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Interpretation of the integration domain symbol

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Nov26-12, 12:48 PM
P: 61
I'm having an interpretation problem with the notation used in physics, under the integration sign.

What is the proper interpretation of the domain of integration symbol, on the integration sign ?

To be more precise, consider a function [itex]F(x)[/itex] of one or several variables. Its integral on a domain [itex]D[/itex] is simply

[itex]I = \int_D F(x) \, dx[/itex].

Now, we procede to a change of variables : [itex]x \rightarrow x' = h(x)[/itex], so the function is now a new function of the new variables : [itex]F(x) \rightarrow G(x')[/itex]. Of course, the integral above gives the same number, but the limits of integration have to change to adapt to the new variables. We can write :

[itex]I \equiv \int_{D'} G(x') \, dx'[/itex].

My question is this : is the symbol [itex]D[/itex] on the integration sign actually an invariant, so [itex]D' \equiv D[/itex] ? Or is it a representation of the values that the variables are taking under the integral (so [itex]D' \ne D[/itex]) ?

In other words : do the domain of integration [itex]D[/itex] change with the coordinates transformation ?
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Nov26-12, 01:18 PM
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In general, D and D' will represent different sets of numbers, so they are different subsets of the real number line.
Nov26-12, 01:54 PM
P: 61
Quote Quote by arildno View Post
In general, D and D' will represent different sets of numbers, so they are different subsets of the real number line.
So you're saying that [itex]D[/itex] isn't invariant : it's a representation of the values taken by a given set of variables (coordinates) ?

I still have doubts on this. For example, when we define a surface integral like this :

[itex]I = \int_{\mathcal{S}} {\bf A} \cdot d{\bf S}[/itex],

the symbol [itex]\mathcal{S}[/itex] is both the domain of integration of the integral and the absolute surface in 3D space on which the integration is performed. So this implies that the domain [itex]D[/itex] is invariant. This is what puzzles me !

Nov27-12, 12:48 AM
P: 756
Interpretation of the integration domain symbol

Why do you think that a domain of integration must be invariant if you change the variable ?
Obviously not ! The domain of integration depends on the way that the function is defined.
For example, consider the function F(x)=a if -1<x<1 and F(x)=b elsewhere.
Let x=x'/2 so, F(x)=F(x'/2)=G(x') then G(x')=a if -2<x'<2 and G(x')=b elswhere.
Integrating F(x) on D=[-1,1] is the same as integrating G(x') on D'=[-2,2]
If you integrate G(x') on [-1,1] instead of [-2,2], you will obtain a different result.

Similary in 3D.
A surface (S) can be defined on many ways, i.e. with many different equations, by change of variables, or by change of coordinates (Cartesian, polar, ...) or by change of axes system. For each one of these definitions of the same surface with same border, the equations which characterize the border are different. If we integrate on (S), of course we have to consider the convenient related equation of the border, not another one, hense not always the same equation (but the border is the same on geometical viewpoint).
Nov27-12, 08:07 AM
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It is "invariant" in the sense that it refers, geometrically, to a specific set of points. When you change variables, you are changing "coordinates" and how those points are given in the new coordinates will change.
Nov27-12, 09:01 AM
P: 61
But then how should we indicate the domain below the integral sign ? As an invariant geometric object ([itex]D' \equiv D[/itex]), or as a coordinates relative set ([itex]D' \ne D[/itex]) ?

Of course, the integration limits are changing with the coordinates change (obvious !), but those limits aren't the same as indicating "[itex]D[/itex]" below the integration sign.

the question can be asked differently : What is an integration domain ? Is it an invariant geometic object on which the integration is performed, or is it the set of values of the integration variables ? I suspect it is the first one.
Nov27-12, 08:01 PM
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P: 820
Barnak, could you tell us what you mean by "invariant geometrical object"?
Nov27-12, 11:03 PM
P: 61
Quote Quote by pwsnafu View Post
Barnak, could you tell us what you mean by "invariant geometrical object"?
Well, I badly expressed myself (sorry, English isn't my primary language).

I mean an "absolute" or coordinates independant object, like a surface in 3D space. Or a region [itex]D[/itex] in space.

To me, it is almost clear (? not sure yet) that the integration domain [itex]D[/itex] is a coordinates independant thing. Of course, to explicitely perform the integration, we need some coordinates and specify the limits on the integral sign. But there should be a difference between the following two notations :

[itex]\int_{D} F \, d^3 x = \iiint_{-\infty}^{+\infty} F(x,y,z) \, dx \, dy \, dz.[/itex]
Nov28-12, 01:24 AM
P: 756
Hi !

The explanation of HallsofIvy (post #5) is clear and suffisant. This should close the discussion. The question was pertinent and the answer is obvious.
It seems to me that it would therefore be misplaced to discuss ad infinitum the gender of angels.

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