Abdullah Almosalami]

Is there such a thing as an antiderivative of a multivariable function? I haven't put too much thought into this yet but I wanted to ask anyways. Sticking for now just to two variables, I was observing that double integrals are always definite integrals, whereas in the single-variable case, we have both definite and indefinite integrals. I suppose I am asking given a function $f(x,y)$, is there any meaningful relationship between a function $f(x,y)$ and some other function $F(x,y)$ such that $\frac {\partial^2 F} {\partial x \partial y} = f(x,y)$? Moreover, what I am getting at after that is is there some kind of analogous Fundamental Theorem of Calculus for two-variable functions that goes something like $\int_{y=c}^d \int_{x=a}^b f(x,y) \, dx \, dy = F(P_1) - F(P_2)$ where $P_1$ and $P_2$ are some points in the domain of $F(x,y)$? I'm not sure if this is down the line my Calculus textbook, so it might literally be in the next section.

 

[mathwonk]

have you read fubini's theorem? ( reducing two variable integrals to repeated one variable integrals.)

 

[Abdullah Almosalami]

have you read fubini's theorem? ( reducing two variable integrals to repeated one variable integrals.)

Actually, to add to my question, I would even make it more general. Instead of integrating over a rectangular region, continuing off of my question, $\iint_D f(x,y) \, dA = F(P_1) - F(P_2)$, where $P_1, P_2 \in D$. And @mathwonk, yes, I have read and used Fubini's theorem, and am also aware that over non-rectangular domains, the double integral becomes a lot more complicated. So I suppose it might make the double integral simpler if we can define $F(x,y)$ and $P_1$ and $P_2$. But like I said, I am still early in the game.

 

[Abdullah Almosalami]

Well, holy smokes, found the part of the answer that corresponds to a rectangular domain of integration:

This is from Calculus - Early Transcendentals, 3rd Edition, Rogowski, Adams, Section 15.1 Exercise 50.

I found a proof here: https://math.stackexchange.com/ques...undamental-calculus-theorem-for-two-variables

 

[FactChecker]

About the first thing that you would want is that the integral is path-independent. Here is a discussion that may be what you are looking for to answer your question. http://tutorial.math.lamar.edu/Classes/CalcIII/FundThmLineIntegrals.aspx

However, this is not the same as the multiple integrals that you are asking about. I do not immediately see the answer to that. But the line integral might be closer to what you really are looking for. The path independence of line integrals has some profound consequences.

 

[Abdullah Almosalami]

About the first thing that you would want is that the integral is path-independent. Here is a discussion that may be what you are looking for to answer your question. http://tutorial.math.lamar.edu/Classes/CalcIII/FundThmLineIntegrals.aspx

However, this is not the same as the multiple integrals that you are asking about. I do not immediately see the answer to that. But the line integral might be closer to what you really are looking for. The path independence of line integrals has some profound consequences.

Well, that might be related, and I have caught glimpes on the fundamental vector theorems stating equivalences between double and triple integrals and path and surface integrals (have no idea yet), but I should add that the path integral is only "path-independent" if you are integrating over a conservative vector field.

 

[mathwonk]

Re: posts #3 and #4, note that the proof you linked uses Fubini to reduce to the one variable FTC.

 

[Abdullah Almosalami]

Re: posts #3 and #4, note that the proof you linked uses Fubini to reduce to the one variable FTC.

Yes indeed, and I just realized my initial reply to your post was an incomplete thought, and for some reason it totally did not sink in what you said that Fubini's theorem reduces a double integral into two single integrals for which FTC may be applied. That was dumb of me.

 

[mathwonk]

not at all. my hint was not very detailed. In fact your version of it was much clearer.

 

[mathwonk]

The full answer you want is probably the general Stokes' theorem relating the integral of an n form dw over a region in n space, to the integral of its antiderivative (n-1) form w, over the boundary of the region. Note however that in this higher dimensional setting, it is not always true that every n form has an antiderivative, i.e. not every n form has the form dw. Whether or not this is true is related to the question raised above of when a path integral is independent of the path. E.g. in a convex region, the path integral of a one form w over a path joining two given points is independent of the path, iff the integral is zero over any closed path, iff dw = 0, iff w is itself equal to df for some function f, i.e. iff w has an antiderivative. More generally, an n form F equals dw for some (n-1) form w, iff the (n+1) form dF = 0, at least in a convex region. This is called the Poincare' lemma, and leads to a ("sheaf theoretic") proof of the deRham theorem. In general regions, the extent to which there can exist forms w with dw = 0 but which still have no global antiderivative, is a measure of the failure of the region to be convex, i.e. it measures the "homology" of the region. E.g. if we remove n points from the plane, the complementary open set will support an n dimensional vector space of 1-forms w satisfying dw = 0, but not of form df. Thus it has the same 1st homology group as the wedge of n circles.

By the way the example you found in a rectangle is entirely general in the sense that the general stokes theorem can be proved by patching together this result in a covering family of rectangles, in each of which it is reduced just to Fubini's theorem plus the usual one dimensional fundamental theorem of calculus. This is perhaps clearest in Lang's Analysis I, chapter XX, but also appears in Spivak's Calculus on Manifolds, and Guillemin and Pollack's Differential Topology. I think Lang does the best job of convincing you that it really only amounts to jazzing up the basic vertsion you ghave above, and hence should not be so scary complicated as it often looks. It may be useful as well to read up classical versions such as Green's theorem, Gauss' theorem, and classical Stokes theorem, but I would start with Green's theorem, i.e. stokes theorem in the plane.