The $d$ in Leibniz's Notation

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Usage of dx in Integrals

I have read some tutorials about Leibniz's notation, and I am still wondering about the $d$ beside variables, and when you can de-attach the $x$ from $dx$ (some sites said $\frac{d}{dx}$ is an operator), and the differential equations make the things more complicated.

What prevents you from writing $\int x dx$ to $\int x^2 d$? and what is the meaning of $\int 1$? Are these following reduction steps valid? $\frac{d \int f(x) dx}{dx} = d\int f(x) = f(x)$

Concerning dx in integrals see this question http://math.stackexchange.com/questions/1068/usage-of-dx-in-integrals — Américo Tavares, Apr 25 '11 at 22:14
For the derivatives see http://math.stackexchange.com/questions/21199/dy-dx-is-not-a-ratio — Américo Tavares, Apr 25 '11 at 22:17

score 1 · Answer 1 · edited Apr 26 '11 at 00:04

Corrections are welcomed. I don't feel too well at home with the content of the last two paragraphs yet. There are also most likely more stories than these three.

Short story; in classical calculus $dx$ is simply a formal symbol. When paired with an integral sign $\int$ it means integrate with respect to $x$. When paired with another formal symbol $df$ with $f$ a function of real numbers (that's right; $df/dx$), it means the derivative of $f$ with respect to $x$. This short story is often peppered (sometimes confusingly) with the intuition that these correspond to so-called infinitesimal quantities.

The notion that these are infinitesimal quantities is made rigorous in synthetic differential geometry. This however supposes a somewhat different logic (e.g., the axiom law of excluded middle will not in general be true). This somewhat different logic yields a different real number object; an infinitesimal is then a non-zero real number $e$ such that $e^2=0$.

In the machinery of real manifolds $dx$ is the image under the (universal) derivation $d:CM\to\Omega^1 M$ of a coordinate function $x:M\to\mathbb{R}$, where $CM$ is the algebra of smooth functions $M\to\mathbb{R}$, and $\Omega^1M$ the vector space of $1$-forms on $M$.

score 1 · Accepted Answer · answered Apr 25 '11 at 23:50

The short answer is, $dx$ is an indivisible symbol, so you can't split it up like that. Similarly, $\int \cdots \, dx$ is a fixed arrangement of symbols, not unlike a pair of brackets $( \cdots )$, so you can't rearrange them at will.

The long answer is $dx \ne xd$, so you can't rearrange $x \, dx$ as $x^2 d$. There is a way to define $d$ so that it has a meaning on its own and so that $dx$ means the same thing as it traditionally does. However, I don't know of any interpretation of $\int$ which gives meaning to $\int f(x)$.

There is in fact an interpretation for such an integral: $\int_{{a}} f(x) = f(a)$. As for the indefinite form, I suppose it would just be $\int f(x) = C$, which is somewhat boring but no more so than you'd expect from such an edge case. — , Apr 23 '13 at 18:22

The $d$ in Leibniz's Notation

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