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I have a basic question. Does it make sense to write $\Delta x$ when one of the endpoint is infinity? For example if I have an interval say, $[x_1 , x_2)$, with both $x_1$ and $x_2$ finite, then it makes sense to write $\Delta x_2 = x_2 - x_1$. What if $x_1$ is finite but now $x_2$ is replaced by $\infty$; i.e $\quad [x_1, \infty)$? How then do I calculate the change?

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I am working on a problem where I have time points say, $x_0 \lt x_1\lt \ldots x_n\lt x_{n+1} = \infty$. I have groped items in such a way that several items falls into successive intervals. For instance If I take the interval $[x_1, x_2]$ to find the change in the number of items, I would do $x_2 - x_1 = \Delta x_2$. Now items may be bigger than than $x_n$ and hence would fall in the interval $[x_n, x_{n+1}) = [x_n, \infty)$. This is where my problem lies.

Tom
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  • Why would you want to "calculate the change"? An example might help. Presumably this is just one step in some larger problem that caused you to want to know this. (Use the "edit" link to edit your question.) – David K Mar 09 '16 at 14:32
  • @ David K I have edited my question. Thanks. – Tom Mar 09 '16 at 15:30
  • It's still a little unclear how $x_1$ and $x_0$ are "time points" but $x_1 - x_0$ is a number of items.Do you mean that at some point in time there are $x_0$ items and at a later time there are $x_1$ items? To find the number of items in the last group it seems you need to know how many items you have at the end; it's not $\infty$ and it occurs before time $\infty$. – David K Mar 10 '16 at 03:40
  • The Riemann integral is only defined for finite intervals. – Chappers Mar 31 '16 at 03:17

2 Answers2

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No, this does not make sense. That's why improper integrals are defined the way they are. One does not define $$\int_0^\infty f(x) \, dx $$ by forming Riemann sums over a partition of the interval $[0,\infty)$. Instead, one first defines $$\int_0^a f(x) \, dx $$ the usual way (form Riemann sums over a partition of $[0,a]$ and then let the mesh of the partition go to zero, in other words a Riemann integral). Then one takes a limit: $$\int_0^\infty f(x) \, dx = \lim_{a \to \infty} \int_0^a f(x) \, dx $$

Lee Mosher
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  • Thanks for your answer. So what do I do in the situation I have? – Tom Mar 09 '16 at 15:31
  • Not knowing the exact nature of your problem, I can only say this much. You do not ever take $x_{n+1}=\infty$; that is nonsense because $\infty$ is not a number. Perhaps instead you can somehow reconfigure your problem to follow the model of the definition of improper integrals: first work with partitions of a finite interval and reach your conclusions in that setting; then let the upper endpoint of the finite interval approach $+\infty$. – Lee Mosher Mar 09 '16 at 15:38
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It does not. The length of an interval of that form is $\infty$. If you want to use $\Delta x$ in a particular calculation, you should do the calculation for $[x_1,x_2]$ and then take the limit as $x_2\to\infty$

Stella Biderman
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  • Thanks for your answer. Could take a look at my edited question. – Tom Mar 09 '16 at 15:32
  • @Tom I stand by my answer in the context of your edited question. Notice the similarity between the integral definition and my comment about calculating functions (of which, the integral is one example) – Stella Biderman Mar 09 '16 at 16:07