Timeline for Seeking Name of Theory for multiple integral
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 12, 2019 at 1:25 | comment | added | Brian Moehring | @KevinNivek Note that I had to add one more condition in order for the iterated integral to make sense. Basically, everything works when all your integrals are finite, but I missed a situation when $I_1I_2$ might make sense in the extended reals while the iterated integral itself is nonsense. | |
| Aug 12, 2019 at 1:17 | history | edited | Brian Moehring | CC BY-SA 4.0 | added 138 characters in body |
| Aug 12, 2019 at 0:34 | comment | added | user679268 | Thanks, I appreciate the clarification. It would seem that Linearity for addition forms a homomorphism of integrals onto itself: i.e. that $f(x + y) = f(x) \oplus f(y)$ if and only if $f(x), f(y)$ and $f(x + y)$ exists. | |
| Aug 12, 2019 at 0:31 | comment | added | Brian Moehring | In order for the equation $$\int_R [f(x) + g(x)]\,dx = \int_R f(x)\,dx + \int_R g(x)\,dx$$ to make sense, we need all the respective integrals to exist and we need the sum on the right-hand side to be meaningful. The easiest way to force this is for both $f(x)$ and $g(x)$ to be "integrable over $R$," which is exactly the prototypical scenario in which we define the integrals. | |
| Aug 12, 2019 at 0:23 | vote | accept | CommunityBot | ||
| Aug 12, 2019 at 0:23 | comment | added | user679268 | Does this linearity axiom under addition as well? As I'm sure you know the following does not hold in all cases: $$ \int_R \left[f(x) + g(x) \right] \:dx = \int_R f(x)\:dx + \int_R g(x)\:dx$$ Or have I confused your statement here? | |
| Aug 12, 2019 at 0:22 | comment | added | Brian Moehring | "Meaningful" here means that the integrals themselves should exist as extended real numbers (i.e. real or $\pm \infty$) and we should be able to take the product of their values (which is true as long as one isn't $0$ while the other is $\pm \infty$) | |
| Aug 12, 2019 at 0:20 | comment | added | Brian Moehring | Yes, linearity holds by definition for all integrals. Even further, if you proposed a new type of "integral" for which linearity doesn't hold, you'll find resistance in the mathematics community to even calling it an integral. | |
| Aug 12, 2019 at 0:16 | comment | added | user679268 | Cheers @Brian. But does linearity hold for all integrals? and if not, under what conditions does it not hold? Also, what is the exact definition of 'meaningful' here? Or is it simply the examples you have? with $-\infty$ included? | |
| Aug 12, 2019 at 0:10 | history | answered | Brian Moehring | CC BY-SA 4.0 |