If a single integral is the area under a curve, and a double

the hypervolume underneath a spacetime

>>8840712
sum of density of volume elements resulting in the total amount

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What about an infinite integral?

>>8840812
you mean a barnett integral?

Related question:
If F(b) - F(a) is the area under the curve between x=a and x=b, what is just a single F(x) under this definition of the integral? The area of an infinitely thin line segment from the x axis to the y value?

>>8840824
f(b)-f(a) is actually the integral of f on the boundary of [a,b] (the two point set {a,b} ) traversed with positive orientation. Since by the generalized stokes theorem that equals the integral of f'(x) over the whole set ( [a,b] ). It makes sense if you think about it.
One single f(b) doesn't really have any special meaning, but it equals the integral itself if f(a)=0. Otherwise it's the integral from p to b, for some p where f(p)=0
Check out the proof of the fundamental theorem of calculus in any analysis book if you're interested in the f(b)-f(a) formula

>>8840812
Actually, yes.
https://en.wikipedia.org/wiki/Bochner_integral

>>8840858
idk senpai. 'double integral' usually refers to the dimension of the domain, not the range. i'd say that weiner measure is closer to the intuitive meaning of "infinite integral".

>>8840870
>weiner measure

>>8840870
Oh, sure. I mean, you can also integrate functionals over Hilbert space

>>8840858
How can we defined an integral on a two point set? I see what you're saying about the generalized stokes, I just don't know how to evaluate the integral of f on a finite set like {a,b}

>>8840712
It depends on what you're integrating.
If you're just integrating 1 wrt volume, it's the volume of some object determined by your bounds of integration.
If you're integrating some function f(x,y,z), which describes a 4 dimensional object, over some region determined by the bounds of integration (the domain of this 4d object) it's the "hypervolume" of that object.
Depending on the function you're integrating, you could interprete the results in different ways. E.g. if you integrate the density function of a 3D object wrt volume, you get it's mass.

>>8840917
Integration over a finite set is just standard summation
The riemann definition is specifically a continuous analog since you've got the whole |(x_n)-(x_n-1)|*|(y_n)-(y_n-1)|*... thing going on in each term, but you can recover the discrete case with the lebesgue definition. Clearly the power set of {a,b} is a sigma algebra, and you just equip it with the counting measure, i.e. m({a,b})=2, m({a})=m({b})=1, m(0)=0. Then integrating some function with respect to m reduces to taking the riemann integral from 0 to inf of a finite step function, i.e. one that is 2 from 0 to min[f(a),f(b)], 1 from min[f(a),f(b)] to max[f(a),f(b)], and 0 else (assuming f(a),f(b)>0), which is just f(a)+f(b). This is a common motivating example for the lebesgue integral, next to integrating p(x)=[0 if x is irrational; 1 if x is rational] over R.

You're "going the other way" over a, though, in the definite integral of f'(x), so it's negative. Think of integrating a gradient over a closed curve, the slope is positive or negative at a point depending on the direction you're going, so that it always comes back to 0

>>8840858
Wow Barnett integral sound so awesome, correct me if I'm wrong: so since the progressive measurable L2 martingales form a Banach space, is it possible to extend a stochastic integral like in a Lebesgue-stieljes sense (as a L2 limit). Like when you find the intensity of a Poisson process defined as the cardinality of the intersection of a random countable set with a borelian?

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Is it possible to integrate a linear operator over a set of functions?

>>8841083
Idk. Haven't really done any stochastic calculus, just the discrete-time asset valuation models

>>8841091
Yes, if and only if it's a Banach operator

>>8841083
Also there's no such thing as a Barnett integral, it's just a meme referencing Jacob Barnett

>>8840719
Wat?
That's the deSitter Space.