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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.

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Are you missing an $i$ in one of your exponentials? Also, how are the integrals and sums supposed to be interpreted if they happen not to converge absolutely? –  Joonas Ilmavirta Aug 23 at 16:16
    
@Joonas Ilmavirta 1 - thanks, fixed. 2 - in the most generalized way, for instance, the integral from $-\infty$ to $\infty$ of an odd function should be considered zero. Anyway if there is a proof for only the strictest case, it would be good. Using Dirac Delta if needed is permitted. –  Anixx Aug 23 at 16:19
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How does your first formula work, even for $s\geq0$ (from your MSE question)? You only evaluate the derivatives at $x$, so the value of $f^{(2)}(x)$, say, can't depend in any way on $f^{(k)}(x)$ for $k\neq2$, but this doesn't seem to be the case. It is also often natural to choose $\binom{-1}{m}=0$, yielding $f^{(-1)}=0$. Can you give references or show that the series does its job for some example function (for any $s$)? –  Joonas Ilmavirta Aug 23 at 17:29
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@Joonas Ilmavirta It is Newton series en.wikipedia.org/wiki/Newton_series#Newton.27s_series where the deltas are taken over the order of derivative rather than the function's argument. –  Anixx Aug 23 at 17:59
    
But still, I don't see how your first formula could hold true. Can you use your series to integrate a monomial, for example? –  Joonas Ilmavirta Aug 23 at 18:12

1 Answer 1

up vote 5 down vote accepted

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.

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What last step do you mean? –  Anixx Aug 23 at 19:38
    
Great! Now it is evident that the Furier formula is a natural consequence of considering integral a -1-th derivative. –  Anixx Aug 23 at 19:54
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By "this last step," I mean proving $(\heartsuit)$ as an identity of multiplication operators, rather than just an identity of analytic functions near $i$. This is important because you ultimately wanted to apply the inverse Fourier transform to both sides of $(\heartsuit)$. For this to work, both sides of the identity have to be well-defined and equal as $L^2$ functions on $\mathbb{R}$, not just analytic functions on the unit disk around $i$. –  Vectornaut Aug 23 at 23:48
    
By the way, the very first identity you wrote is obviously true, it follows from Newton series properties. It is just Newton series expansion of exponential function $g(x)=\hat{f}(\xi)(-i\xi)^x$ taken at $x=-1$. It is obviously true if converges, there is no need for a proof. But what about how the first identity was produced? –  Anixx Aug 24 at 11:02
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@Anixx, yes: if two elements of $L^2(\mathbb{R})$ have the same Fourier transform, then they are the same. In fact, the Fourier transform on $L^2(\mathbb{R})$ is invertible. If this fact is new to you, I think you might gain a lot from reading a book or taking a course that introduces the Fourier transform from a mathematical perspective (Reed and Simon's Functional Analysis is my own favorite). –  Vectornaut Aug 26 at 9:23

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