Factorial-like interpolation problem: looking for a function that satisfies $f(x) - f(x - 1) = \frac{\sin (πx)}{πx}$ I would like a function that should satisfiy the following equality for any real number $x$:
$$ f(x) - f(x-1) = \frac{\sin (πx)}{πx} $$
But the only initial condition I have for sure is $f(0) = 1$. This immediately yields values for all integer numbers:
$$
f(x) = \left\{\begin{aligned}
0, \ \ & x=\ ...,-3,-2,-1 \\
1, \ \ & x=0,1,2\ ... 
\end{aligned}\right.
$$
But leaves no clue for what the values might be at non-integer points. Of course there are infinitely many ways to smoothly interpolate $f(x)$ on the real numbers that will satisfy the equations above, but what I'm trying to find is the most «natural» of them, like the gamma function which interpolates factorials and satisfies $f(x) / f(x-1) = x-1$. In fact, these two problems have a lot in common, so the solution to mine might be somewhat realted.
So, does anyone know / can think of a function that satisfies the aforementioned equality? As I said, the answer might be related to factorials, gamma function, binomials, etc. One of the hints may be that the following identities hold for any $x$:
$$ \frac{\sin (πx)}{πx} = \binom{x}{0}·\binom{0}{x} = \frac{1}{Γ(1-x)·Γ(1+x)} $$
Maybe there is a way to derive the desirable function via some kind of technique similar to the one used when interpolating factorials? Any tips on that would also be of great help!
 A: Consider the function
$$g(x)=\sum_{n=0}^\infty \frac{(-1)^n}{n-x}$$
defined for $x\in\mathbb R\setminus\mathbb Z$. By grouping terms as
$$\tag{1}\sum_{n=0}^\infty \left(\frac{1}{2n-x}-\frac1{2n+1-x}\right),$$
we see that this series converges for all $x\in\mathbb R\setminus \mathbb Z$. Now, define
$$f(x)=-\frac{\sin(\pi x)}{\pi}g(x).$$
We have
$$f(x)=\sum_{n=0}^\infty\frac{(-1)^n\sin(\pi x)}{\pi(x-n)}=\sum_{n=0}^\infty \frac{\sin(\pi(x-n))}{\pi(x-n)},$$
so $f(x)-f(x-1)=\sin(\pi x)/(\pi x)$. You can show that $f(x)$ is meromorphic on $\mathbb C$ with poles only at the negative integers, as $(1)$ converges uniformly for $x$ in any compact subset of $\mathbb C\setminus\mathbb Z$. This gives that $f(x)$ is actually analytic, since the (simple) poles of $g$ cancel exactly with the zeros of $f$.
(One can write
$$g(x)=\Phi(-1,1,-x)$$
where $\Phi$ is the Lerch transcendent.)
A: Here’s another (representation of) a quite natural entire function that satisfies your requirements:
$$f(x) = \frac12 + \frac1{2 \pi} \int_0^{\pi} \frac{\sin((x+\tfrac12) s)}{\sin(\tfrac12 s)} \mathrm ds.$$ The functional equation is easy to check with the addition formula for sine, and $f(0)=1$.
