plz explain kouznetsov slowly

[bo198214]

and does that really converge ?? why ?

It looks so, nobody knows yet why.

so lets say we use dummy variables : f(i) = a , f(2i) = b , f(3i) = c

but approaching an integral from -oo to + oo with a few integer points of dummy variables ?
...
is f(i) , f(2i) , f(3i) , ... a good choice or do you mean suitable dense rather as in " the rationals are dense in the reals " ?

"Dense" in the normal man's understanding, = tight, close, etc.
I mean we want to numerically approximate a function (here the function f on \( i\mathbb{R} \)), so we place enough supporting points and calculate the function values there.
The more exact we want to be the more points we place, the denser the grid.

i even doubt if numeric integration is a good approximation of the whole integral !? afterall it goes from -oo to + oo.

Kouznetsov does it by integrating to a bound of \( \pm 10 \) instead of \( \pm \infty \).

am i correct in assuming you use dummy variables for the values of f(i) , f(2i) etc and then try to solve it by replacing the integral on the RHS with " rectangles " and the dummy variables ?

Ya not at integer points, ususally more dense. Kouznetsov uses Gauss legendre integration which places the supporting points as zero's of the legendre polynomials.

and then further try to control those dummy variables by placing upper and lower bounds on their values ?

No its just assumed that at "i*infinity", i.e. at the value using for approximating i*infinity, the value of f is the fixed point (and at "-i*infinity" the conjugated fixed point).

assuming convergence - hence existance - , how is this necc unique ?

Well if it satisfies a certain uniqueness criterion, which it seems to do.
Which is basically injectivity on the vertical strip \( 0\le \Re(z)\le 1 \).

can you proof the recurrence to have a single unique solution ?

A recurrence is a recurrence. Either it converges then this is a solution or not, then its not a solution.