# If $dx$ is just syntax and not an infinitesimal then why do we apply operations to it?

So apparently my understanding of this concept is either old, outdated, or nonstandard, etc, but I was under the assumption that in an integral, $dx$ represented the "infinitesimal change in $x$", analogous to $\Delta x$, the "width" of an approximating rectangle under the curve, where $f(x)$ is the height of that rectangle.

But now I hear it's just syntax to let you know the variable you're integrating. But if this is true then why, during things like $u$-substitution, do we still manipulate $dx$ as if it were a real quantity and we're just changing the units? It's like as if we were doing dimensional analysis for example.

So what exactly is $dx$ if it's not a real thing but is still something we apparently manipulate in certain situations such as $u$-substitution? And moreover if $dx$ is just syntax when what exactly is the integral doing if not implicitly taking the sum of infinitely many "rectangles" with infinitely small width?

• It's not just syntax. See also this answer: math.stackexchange.com/questions/200393/… – Michael Hardy Jan 28 '18 at 4:58
• @MichaelHardy I saw this other response here: math.stackexchange.com/a/143262/525456 – user525456 Jan 28 '18 at 5:26
• This question is very complicated, look here: math.stackexchange.com/a/21209/471959 this answer explains very good when you can think of $dx$ to be a infinitesimal change and when you can't and why it is useful. More simple explanation is that $u$ is a function of $x$, so we don't really change variables, but only changing the function $f(x)$ to be $g(u(x))$(where $f(x)=g(u(x))\implies du=u'(x)dx$) – ℋolo Jan 28 '18 at 5:36
• @Jared I don't think so, this question is specific about $u$-substitution, my link is only to clarify what $dx$ really means – ℋolo Jan 28 '18 at 5:48
• It is important to remember that there are choices in determining notation. The notation for $dx$ (and for $dy/dx$ in differentiation) were chosen well because they often can be manipulated in a manner one would expect. – davidlowryduda Jan 28 '18 at 12:48

Let's look at $dx$ to be really only syntax.

Now let's say we have $\int_{u(a)}^{u(b)} f(x)dx$, now I am defining $F(u(t))\implies (F(u(t)))'=f(u(t))u'(t)$ by the fundamentals theorem of calculus we have$$\int_a^b f(u(t))u'(t)dt=\int_a^b (F(u(t)))'dt=F(u(b))-F(u(a))=\int_{u(a)}^{u(b)}F'(x)dx=\int_{u(a)}^{u(b)}f(x)dx$$

This is a way to see that even when we look at $dx$ as syntax we can do it, further more it can even be a justification for being able to look at it like Leibniz did.

More logically we can look at it like the following:

When we change variables we changing $f(x)$ to be $g(u(x))$, now we need to change $dx$ to $d(u(x))$ and really saying $f(x)dx=g(u(x))d(u(x))$ is correct! And even is used in physics and in some area in math. But we also know that $d(u(x))=u'(x)dx$ hance we can change it into $g(u(x))u'(x)dx$

Also, $dx$ can be looked as something that is not syntax, look at: https://math.stackexchange.com/a/21209/471959 and What is $dx$ in integration?

And I am sure there are more questions that discuss this over the internet

If one is working in a framework devoid of infinitesimals, naturally one would be led to deny that $dx$ in the formula for an integral is not an infinitesimal, since there is no such entity to begin with. However, when one works with an infinitesimal-enriched continuum one naturally interprets the symbol $dx$ in the formula for an integral as an infinitesimal, as is done for example in Keisler's textbook Elementary Calculus.

The change of variable formula (also known as "u-substitution") is a consequence of the chain rule for derivatives, which admits a more natural interpretation in an infinitesimal-enriched framework.

You ask: "And moreover if $dx$ is just syntax when what exactly is the integral doing if not implicitly taking the sum of infinitely many 'rectangles' with infinitely small width?" Indeed in Keisler's textbook that's precisely the viewpoint adopted. One only needs to declare $dx$ to be "just syntax" in a framework where infinitesimals aren't available.