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The integration by substitution technique is dervied from the following statement: $$\int _{a}^{b}f(\varphi (x))\varphi '(x)\,dx=\int _{\varphi (a)}^{\varphi (b)}f(u)\,du$$ Now almost all the explanations I saw about the actual technique start playing around with the Leibniz notation and $dx$, $du$ to get to the new form, e.g stating $du=\varphi '(x)\,dx$ from ${\frac {du}{dx}}=\varphi '(x)$.

Now the problem is that $dx$ and $du$ are said to have no actual meaning, at least in elementary calculus, and yet we're playing around with them as if they were actual variables forming actual quotients. As far as I am aware this wouldn't be a problem in nonstandard analysis, but how should one make sense of it in the elementary settings? How can we apply this technique without "heuristic" methods (which actually seems to be used almost all the time for different techniques)?

Note: My problem is actually applying the method, not the proof, as there are many proofs available not using the Leibniz's notation.

zareami10
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    Proof on Wikipedia that just uses the chain rule and the Fundamental Theorem of Calculus twice; no funny business with Leibniz notation. – angryavian Dec 28 '20 at 19:05
  • @angryavian Thanks, but I don't have a problem with the proof, but rather actually applying the method. – zareami10 Dec 28 '20 at 19:06
  • There is some prior discussion on this on this site. My understanding is that Leibniz notation is intuitive, but informal; there are ways that properties of the derivative end up working the same way as fractions, but other ways where it can lead you astray. Personally, I would rely on the actual definitions and theorems over Leibniz notation, while understanding that Leibniz notation can be helpful for intuition/memorization. – angryavian Dec 28 '20 at 19:18

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Here is one way to look at it. Let $F$ be an anti-derivative of $f$. Then $F(\phi(x))$ is an anti-derivative of $f(\phi(x)) \phi'(x)$ (by the chain rule). So by the fundamental theorem of calculus, $$ \int_a^b f(\phi(x)) \phi'(x) \, dx = F(\phi(x)) \bigg \rvert_a^b = F(\phi(b)) - F(\phi(a)). $$ But also, by the fundamental theorem of calculus, $$ \int_{\phi(a)}^{\phi(b)} f(u) \, du = F(u) \bigg \rvert_{\phi(a)}^{\phi(b)} = F(\phi(b)) - F(\phi(a)). $$ This reveals that $$ \int_a^b f(\phi(x)) \phi'(x) \, dx = \int_{\phi(a)}^{\phi(b)} f(u) \, du. $$

We should not throw "infinitesimal intuition" out the window, though. The following intuitive argument is quite illuminating. Suppose we chop up the interval $[a,b]$ into tiny pieces and the $i$th piece is $[x_i, x_{i+1}]$. Then correspondingly the interval $[\phi(a), \phi(b)]$ is chopped up into tiny pieces and the $i$th piece is $[\phi(x_i), \phi(x_{i+1}]$. To evaluate $\int_{\phi(a)}^{\phi(b)} f(u) \, du$, we must add up the contributions of each tiny piece: $$ \int_{\phi(a)}^{\phi(b)} f(u) \, du \approx \sum_i f(\phi(x_i)) (\phi(x_{i+1}) - \phi(x_i)). $$ But notice that $\phi(x_{i+1}) - \phi(x_i)) \approx \phi'(x_i)(x_{i+1} - x_i)$. So, $$ \int_{\phi(a)}^{\phi(b)} f(u) \, du \approx \sum_i f(\phi(x_i)) \phi'(x_i)(x_{i+1} - x_i) \approx \int_a^b f(\phi(x)) \phi'(x) \, dx. $$

littleO
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  • Thank you, though my problem is about actually applying the technique and not the proof. I added a note to my question, sorry if it was unclear. – zareami10 Dec 28 '20 at 19:11
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    To apply it you can just use the formula, right? – littleO Dec 28 '20 at 19:15
  • Yes but that's where my problem lies. When we use the formula we first say $\phi(x)=u$ and then $\phi'(x)dx=du$, and that's where I get lost. How can we play around with $dx$ and $du$ as if they were real variables? – zareami10 Dec 28 '20 at 19:31
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    @zareami10: It's because you proved in another way that the correct formula is the one that you get if you just substitute $\phi'(x) , dx=du$. If you say that you understand that proof, how can you doubt that formula works? There's no need to assign any meaning to $dx$ and $du$ beyond being mnemonics for writing down the correct formula. – Hans Lundmark Dec 28 '20 at 20:05
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    Just don't say "$du = \phi'(x) dx$" and just use the formula, which states that a certain definite integral is equal to a certain other definite integral. – littleO Dec 28 '20 at 20:37
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It's quite simple once the notion of differential of a function has been defined. For a differentiable function $\varphi$; its differential $\mathrm d\varphi$ at a point $x$ is its best linear approximation at $x$, which is shown to be the map of the variable $h$: $\enspace h\longmapsto f'(x)h$. All linear maps ar clearly their own differentials.

In particular, the differential of the identity map is itself, since it's a linear map, and, as by abuse of language, this identity map is often denoted $x$, its differential is denoted $\mathrm dx$.

In this context, if in the integral $\;\int_a^b f\bigl(\varphi(x)\bigr)\varphi'(x)\,\mathrm dx$, you set $u=\varphi (x)$, its differential is precisely $\mathrm du=\varphi'(x)\,\mathrm dx$, and the corresponding bounds for $u$ are clearly $\varphi(a)$ and $\varphi(b)$, so that the integral can quite naturally be written as $$\int _{a}^{b}f(\varphi (x))\varphi '(x)\,\mathrm dx=\int _{\varphi (a)}^{\varphi (b)}f(u)\,\mathrm du.$$

Bernard
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  • After reading about differentials of functions, this does make more sense to me. Though I should also say that it also creates quite a few other questions. Are there any sources I could refer to for a more detailed explanation? – zareami10 Dec 28 '20 at 22:01
  • I just wrote what I learnt as a student, mainly with Henri Cartan's book Differential Calculus. I don't if it's always available. But feel free to ask questions if some points are unclear. – Bernard Dec 28 '20 at 22:16