The following formulas give natural differintegral (that is one with naturally fixed integration constant):

$$f^{(s)}(x)=\sum_{m=0}^{\infty} \binom {s}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$

$$f^{(s)}(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}(-i \omega)^s \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega = \mathcal{F}^{-1}\left((-ix)^s \mathcal{F}(f(x))\right)$$

When $s=-1$ the expressions produce the antiderivative of the function $f(x)$. I wonder, whether there are any practical applications where the integration constant fixed this way is useful? For instance, unsurprisingly it gives $\sin(x+\frac{\pi s}2)$ as differintegral of sine, but non-trivially gives $\ln(x)+\gamma=\ln(xe^{\gamma})$ as integral of $1/x$

I also wonder whether $\ln(xe^{\gamma})$ has any intuitive meaning (especially given the analogy between $e^{-\gamma}$ and $\frac{\pi}4$).

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    $\begingroup$ What you are witnessing here is that, for the $1/x$ case, the Fourier transform 'see's the branch cut of ln on the negative real axis and its jump (as arising from circling around 0) and the divergence of the area of 1/x with regular part $\gamma$. $\endgroup$ – Jacques Carette Nov 12 '19 at 16:02

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