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I am having trouble with notation, specifically how certain substitutions are rigorously made when going back and forth between a line or contour integral and a definite integral.

For example if:

$$ \phi(\tau) = t \\ \alpha \leq \tau \leq \beta $$

then:

$$ \int_a^bw(t)dt = \int_{\alpha}^{\beta}w(\phi(\tau))\phi'(\tau)d\tau $$

It appears that $dt = \phi(\tau)d\tau$

But I always thought $dt$ was just a symbol that told us which variable we are integrating with respect to. So it seems very wishy-washy to make this substitution.

Question:

How can I understand this rigorously enough that I don't feel guilty making the substitution when I am working problems? Also, how do I justify the change of variables on the bounds?

To clarify my question, I am seeking a rigorous algebraic explanation. I vaguely understand the larger idea behind the paramaterization, but I want to see all the substitutions made rigorously, step by step with no hand waving. This seems to have been neglected in my classes, and it's killing me.

1 Answers1

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There are various ways to make this rigourous. The one that requires the less machinery is to simply use the fundamental theorem of calculus as it is done here. In this approach, the $dt$ is purely notation for the integration variable. The proof goes

Let $f$ and $\phi$ be two functions satisfying the above hypothesis that $f$ is continuous on $I$ and $\phi'$ is integrable on the closed interval $[a,b]$. Then the function $f(\phi(x))\phi'(x)$ is also integrable on $[a,b]$. Hence the integrals

$\displaystyle\int _{\phi (a)}^{\phi (b)}f(u)\,du\hspace{10mm}$ and $\hspace{10mm}\displaystyle\int _{a}^{b}f(\phi(x))\phi '(x)\,dx$

in fact exist, and it remains to show that they are equal.

Since $f$ is continuous, it has an antiderivative $F$. The composite function $F\circ\phi$ is then defined. Since $\phi$ is differentiable, combining the chain rule and the definition of an antiderivative gives $$(F\circ \phi )'(x)=F'(\phi (x))\phi'(x)=f(\phi (x))\phi'(x)$$ Applying the fundamental theorem of calculus twice gives $$\begin{aligned} \int_{a}^{b}f(\phi (x))\phi'(x)\,dx &=\int _{a}^{b}(F\circ\phi)'(x)\,dx\\ &=(F\circ\phi )(b)-(F\circ\phi)(a)\\ &=F(\phi(b))-F(\phi (a))\\ &=\int_{\phi (a)}^{\phi(b)}f(u)\,du \end{aligned}$$ which is the substitution rule.

In the formalism of differential forms, the substitution can also be thought of as a pullback along a diffeomorphism. Here the integral can be written $$\int_{[a,b]}w\,\mu$$ where $\mu$ is the volume 1-form on $[a,b]$. I.e. your integral is thought of as the integral of a differential 1-form along a 1-chain. Then you use a diffeomorphism $[\alpha,\beta]\xrightarrow{\phi}[a,b]$ to rewrite this as a pullback $$\int_{{\phi}([\alpha,\beta])}w\,\mu=\int_{[\alpha,\beta]}\phi^{*}(w\,\mu)=\int_{[\alpha,\beta]}(w\circ\phi)\,\phi^{*}\mu=\int_{[\alpha,\beta]}(w\circ\phi)\,\phi'\,\nu$$ where $\nu$ is the volume form on $[\alpha,\beta]$.

There is also the formalization of the change of variable used in measure theory. That is: you define the pushforward measure from a measure space to a measurable space along a measurable function. You can read about it here.

And so on.

Jackozee Hakkiuz
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