Hi.
I have this idea about developing what I call a "continuum sum", that is, a method to "add up a non-integer number of terms", i.e. to see if there is a "natural" way to assign a meaning to the expression
$\sum_{n=a}^b f(n)$
where $a$ and $b$ are non-integer fractional, real, or even complex numbers, for a target function $f$ defined on one of those domains.
What does that mean, exactly? Well, I consider this the problem of constructing a "summation operator" $\Sigma$ which is an "inverse" of the unit forward difference operator $\Delta$, but with both acting on functions on the real or complex continuum instead of just on the integers. This relationship is the same as how the integral is the inverse of the derivative. That is, applying $\Sigma$ to a function $f$ is equivalent to solving the functional equation $F(z + 1) - F(z) = f(z)$ for $F$. With such an $F$ in hand, we can then say $\sum_{n=a}^b f(n) = F(b+1) - F(a)$.
Solutions to this equation, however, are not unique. If we fill any unit real interval with some function, then the equation entails the function at the whole real line (provided $f$ is defined there.). So there are as many solutions as there are functions on the real numbers -- an uncountable $\beth_2$ possible solutions (which is even bigger than the continuum itself at $\beth_1$). In general, given any solution $F$, we can express any other $G$ as $G(x) = F(x) + \theta(x)$, where $\theta(x)$ is a 1-periodic "wobble" function.
And thus our problem is: is there some solution of this equation for a given function $f$, which is more "good" or "natural" than others? To attempt to ease the problem, it seems best to impose some restrictions on the candidate pool of input functions $f$ to consider, as in general, the tamer the function, the easier the analysis of the equation and the more techniques are available. But we don't want too much up-front restriction, or the methods available may then become relatively useless (e.g. a method that only summed linear functions and nothing else would have little to no use.).
There was a method proposed by Markus Mueller in an article called "Fractional sums and Euler-like identities" that attempts to do something like this, but its restrictions seem a little too stiff. For example, it does not appear the method is of any use in summing, e.g.
$\sum_{n=a}^b n! = \sum_{n=a}^b \Gamma(n+1)$
though a solution does exist, namely
$\sum_{k=1}^{n} k! = \frac{-e + \mathrm{Ei}(1) + \pi i + E_{n+2}(-1) \Gamma(n+2)}{e}$
using the exponential integral and En-function.
What is wanted here is to come up with some kind of "grand unified theory" of these kinds of sums: a solution for the functional equation that would be able to recover most if not all of these kind of formulae, and also enable the summation of many other functions.
For my approach, I suppose that $f$ be at least a holomorphic function of a complex variable, and that $F$ is to be as well. This is still pretty broad and does not uniquely determine a solution, but is tight enough to maximize the available tools for analysis. At this point, though, I'll suppose even further that $f$ and $F$ be entire. Then we can see if the techniques generalize to non-entire cases.
The approach I settled on (how I got to this is omitted here for the sake of brevity), is using Fourier series. If $f(z)$ is a periodic function with period $P$, then
$f(z) = \sum_{n=-\infty}^{\infty} a_n e^{\frac{2\pi i}{P} nz}$.
Now we have a simple formula for the sum of an exponential: $\sum_{n=0}^{z-1} e^{un} = \frac{e^{un} - 1}{e^u - 1}$. We can apply this to the above to get
$\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} \frac{a_n}{e^{\frac{2\pi i}{P} n} - 1} \left(e^{\frac{2\pi i}{P} nz} - 1\right)$
with the term at $n = 0$ on the right (for which the given expression fails directly with a division by 0) interpreted as $a_0 z$. This series can converge even when the given function fails some of the mentioned requirements, however it does not work for functions with harmonics of period 1.
My idea, then, was to consider a sequence of periodic entire functions $f_i$ that converge to a given function $f$ that's entire and aperiodic, but not necessarily of exponential type less than $2\pi$. Then take their continuum sums by the above formula and take the limit. If this limit exists, call it the continuum sum of $f$ itself. The questions I have, then, are, what conditions are needed on $f$ for this to work, and also, more importantly, is the limit independent of the chosen sequence of functions, and if so, what is the proof, and if not, what is a counterexample? This is why I mean by it being "viable" or not. If the limit doesn't work, this isn't of much use. I'm not necessarily interested in a complete proof but more on advice about how one would go about approaching a proof of this, useful reference material, etc. as I'd like to do some of it myself. However, if the hypothesis is false, I'd like a full counterexample.
Add: The justification for considering this approach as "natural" is based on two approaches. One is Faulhaber's formula, which gives a sum of powers by the Bernoulli polynomials, and this sum has a simple uniqueness criterion: it sends polynomials to polynomials. One can then apply this to Taylor series. The trouble is, that such a method looks only to work on an extremely limited set of analytic functions: entire functions of exponential type less than $2\pi$. This limit seems a bit too onerous. This is one of the "some methods" I mentioned as having experimented with for defining the continuum sum. It is somewhat long to give the whole derivation, but for $e^{uz}$, summed from $0$ to $z-1$, and $|u| < 2\pi$, it yields $\frac{e^{uz} - 1}{e^u - 1}$. Another justification is much simpler. We know that $\Delta e^{uz} = \left(e^u - 1\right) e^{uz}$. Thus, it'd seem sensible to presume $\Sigma \left(e^u - 1\right) e^{uz} = e^{uz}$. This leads to (assuming $\Sigma$ is linear), $\Sigma e^{uz} = \frac{e^{uz}}{e^u - 1}$, and then the sum from $0$ to $z-1$ is $\frac{e^{uz} - 1}{e^u - 1}$. I suppose one could get a third justification in that both of these methods give the same result. Finally, because $\Sigma$ is linear, it is no big step to obtain the result for periodic functions.
