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When using the subsitituion rule in integration of an integral $\displaystyle \int f(x)\,\mathrm dx$, one turns the integral from the form $$\displaystyle \int f(x(t))\,\mathrm dx \quad(*)$$ into the form $$\int f(x(t))\,\frac{\mathrm dx}{\mathrm dt}\mathrm dt \quad(**)$$. This transform is usually accomplished by means of differenting the subsitution $x = x(t)$, such that $\dfrac{\mathrm dx}{\mathrm dt} = \dfrac{\mathrm d x(t)}{\mathrm dt}$. Now, at this point, one turns this into a differential form by means of magic, s.t. $\mathrm dx = \dfrac{\mathrm dx(t)}{\mathrm dt}\mathrm dt$. This now substitutes the differential term $\mathrm dx$ in the original expression $(*)$ to the one in the transformed expression $(**)$.

I'd like to learn that magic step a bit more rigorous – so that I can better understand it. It is often explained by "multiplication" of $\mathrm dt$, which do make sense, but it does not explain the nature of differentials; when is "multiplication" allowed? It seems there should be a more rigorous way of explaining it, perhaps by defining the "multiplication.

So, in what ways can differentials like $\mathrm dx$ and $\mathrm dt$ be formalized in this context? I've seen them being compared to small numbers, which often work, but can this analogy fail? (And what are the prerequisites needed to understand them?)

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marked as duplicate by suomynonA, Namaste integration Dec 7 '16 at 1:31

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  • $\begingroup$ Under the definition of integrals you can prove this, in particular it is easier to prove this for Darboux integral. $\endgroup$ – Masacroso Dec 6 '16 at 15:16
  • $\begingroup$ Many related questions on this site, some of which may help. See math.stackexchange.com/questions/1991575/ $\endgroup$ – Ethan Bolker Dec 6 '16 at 19:52
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Here's one way:

Consider $x$ and $t$ are coordinate systems of $\mathbb{R}$. If we wish to change coordinate systems, we have to look at how they transform into one another. If we consider $t$ to be a reference coordinate system and let the coordinate transformation be defined as $x(t) = 2t$ then for any $t$ element, $x$ is twice that (under $x$ view).

Now, since $(\mathbb{R}, + , \cdot)$ is a vector space, it has a dual $\mathbb{R}^*$. Using this space, we can start defining the elements $dx, dt$. Specifically, $dt$ will be a basis for $\mathbb{R}^*$ if $t$ is the basis vector for $\mathbb{R}$ . The elements of the dual space are called 1-forms. 1-forms of $\mathbb{R}^*$ "eat" vector elements of $\mathbb{R}$ and return a measure along that direction (only 1 dimension, so one direction). In this case you can consider elements of $\mathbb{R}^*$ as "row" vectors and multiply column vectors in $\mathbb{R}$ (which is the dot product of two vectors).

We can define a different basis for $\mathbb{R}$ and $\mathbb{R}^*$ with a coordinate change. For this example, if $dt$ eats a one dimensional vector $a$, it will return $a$. But when $dx$ eats $a$ it returns $2a$ in the $t$ coordinate system. That is $dx = 2dt$. For a general coordinate transform, a 1-form can be describe by $dx = \frac{dx}{dt} dt$.

This provides us with a way to talk about $dx$ and $dt$ meaningfully. Since $f: \mathbb{R} \to \mathbb{R}$ then $f(x)dx$ is $dx$ "eating" the vector $f(x)$ with regards to the $x$ coordinate system. Sometimes $f$ is easier to think of in a different coordinate system and so we wish to change it. $f(x)$ then becomes $f(x(t))$ and $dx$ becomes $\frac{dx}{dt}dt$. Now $dt$ is eating vectors $f(x(t))$ in its own coordinate system.

Consider how the uniform subdivide interval $(a,b)$ looks in a new coordinate system. For example $\{(0,\frac{1}{2}), (\frac{1}{2},1), (1,\frac{3}{2})\}$ in $t$ looks like $\{(0,\frac{2}{3}), (\frac{2}{3},2), (2, \frac{6}{2})\}$ in $x$ in the example coordinate transform. $\frac{dx}{dt}$ tells us precisely how the intervals change under our transformation.

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This analogy doesn't fail. Actually $dx$ indicates a small, rather infinitesimal change in the value of $x$ with change (no matter, however small) in the parameter, say $t$, on which $x$ depends. Looking at in another way, we can comment that the infinitesimal change in $x$, represented here by $dx $ can be considered equivalent to the rate of change of $x$ with $t$ multiplied by the change $dt $ in the value of the parameter $t$.

Mathematically it is written as: $$\mathrm dx = \dfrac{\mathrm dx(t)}{\mathrm dt}\mathrm dt$$

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There are a lot of books on non-standard analysis I think that really goes into this, the one I have is Infinitesimal Calculus by James M. Henle and Eugene M Kleinberg. Basically, what I think the confusion is that this dx isnt actually going to be a constant, as it is dependent both on t and the infinitesimal change of t. However, what they show is that it doesn't matter the value of dx (or even if it is non-constant) but rather that it is always infinitesimal. If it is always infinitesimal then the standard part of the integral will always be the same no matter our choice of dx.

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