I saw this question and found the accepted answer to be very satisfying. However I thought about indefinite integrals and the differential in them (eg. $dx$). And a central part of the answer's argument is that $\dfrac{dy}{dx}$ is no longer a ratio, but a limit of a ratio. Seeing as there is no limit form of the antiderivative, I thought "what justifies the differential in the integral then". If $y = F(x)$ and $f(x) = F'(x)$ then $$\int f(x)dx = \int \dfrac{dy}{dx} \cdot dx$$ Obviously if we treat $dx$ like a fraction then they cancel out, and we end up summing $dy$, which is what we actually do when we take an antiderivative. This is a very compelling argument, but only in the context of hyperreals, where treating infinitesimals like fractions is justified. However, seeing as calculus was developed and made rigorous by people like Euler and the Bernoulli's much before hyperreals were made rigorous (by people like Abraham Robinson), what was the original justification for the differential? (I'm looking for an answer that doesn't delve into hyperreals or non-standard analysis)


Statements like "variable of integration" and "with respect to $x$" are irrelevant. Let me put that into context. When we do u-substitution, a central part is treating differentials like fractions. That is, Let $ u = y(x)$. Then $\dfrac{du}{dx} = y'$ Then we take $du = y'dx$ or $dx = \dfrac{du}{y'}$. You see where I'm getting at.

Further Note:

I know that there have been similar questions asked before, but I feel like their context wasn't exactly identical to mine, and furthermore, none of the answers were as solid or convincing as Arturo Magidin's.

  • $\begingroup$ The "reason" for the symbolism is mainly historical (as per Arturo's answer); the success of Leibniz's notation was exactly due to his "algebraic" nature, that allowed for "easy manipulations". Of course, those manipulations must be justified, and Leibniz - as well as the otehrs "founding fathers", like Euler, Bernoulli, l'Hopital - was well aware of this. $\endgroup$ – Mauro ALLEGRANZA Mar 19 '16 at 15:54
  • $\begingroup$ @MauroALLEGRANZA That is in part the motivation of my question. As you say his notation was successful due to its algebraic nature, that must mean that later mathematicians agreed to some extent with his algebraic notation. I want to understand how said mathematicians justified the use of algebraic manipulations in integrals. Arturo's answer has given me a thorough understanding for derivatives, namely because of his discussion on the limit. $\endgroup$ – Airdish Mar 19 '16 at 15:56
  • $\begingroup$ I think the reason is because it happens to work just like a fraction. In reality, we cannot directly say that $\int \frac{dy}{dx}dx=\int dy$ if we didnt have the fundamental theorem of calculus. $\endgroup$ – lEm Mar 19 '16 at 16:16

About Euler's definition of integral, see:

Definitio I. Calculus integralis est methodus ex data differentialia relatione inveniendi relationem ipsarum quantitaum: et operatio, qua hoc prestatur, integratio vocari solet.

Thus, the starting point is the (already known; see: Leonhard Euler, Institutiones calculi differentialis (1755), page 100) Leibnizian notion of differential, i.e. "differentia infinite parvas".

Coroll.I. Cum igitur calculus differentialis ex data relatione quatitate variabilium, relatione differentialium investigare doceat: calculus integralis methodum inversam suppeditat.

Again, the "conputational" aspect of the Leibnizian method is stressed: integration is the "inverse operation" withe respect to differentatition.

Coroll.III. Proposita relatione quacunque inter binas quantitates variabiles $x$ et $y$, in calculo differentiali methodus traditur rationem differentialium $dy : dx$ investigandi: sin autem vicissim ex hac differentialium ratione ipsa quantitaum $x$ et $y$ relatio sit definienda, hoc opus calculo integrali tribuitur.

Thus, having set up in the "calculo differentiali methodus" the way to investigate the behaviour of the rationem differentialium $dy : dx$ generated by two "related" variables quantities $x$ and $y$, the calculo integralis is aimed at "retrieving" from the rationem differentialium the relation between the two original quantities $x$ and $y$.

Definitio 2 [ page 4 ] Cum functionis cuiuscunque ipsius $x$ differentiale huiusmodi habet formam $Xdx$, proposita tali formam differentiali $Xdx$, in qua $X$ sit functio quaecunque ipsius $x$, illa functio, cuius differentiale est $=Xdx$, huius vocatur integrale, et prefixo signo signo $\int$ indicari solet, ita ut $\int Xdx$ eam denotet quantitatem variabilem, cuius differentiale est $=Xdx$.

In other words, if $F(x)$ is a function whatever of $x$ whose differential is $f(x)dx$, the said $F(x)$ is said the integral of the above differential and we denote it with $\int f(x)dx$.

There is a basic difference here: for us (from Weierstrass and on) $\dfrac{dy}{dx}$ is not a fraction, but one symbol.

For Euler, following Leibniz, $dy \div dx$ is a ratio between differentials.

  • $\begingroup$ "whose differential is $f(x) d x$" -- this point arises because $$y = F(x) \Longleftrightarrow F'(x) = \dfrac{d y}{d x} = f(x) \rightarrow d y = f(x) d x$$ I know that the explanation behind the differential is intuitive, but what is the formality behind it. In other words, what justifies our taking $dx$ to the other side, since we can't treat it as an algebraic quantity? $\endgroup$ – Airdish Mar 20 '16 at 14:47
  • $\begingroup$ Let me put my last statement into context. $$\dfrac{dy}{dx} = \lim_{h \rightarrow 0} \dfrac{F(x+h) - F(x)}{h}$$ If we take $dx$ to the other side as illustrated in my comment above, then $$\lim_{h \rightarrow 0} (F(x+h) - F(x)) = f(x) \lim_{h \rightarrow 0} h$$ This clearly cannot be done, because everything evaluates to $0$. $\endgroup$ – Airdish Mar 20 '16 at 14:54
  • $\begingroup$ @TheOddbodNumber - Euler does not "take $dx$ to the other side"... He consider a function $F$ of $x$ whose differential is expressible as $Xdx$. He does not state -in modern terms - the conditions for a function to be "integrable", but his assumption holds of course for many known functions, like polynomials. $\endgroup$ – Mauro ALLEGRANZA Mar 20 '16 at 16:00
  • $\begingroup$ My last comments were outside of the context of Euler. Sorry if I misunderstand (I'm a very novice student of math) but in modern times, $$F'(x) = f(x)$$ implies $$d(F(x)) = f(x)dx$$ right? $\endgroup$ – Airdish Mar 20 '16 at 17:47
  • $\begingroup$ @TheOddbodNumber - in your previous to last comment, you cannot do it; you cannot take $h$ outside od $\lim$, bcause it is not an algebraic expression. $\endgroup$ – Mauro ALLEGRANZA Mar 20 '16 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.