# Integral Notation

1. Jul 21, 2008

### krcmd1

Is there a substantive difference (not merely change of convention) between

$$\int{a_b}{ (f - \lambda * g)^2} = 0$$

and

$$\int{a_b}{(f - \lambda * g)^2}*dx= 0$$

Last edited by a moderator: Jul 23, 2008
2. Jul 21, 2008

### krcmd1

still not getting Latex to work.

3. Jul 21, 2008

### Defennder

I assume this is what you mean? Because they look pretty much the same to me. When using the "tex" code, do it without the space in between the word "tex" and the square brackets:

4. Jul 22, 2008

### HallsofIvy

Staff Emeritus
The only difference I can see is that the second formula has "*dx" rather than just "dx". Since multiplication is normally represented by "juxtaposition", the * is not necessary and these are identical.

5. Jul 22, 2008

### matticus

i don't see a dx in the first one...

6. Jul 22, 2008

### krcmd1

The question is: is the presence of "dx" in the integral simply a matter of convention, or does the expression change value w/ and w/o the "dx"

Thanks to all.
Ken

7. Jul 23, 2008

### matticus

not putting in the dx seems lazy and produces ambiguity. without it we must assume that x is the variable of integration.

8. Jul 23, 2008

### Integral

Staff Emeritus
I removed the spaces from your post and changed the direction of the closing slash from \ to/

9. Jul 23, 2008

### Ben Niehoff

The dx is actually meaningful. The integral sign is actually a long letter S, for "summa". The expression

$$\int_a^b f(x) \, dx$$

means "find the sum, over the interval [a,b], of the product of the value of the function f at x, times the infinitesimal width dx of the rectangular column sitting beneath f(x) at x."

10. Jul 23, 2008

### krcmd1

I asked because I'm trying to work my way through Spivak's "Calculus On Manifolds" which was highly recommended, and the integral w/o the "dx" appeared in the first set of exercises w/o an explanation.

So w/o the "dx" it isn't something like the integral of a "form?"

11. Jul 23, 2008

### Ben Niehoff

Hmm...it could be a typo, but I don't have the book, so I can't be sure.

When integrating differential forms, you do end up with expressions like

$$\int \mathbf \alpha$$

where $\mathbf \alpha$ is a differential form. Note, however, that differential forms generally consist of expressions involving dx, dy, etc. You could have $\mathbf \alpha = 4 \, dx + B \, dy + f(\rho) \, dz$, in which case the expression above simply means

$$\int 4 \, dx + B \, dy + f(\rho) \, dz$$

Note that "dx" itself is a differential form. In fact, "f(x) dx" is a differential form. In a very broad sense, you could consider a differential form to mean "a thing that can be integrated", but there are more specific rules about exactly what that means.

12. Jul 24, 2008

### krcmd1

thank you. Don't think it's a typo.

13. Jul 24, 2008

### Werg22

The dx is a remnant of the time when mathematicians did not yet develop solid foundations to the (real) number system and conceived of "numbers", called infinitesimals, that are infinitely small but, ridiculously enough, not equal to 0. When you compare the usual notation for a Riemann sum and Leibniz's notation for the integral, you see that the process of integration was viewed as being parallel to that of calculating the area of a step region. In short, integration was viewed as the infinity's equivalent to the calculation of the area of a step function. Today integration is regarded as a limiting process rather than anything that stands alone.

We still use the notation introduced by Leibniz, however, because the algebra of infinitesimals lead to results that, when reinterpreted, are actually valid under the modern-day definitions. It's also still useful to think in terms of infinitesimals in some applications - in physics, this is almost always the case.

Also, there's a branch of mathematics, called non-standard analysis, developed by Abraham Robinson and completed in 1960, that gives a solid logical foundation to the theory of infinitesimals by adopting a different structure than that of the complete ordered field as the foundation of analysis.

Last edited: Jul 24, 2008
14. Jul 25, 2008

### Freyster98

Just thought I'd throw in my \$.02....

The "dx" seems pointless until you start doing multivariate, which will include "dxdydz", or "dzdxdy", etc. This shows you the order in which you should be integrating the function and which bounds to use for that variable. For a single variable function, however, this is basically meaningless.

15. Jul 26, 2008

### zpconn

I haven't read the book, but if it's calculus on manifolds, the integrals you're encountering in the book are probably integrals over more general and higher dimensional domains. Most likely you saw an integral of a differential form over some domain.

With differential forms, the symbols "dx," "dy," and so on take on a new meaning and represent a local basis for 1-forms. The notation is designed so that the notation for differential forms integrates with and seamlessly generalizes the notation for ordinary integrals.

One consequence is that, if we write $$\omega = f(x) dx$$ for a 1-form, then the integral of this will just be

$$\int_D{\omega}$$

because the "dx" is part of $$\omega$$. D in this case is the domain over which the form is integrated.