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Consider integral $\int_{0}^{\frac{\pi}{2}}\sin x\cos xdx$, I substitute $u=\sin x$, $\frac{du}{dx}=\cos x$, $du=\cos xdx$.

How can we prove last step, multiplying by $dx$. Everywhere I see everyone says it can be justified by differential forms. But how? Where can I read proof and not just sentence that it can be justified by differential forms?

math-physicist
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romperextremeabuser
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    Another way: Ignore $dx$ and $du$, simply viewing them as computational aids. – Andrew Jun 22 '23 at 13:19
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    The simplest thing to do is to rephrase the argument to avoid that last step. There is no need for differential forms here. We are just using the chain rule in reverse to find an antiderivative of the function we are integrating. – littleO Jun 22 '23 at 13:19
  • https://math.stackexchange.com/questions/174871/does-multiplying-by-dt-have-any-meaning?rq=1 – Mostafa Jun 22 '23 at 13:21
  • Does this answer your question? Does multiplying by $dt$ have any meaning? – Mostafa Jun 22 '23 at 13:22
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    Any book on differential geometry will rigorously define a differential form. – John Douma Jun 22 '23 at 13:31
  • Yes, it is a fiction in the notation that $\frac{du}{dx}$ looks like a fraction of numbers, when $du$ and $dx$ are not numbers. It is a useful fiction - for example, the chain rule can be written as $$\frac{du}{dt}=\frac{du}{dx}\frac{dx}{dt}.$$ There are reasons this fiction works, but it can be tough to explain in intro class. The question might even be, "What does $\cos x,dx$ mean?" – Thomas Andrews Jun 22 '23 at 14:09
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    I advise against thinking about this in differential forms terms. That's not the point. Here I totally agree with @AndrewZhang: . For a concrete example I recommend this question and its answers. – Giuseppe Negro Jun 22 '23 at 15:49

2 Answers2

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That operation is the exterior derivative.

Essentially, a smooth function can be understood as a zero form. Then you have the exterior derivative which is an operator that takes $k-$forms into $(k+1)-forms$. In local coordinates (to avoid the generality) if $\omega = \sum_I a_Idx_I$ is a $k-$form where $a_I$ are smooth functions on a manifold, then its exterior derivative is $$d\omega = \sum_{j}\sum_I\frac{\partial a_I}{x_j}dx_j \wedge dx_I.$$

In your case, $f$ is a smooth function, hence a $0-$form on $\mathbb{R}$. So, $df = \frac{\partial f}{\partial x}dx$

user57
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    Worth mentioning that we don’t need any of this to evaluate the given integral . When textbooks such as Calculus by Spivak or Understanding Analysis by Abbot develop calculus rigorously, they simply avoid “infinitesimal” quantities like $dx$. – littleO Jun 22 '23 at 13:29
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Yes, it is a fiction in the notation that $\frac{du}{dx}$ looks like a fraction of numbers, when $du$ and $dx$ are not numbers.

It is a useful fiction - for example, the chain rule can be written as $$\frac{du}{dt}=\frac{du}{dx}\frac{dx}{dt}.$$ There are reasons this fiction works, but it can be tough to explain in an intro class. The question might even be, "What does '$\cos x\,dx$' mean in isolation? What does it mean for it to be equal to $du?$"

And yes, the ultimate answer to that question is in differential forms. But that is a complex topic, which I'll skip here.

In the end, at an intro level, it is better to phrase this result in terms of functions. Treat this as a substitution rule, and prove separately:

If $$F(u)=\int f(u)du\tag1$$ then $$F(g(x))=\int f(g(x))\,g'(x)\,dx\tag 2$$

But $(1)$ means $F'(u)=f(u)$ and $(2)$ means $(F\circ g)'(x)=f(g(x))g'(x).$ And this is just the chain rule.

Your case is $f(u)=u$ and $g(x)=\sin x.$

An additional notation wrinkle is to use Riemann-Stieljes integrals. We can show, if $u(x)$ is differentiable, then:

$$\int_{a}^{b}g(x)u'(x)\,dx=\int_{a}^{b} g(x)\,du(x),\tag3$$ where the right side is a Riemann-Stieljes integral.

In this sense, $du(x)=u'(x)\,dx$ as distributions.

A distribution is a linear map which takes a function $g$ to a number. In this case, $du(x)$ takes $g$ to $\int_{a}^{b} g(x)du(x)$ and $u'(x)\,dx$ takes $g$ to $\int_{a}^b g(x)u'(x)\,dx.$ And we see from $(1)$ that both are equal.

When $f$ is a function whose domain includes the range of $u,$ then we can treat these as distributions on such $f$ by applying the distributions to $f\circ u.$ Then we get from the first part of this answer:

$$\int_a^b f(u(x))du(x) =\int_{u(a)}^{u(b)} f(u)\,du$$

Ultimately, you can think of Stieljes integrals as an introduction to the idea of general distributions, which will ultimately lead to differentiable forms.

Thomas Andrews
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