# How to define an 'infinite dimensional integral'

1. Jan 9, 2007

### 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?

2. Jan 9, 2007

### HallsofIvy

Staff Emeritus
3. Jan 9, 2007

### dextercioby

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

Daniel.

4. Jan 9, 2007

### 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 .

Last edited: Jan 9, 2007