How to define an 'infinite dimensional integral'

[Kevin_spencer2]

Hello, my question is how could we define integration on infinite dimensional spaces?, my idea is, let be the multiple integral.

$$\int_{V}dVf(X)$$ where $$X=(x_1 ,x_2 , x_3 ,...,x_n )$$

then i define a family of trial functions, in my case they are just 'step functions' so $$H(X)=H(x_1) H(x_2 ) H(x_3 )...H(x_n)$$ and define a some kind of axiomatic integral for htem (i don't know how unfortunately) then i try to apply integration by parts so integral hold to and make n -> infinite so we define an infinite dimensional integral.

By the way, is there an analogue to Euler-Mc Laurin sum formula for infinite dimensional spaces?

 

[HallsofIvy]

One problem you will have is that the various plausible "metrics", which give exactly the same topology in finite dimensions and so are equivalent, give different topologies in infinite dimensional space. That is, before you can define integrals or derivatives in infinite dimensions, you will need to specify the topology.

You might want to look at this
http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=528446
or this
http://adsabs.harvard.edu/abs/1979PhLA...73..287B

 

[dextercioby]

Unfortunately the second article is not free, unless you have a paid subscription to Elsevier.

Daniel.

 

[Kevin_spencer2]

But the infinite dimensional derivative (functional derivative) can be defined for a functional in 2 ways:

$$\frac{F[\phi +\epsilon \delta (x-y)]-F[\phi]}{\epsilon}$$ or

$$\frac{dF[\phi +\epsilon (x-y)]}{d\epsilon}$$

for epsilon tending to 0, and it yields to Euler-Lagrange equation, the question is why can't we define the integral by means perhaps of the sum, with epsilon tending to 0 in the form?

$$\sum_{n=0}^{\infty}\epsilon F[\phi +n\epsilon \delta (x-y)]$$ or if we denote the functional derivative operator $$\delta$$ then its inverse is just the functional integral operator .