# Does this integral make sense in some way?

I have a very simple, possibly silly question...

Can this integral make sense in some way? $$\int \frac{dx}{dx}$$ And does it actually mean something to write things like $$\int f(x)$$ without the differential?

I just got an expression of this type and I probably made some mistake along the way, but I'm still curious... my intuition from my limited knowledge of differential forms suggests that the last integral should simply evaluate to $f(x)$. How much sense does this make? I pretty much have no idea what to do here, so any help would be appreciated, especially if it includes intuitively pleasing explanations! :)

EDIT: To clarify my question some more, consider an ordinary integral $$\int f(x) dx = F(x) + c$$ where $\frac{dF}{dx} = f$(x).

We could write it as $$\int \frac{dF}{dx}dx = \int dF = F +c$$ Then, going back to my original integral, I could write it as $$\int \frac{1}{dx} dx = \int 1$$ Or for an arbitrary function: $$\int f = \int \frac{f}{dx} dx$$ But does that mean anything or is it just (incorrect) notation gymnastics? Does it make sense for "$1$" to be a differential of something? What about $f$?

• I strongly suggest to look for symbols that represent an idea, rather than the other way around. Oct 6 '13 at 17:12
• What do you mean?
– lel
Oct 6 '13 at 17:16
• It seems to me you are just putting together a few symbols that do have a meaning separately. I could write $$\int^{-1} \frac{dx}{\sum _x},$$ but why should we give this any meaning? Oct 6 '13 at 17:19
• I'm not asking should we give it meaning, I'm asking if it means anything, there's a difference between those two!
– lel
Oct 6 '13 at 17:26

In most fields the integrand needs a $\mathrm{d(something)}$ for it to make sense. This is what makes it in a hand-wavy way "infinitesimal".

However, in some fields of maths you are allowed to say things like $\omega = \sum_a w^a{\rm d}x_a$ and then you can have $\int \omega$. But this only works in fields where people carefully explained just that particular kind of integration means. One such field is differential geometry, and one important use of the notation is Stokes' Theorem

$$\int_{\partial R}\omega = \int_R {\rm d}\omega$$

• Yes, that's what I had in mind... so, could this make sense if $f$ is on a $0$-manifold? What would its boundary be? I don't know too much about these things, I had only superficial encounters with them...
– lel
Oct 6 '13 at 18:02
• @Adrian: But the important point of this is that $\omega$ already has 'd's in it.
– user14972
Oct 6 '13 at 18:03
• Does it have to have a 'd' on a $0$-manifold?
– lel
Oct 6 '13 at 18:04
• @Hurkyl only in a sense. $\omega$ is a 1-form (field) and therefore a proper target for integration. It's true that many $n$-forms are constructed from the $d$ operator, but this is not an essential thing. Oct 6 '13 at 18:46
• @SchlomoSteinbergerstein That's an interesting point. If $\omega$ was a scalar, you could still "integrate" it over a 0-dimensional manifold, but that would just be evaluating $\omega$ (and perhaps summing over a countable set of points?). I don't know what the boundary is. Oct 6 '13 at 18:48

The symbol $\int f(x)$ is usually just an abbreviation for $\int f(x)dx$. Using this I would understand the first symbol as $\int\frac{d(x)}{dx}dx$ where $\frac{d(x)}{dx}$ is the derivative of the function $g(x)=x$.

• This is not what I'm asking.
– lel
Oct 6 '13 at 17:27
• According to your comment in the comment section I believe you are asking if the symbols means anything? Did I make a mistake? My answer is that the symbols mean an abbreviation. Oct 6 '13 at 17:29
• I wasn't referring to any abbreviations, here's an update to clarify it! :)
– lel
Oct 6 '13 at 17:36
• I'm afraid those symbols appear to be notation gymnastics. Oct 6 '13 at 17:45
• It might be so, but I've seen quite a few things with differentials which make sense although they look like complete gymnastics and abuse of notation, so I don't know what to think and I can never say I'm sure, I have to ask :D
– lel
Oct 6 '13 at 17:49