Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've noticed that in analysis we often treat the unit-interval $[0,1]$ differently from $[0,\infty)$, particularly in improper-integration (but certainly not limited to).

By lieu of example, consider proving that the Gamma function converges; i.e., the integral exists. The Gamma function is defined as follows:
$$ \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt\,. $$

Typical proofs I've encountered consider the two cases over the interval $[0,1]$ and $[0,\infty)$. This is because of the properties of $t^{z-1}e^{-t}$ over $[0,\infty)$. (Sorry, I'm not going to go in detail here; it's just an example.)

However, this has me wondering: what makes the unit-interval $[0,1] \subset \mathbb R$ so special? Although I'm mindful that I may be splitting-hairs, I'm seeking to understand if there's some concept that generalizes the properties of the unit-interval; perhaps this suggests why we may often have to treat it differently in, for example, integration-problems? I'm looking for something related to the closed multiplicative (group under multiplication?) of $[0,1]$.

As a start, I know that from elementary calculus, $ \lim_{x -> \infty} a^x = 0 $ if $a \in [0,1)$, $a^x=1$ if $a=1$ and $a^x=\infty$ if a>1. I'm thinking my answer lies somewhere in field-theory / group theory under the operation multiplication. Obviously the question is open-ended, but I'm hoping there is some general property about the interval between the multiplicative and additive identities of $\mathbb R$ that perhaps explain why we may often have to treat it differently.

(Again, I apologize for being so vague; just looking for someone to direct me to further reading/subjects/theorems.)

share|improve this question
1  
Well, you may think as $[0,1]$ as the "generator" of any other finite interval, formally I'd think about it as $a\cdot[0,1]=[0,a]$. Similarily, you can think about $[a,b]=b\cdot[0,1]-a\cdot[0,1]$ so usually when one proves some important properties of a function or expression over $[0,1]$ it is usually the case we can extend its properties by a "dilatation" of the unit interval. I can give you an example precisely for the $\Gamma$ function, if you want to. –  Pedro Tamaroff Apr 18 '12 at 19:27
    
Another thing to note is that (as stated above) $[0,1]$ essentially generates all compact intervals, i.e., those closed intervals that have finite endpoints. On the other hand $[0,\infty)$ is a simple non-compact closed interval, since it is not bounded. It can also be used as generator for the closed non-compact intervals in a similar way. –  Johannes Kloos Apr 18 '12 at 19:48
add comment

5 Answers

I think there are a variety of reasons why $[0,1]$ is special. Most will be related to picking $0$ and $1$ as the simplest example distinct numbers, for instance when making an interval. But there may be additional requirements selecting $0$ or $1$ as endpoints for your interval (e.g. positive number, smallest positive integer, for $1$, smallest nonnegative, for $0$).

Now regarding your example and remarks more precisely.

There are 2 different splittings when treating the convergence of the $\Gamma$ function. One for the $x$ variable (either $(-\infty,1)$ or $(1,\infty)$) and one for the $t$ variable when $z<1$. The first splitting must be made at $1$ while the second can be made anywhere in $(0,\infty)$. 1 is most often chosen in examples where it is possible to choose an arbitrary positive number because it is the simplest such number. 2 is defined as 1+1, 1 is the generator of $\mathbb N$ as a monoid (almost a group, just not requiring inverses). But you can pick any $a>0$ to divide your improper integration interval $(0,\infty)=(0,a)\cup (a,\infty)$ for $z<1$.

Regarding your intuition on the structure of $[0,1]$, the correct statement is that it is a monoid under multiplication, not a group. ($(0,1]$ is also a monoid and $(0,1)$ a semigroup, because it lacks an identity, $1$).

This observation relates to the existence of an order on $\mathbb R$ (and $\mathbb Q$, $\mathbb Z$, $\mathbb N$) which respects the additive and multiplicative structure.

1 has special properties with respect to this order, and so does the operation of taking inverses: for all $0<x<1$ we have $x^{-1}>1$ because $xx^{-1}=1>x=x\cdot 1$ and multiplication by a positive number preserves order (this is where compatibility with the algebraic structure enters).

As a final remark, if you relax the requirement of respecting the algebraic structure of $\mathbb R$ you may have a more symmetric notion of sizes of numbers, and of distances between them. You may choose the metric the same on $[0,1]$, but set distances between points in $[1,\infty]$ equal to distances between their inverses. So the distance between $1$ and $2$ is the same as that between $1/2$ and $1$, it is $1/2$, and $1$ is at the same distance from $\infty$ as it is from $0$. This is related to seeing the real line together with $\infty$ as a circle, and the complex plane with $\infty$ added as the "Riemann sphere". This is also related to the Fubini-Study metric on projective spaces.

share|improve this answer
add comment

Maybe you are not looking in the right direction here. $[0,1]$ is not a group under multiplication (which is the inverse of $0$? of $1/2$?) It is however a "normalized" (of length 1) closed interval, every closed interval of $\mathbb{R}$ is homeomorphic to it (under a map which preserves the natural order, if you want) and this implies that many of the properties of a closed interval you can wonder about are proved true once checked for $[0,1].$ For instance, closed paths (continuous functions from a closed interval to any topological space) can be assumed to be defined on $[0,1]$ without loss of generality. And so on. But you can choose any closed interval to represent all of them in this sense.

Concerning your initial motivation, any integral on $[1,\infty)$ is by definition improper (Riemann-wise, because this distinction sort of dissapears if you switch to Lebesgue integration). An integral on $[0,1]$ might be improper or might not be. If the integrand is continuous then it is not improper (meaning: we don't have to worry about convergence of the integral). But consider the function $f(x)=0$ if $x=0$ and $f(x)=1/\sqrt{x}$ if $x\in (0,1]$. It has a finite integral on $[0,1]$, and it is an improper one (it involves a limit) because $f$ has an asymptote (only a vertical one, instead of a horizontal one as in your example).

Continuous functions on (arbitrary) closed intervals are always integrable, which is not true for noncompact ones. But I would not call this a feature of $[0,1]...$

share|improve this answer
add comment

The intuitively simplest notion of integral is the Riemann integral. It deals with bounded functions on sets $B$ of finite total measure, e.g., continuous functions on the interval $[0,1]$. The exact definition via arbitrary partitions of $B$ into small "blocks" is already quite involved, but in the end you have a well determined value of such an integral. When the function under consideration is not bounded or the domain $B$ over which we want to integrate has infinite measure, as in the case $t\mapsto t^{x-1} e^{-t}$ $\ (0<t<\infty)$, a second limiting process is needed to get hold of the integral: $$\int_0^\infty t^{x-1} e^{-t}\ dt\ :=\ \lim\nolimits _{a\to 0+,\ b\to\infty}\int_a^b t^{x-1} e^{-t}\ dt\ .$$ That's why the expression defining the Gamma function is called an improper integral.

share|improve this answer
add comment

Thanks for the thoughts! Definitely some help! However, I suppose I misdirected the potential answer by discussing improper integration; I really am more concerned with $[0,1] \subset \mathbb R$ itself.

So if I'm correct with my algebra, $(0,1]$ is the set of units for $\{x ~| ~ x \in \mathbb R, x\geq 1\}$. I suppose that as a further development on my initial question is understand this distribution: why are all the units organised in this manner in the set $\mathbb R$? Can we generalise this result?

Perhaps I don't know enough field-theory, but is there some aspect of certain fields that posit this result? I.e., are we to expect the units to be contained between the multiplicative and additive identities in a field (or just true for the complete, totally-ordered field)?

Finally, I suppose I could casually define $1 \in \mathbb R$ as a "pivot point". (Sorry that what is about to follow is non-rigorous.) That is, if we consider the set $\{\displaystyle \frac{1}{(0,1]}\}$, we see have the elements $x \in \mathbb R, x\geq 1$. I find this distribution fascinating (and I suppose it could be just by construction of $\mathbb R$ -- or really because of rational fields $\mathbb Q$)

Thus, is there a theorem/result that would indicate that in (certain) rational fields we might expect this location of units? Is there any generalisation for why we find them between the additive and multiplicative identities? Now I'm repeating myself...so I I hope I've clarified what my mind is trying to convey. Any thoughts are appreciated!

(Ultimately, I think that this notion of a "pivot" is why my mind believes we, in general, treat [0,1] differently in problems like improper-integration.)

share|improve this answer
    
I'm still struggling to be clear. But I suppose that in a field like $R$, 1 is the only unit that equals it's own inverse. However, I guess as a generalised result, if $x \in R$ is greater than the multiplicative identity 1 in (in a field with such an order), we can expect to find it's inverse less than the multiplicative identity. Thus, if we're integrating from $(0,\infty)$, we are integrating over the units in R. Thus, perhaps we have to treat the elements and their inverses differently. –  A.G. Apr 19 '12 at 18:21
    
Moreover, the best I can say is that if we define the hyperbolic function $y= \frac{I}{x}$ in some field like $\mathbb R$, where $I$ is multiplicative identity in $\mathbb R$, we essentially understand why this "symmetry" occurs. However, I'm particularly interested in the generalisation of this concept: what implications does this symmetry have? Is it mandatory for fields, or can we modify it? –  A.G. Apr 19 '12 at 18:48
    
A unit in a ring is an invertible element, so in a field all nonzero elements are units. –  plm Apr 21 '12 at 19:30
    
In a field (or more generally a ring without zero-divisor) $1$ and $-1$ are the only elements equal to their inverse, we see this as follows: $x=x^{-1}\Rightarrow x(1-x)=x(xx-x)=x^2(x-1)=(x-1)=-(1-x)$, then either $x=1$ and $1-x=0$, or $x=-1$ if we can divide by $(1-x)$. And you may look at my answer for comments regarding ordered fields and the role of $1$ and the operation of taking inverse. –  plm Apr 21 '12 at 19:39
    
we, in general, treat [0,1] differently in problems like improper-integration... We do not. –  Did Sep 3 '12 at 7:08
add comment

Sorry but what you ask for seems to be based on misconceptions and is essentially a dead-end.

what makes the unit-interval $[0,1] \subset \mathbb R$ so special?

Nothing, at least as regards integration purposes. In particular, $[0,1]$ has nothing special in the examples you mention.

perhaps this suggests why we may often have to treat it differently in, for example, integration-problems?

We do not (have to treat it differently).

I'm looking for something related to the closed multiplicative (group under multiplication?) of $[0,1]$.

The interval $[0,1]$ is closed under multiplication (as many others, like the interval $[-\frac14,\mathrm e^{-1}]$). It is not a group under multiplication.

Finally:

By lieu of example, consider proving that the Gamma function converges

Here, you might be taking niceties of some presentations of the analysis of this function for deep properties of the function. To show that $\Gamma(x)$ converges for some given $x$, since the integral restricted to every compact subinterval of $(0,+\infty)$ obviously exists, one just has to look at the behaviour of the function $t\mapsto t^x\mathrm e^{-t}$ when $t\to0^+$ and when $t\to+\infty$. Thus, one could split the integral at $\pi+7$ instead of $1$ and proceed with the analysis exactly in the same way--which brings us back to the very first part of my answer.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.