First consider the function inside the integrand without the f(x) as 2*n increases:

See that as 2*n gets larger this function becomes more and more narrow around the integers. It ends up that there is an exact formula for the area over an interval like 1/2..3/2, which is:

for example:

So to see what the formula I first posted does, first consider this subformula:

This should equal f(1) as n increases, because the narrow spike of the sin function multiples with f(x) mostly in the neighborhood of x=1 and the area of that spike is normalized to 1 by dividing by it's actual area using the 4^n/binomial(2*n, n)

So for example if f(x) is 1/x^2 and we look at the area around x=3:

Let's look at when n = 10:

So we can find expressions like this for n = 20:

which says that 1/x^3 or zeta(3) should be roughly equal to 1.217707604... summed to infinity via these Ci functions which are equal to:

Unfornutaely I don't know exactly how maple figures these integrals as n gets larger but it looks like there will be a way to write a generalized sum for any n.

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