Prove that these two definitions of "natural" integration constant coincide when both converge These are two possible definitions of antiderivative (integral) incorporating a supposedly natural choice of an integration constant (see this question for further details).
The first one is based on Newton series, interpolated over consecutive derivatives:
$$f^{(-1)}(x)=\sum_{m=0}^{\infty} \binom {-1}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$
The second one is based on Furier transform:
$$f^{(-1)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} \frac{e^{- i \omega x}}{\omega} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$
The question is for a proof that the both definitions coincide exactly (i.e. their values at all points and not only up to a constant) when the both converge.
Without losing the generality it is possible to consider only one point, say, $x=0$, since equality at this point guarantees equality elsewhere.
Note. $f(x)$ is required to be alalytic.
 A: If $f$ is an $L^2$ function, with Fourier transform $\hat{f}$, then the identity you're trying to prove is
$$\sum_{m = 0}^\infty \binom{-1}{m} \sum_{k = 0}^m \binom{m}{k} (-1)^{m - k} (-i\xi)^k \hat{f}(\xi) = \frac{1}{-i\xi} \hat{f}(\xi).$$
In other words, you want to show that
$$\sum_{m = 0}^\infty \binom{-1}{m} \sum_{k = 0}^m \binom{m}{k} (-1)^{m - k} (-i\xi)^k = \frac{1}{-i\xi}$$
as multiplication operators on some large subspace of $L^2(\mathbb{R})$. Let's call this identity $(\heartsuit)$. The left-hand side can be rewritten as
$$\sum_{m = 0}^\infty \binom{-1}{m} (-1 - i\xi)^m.$$
Since $\binom{-1}{m} = (-1)^m$ (Kronenburg 2011), this simplifies to
$$\sum_{m = 0}^\infty (1 + i\xi)^m.$$
This is the Taylor series of $\frac{1}{-i\xi}$ at $i$, so we've proven $(\heartsuit)$ as an identity of analytic functions on the unit disk centered at $i$. I think there should be a way to extend this to an identity of multiplication operators, like you wanted, but I don't know what it is.
Readers, if you come up with a way to do this last step, please let me know! I'll make the answer community wiki so you can add it.
