Clarification of the $u$-substitution theorem

Question

I came across this phrasing of the theorem justifying u-substitution:

Let $F(x)$ be an antiderivative of $f(x)$ in an interval $I.$ Let $\phi$ from $J$ to $I$, $\phi(t) = x$ be a differentiable function. Then $\int f(x)dx=\int f(\phi(t))\phi^{\prime}(t)dt.$

I am confused about the assumptions part - first of all, why can we assume that there exists a function $\phi$ such that $\phi(t)=x$?

Secondly, we know that the image of $\phi$ over $J$ is a subset of $I$. Why aren't we demanding that the image of $\phi$ will be equal to $I$, and not just a subset? In my mind, we are "losing" some $x$ values that are not given by $\phi(t)$ if it is strictly contained.

Lastly, in the final steps of the proof of this theorem, we said that $F(\phi(t))+c=F(x)+c=\int f(x)dx.$ Why can we treat $x$ just like a "dummy" variable? We assumed it equals a function $\phi(t)$ after all.

Answer 1

Let $F(x)$ be an antiderivative of $f(x)$ in an interval $I.$ Let $\phi$ from $J$ to $I$, $\phi(t) = x$ be a differentiable function. Then $\int f(x)dx=\int f(\phi(t))\phi^{\prime}(t)dt.$

I am confused about the assumptions part - first of all, why can we assume that there exists a function $\phi$ such that $\phi(t)=x$?

The sentences starting with "let" specify the notation for the objects whose relationship is displayed after the word "then". Your question is akin to asking why in the statement of Pythagoras' Theorem, we can assume that a right-angled triangle has a longest side of length $c.$ If your integrand can be expressed in terms of some function $\phi$ in a way that satisfies the theorem's conditions, then the theorem is applicable.

Secondly, we know that the image of $\phi$ over $J$ is a subset of $I.$ Why aren't we demanding that the image of $\phi$ will be equal to $I$, and not just a subset?

$I$ is in fact meant to be the image, not necessarily the codomain, of $\phi.$

Lastly, in the final steps of the proof of this theorem, we said that $F(\phi(t))+c=F(x)+c=\int f(x)dx.$ Why can we treat $x$ just like a "dummy" variable? We assumed it equals a function $\phi(t)$ after all.

An indefinite integral's variable is not dummy, because its scope extends outside the integral; in this case, variable $x$ is related to variable $t$ via the function $\phi.$

By the way, contrary to Jorge's assertion, your theorem statement is correct: the indefinite-integral version has more relaxed conditions than the standard versions of the change-of-variable theorem. You may be interested to read my stronger (i.e., slightly more general) statement & proof of the theorem and explanation there regarding dummy variables.

Answer 2

First, there is no a special meaning in the expression $\phi(t) =x$, is just a way to write that the function goes from $J$ to $I$, probably the author wrote it in this way to make more intuitive that we are substituting $x$ for $\phi(t)$ and there is no a special reason if you consider that $\phi(J)$ is a subset of $I$ or equals $I$, you can assume any and the theorem remains true, we only require that the composition $f\circ\phi$ makes sense. Now, you have to notice that the Theorem (as you wrote it) is not true, some hypotheses are missing, what we are really doing is to use the following theorem:

Let $J=[\alpha, \beta]$, $I$ be an interval and let $\phi : J\rightarrow I$ have a continuous derivative on $J$. If $f: I\rightarrow \mathbb{R}$ is continuous function. Then $(f\circ \phi)\phi'$ is integrable over $J$ and $$\int_{\alpha}^{\beta} f(\phi(t))\phi'(t) \ dt = \int_{\phi(\alpha)}^{\phi(\beta)} f(x) \ dx$$

Finally, at the final step, we are not treating $x$ as a dummy variable, the reason of why the eaquality holds, is that if $F$ is an antiderivative of $f$ and $H = F\circ \phi$, then $H$ is differentiable and $H'(t) = f(\phi(t))\phi'(t)$, this implies that $H$ is an antiderivative of $(f\circ \phi)\phi'$ and by the Fundamental theorem of calculus we have

$$\int_{\phi(\alpha)}^{\phi(\beta)} f(x) \ dx = F(\phi(\beta)) - F(\phi(\alpha)) $$

$$\int_{\alpha}^{\beta} f(\phi(t))\phi'(t) \ dt = H(\beta) - H(\alpha)$$

but notice that by definition of $H$ you have that $H(\beta) = F(\phi(\beta))$, this is the equality that you have. The theorem and the proof you mentioned looks like an engineering proof, probably that is the reason the author omit some details that a priori can cause confusion.