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In the single variable calculus, when doing a u substitution, we have

$\qquad\begin{align}\int_a^b{f'(x)}\,\mathrm dx &= \int_a^b{(h \circ g)'(x)\,\mathrm dx}\\& = \int_a^b{(h' \circ g)(x)\,g'(x)\,\mathrm dx} \\&= (h \circ g)(x) \Big|_a^b \\&= h(g(b)) - h(g(a)) \\&= \int_{g(a)}^{g(b)}{h'(u)\,\mathrm du}\end{align}$

Here, there's a nice connection between the chain rule and the Jacobian, where the $g'(x)dx$ term comes out of the wash without any appeals to geometric reasoning, like is often the case when introducing the Jacobian term for the multivariable change of variables. Is it possible to make an analogous derivation for the multivariable case?

What confuses me is that the multivariable chain rule has the form of a sum, but the Jacobian term has the form of a difference, so at first glance it seems not possible.

Graham Kemp
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Tomek Dobrzynski
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    Relevant thread about Jacobian matrix: https://math.stackexchange.com/a/1127350/40119 Relevant post about change of variables theorem and Jacobian determinant: https://math.stackexchange.com/a/464972/40119 – littleO Aug 03 '22 at 21:07
  • There needs to be an integration by parts somewhere in the derivation, that's where something in the form of a difference will come from. – Tomek Dobrzynski Aug 04 '22 at 19:54
  • Here’s a fundamental fact: Suppose that $A$ is a real $n \times n$ matrix, $b \in \mathbb R^n$ and $T:\mathbb R^n \to \mathbb R^n$ is defined by $T(x) = Ax + b$ for all $x \in \mathbb R^n$. If a set $S \subset \mathbb R^n$ has volume (or Lebesgue measure) $V$, then the volume (or Lebesgue measure) of $T(S)$ is $| \det A | V$. That is where the Jacobian determinant comes from in the change of variables formula for integration. More details are given in the posts in linked above. – littleO Aug 04 '22 at 20:12
  • Yeah, I know about that, but I'm after something else. Supposedly, there's a derivation like this in a proof by PD Lax published in an article titled Change of Variables in Multiple Integrals II, but I I don't have access to Jstor, so I can't verify. – Tomek Dobrzynski Aug 04 '22 at 23:57
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    I took a look at the article you mentioned, and the proof there is nothing like the 1D case (the article is very technical, as expected). The typical proof in the 1D case is rather special in that it exploits the FTC and the chain rule. By using the FTC, one by-passes a lot of the geometry regarding how a mapping $g$ distorts lengths/volumes. Also, 1D is trivial because $g$ maps intervals to intervals. The above links by littleO are, I think, the best ways to think about the theorem: always start with the linear case, and understand how determinants relate to volumes, and then generalize. – peek-a-boo Aug 05 '22 at 21:21
  • Thanks for reviewing the article. Ifound the best insight in this post https://math.stackexchange.com/questions/2094567/origin-of-jacobian-determinant. The key feature of multivariable integrals is the term dxdy, which is not an ordinary product, but a wedge product dx $\wedge$ dy, which is anti-commutative, and this property is ultimately the source of the differences of products of partial derivatives that appear in the Jacobian term. So it seems to be possible to make the derivation without appealing to geometry by using the chain rule and the anti-commutativity of the wedge product. – Tomek Dobrzynski Aug 05 '22 at 23:35
  • sure, but the reason why we use differential forms for integration is because determinants and volumes are very closely related to anti-commutativity. So, you're not really by-passing the geometry; rather the geometry is 'encoded' into the algebraic properties of wedge-products. But as a cautionary remark: note that even after you develop the notion of wedge products and differential forms (and pullbacks), you still have to prove the change-of-variables theorem (which is highly non-trivial to prove in general; though it's very easy to state). – peek-a-boo Aug 06 '22 at 05:23
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    See page 207 of http://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/analmv.pdf. There the theorem is proved using the chain rule rather than standard geometric proof that estimates volumes of images of cubes that the above comments describe. – Mason Aug 18 '22 at 00:36

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