Is there a good unambiguous formula for the linear operator $\frac D{e^D-1}$?
I mean,
$$x^a\to B_a(x)$$ $$\ln x \to \psi(x)$$ $$e^x\to\frac{e^x}{e-1}$$
etc.
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You mean the operator $D(e^D-1)^{-1}$? – Cameron Williams Nov 30 '20 at 18:15
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@CameronWilliams yes – Anixx Nov 30 '20 at 18:15
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This is the derivative of the operator that gives the Euler-Maclaurin Sum Formula. – robjohn Nov 30 '20 at 18:16
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@robjohn yes, we first differentiate, then discrete integrate. The problem is, discrete integration is ambiguous (needs fixing constants etc). I need them fixed as in the examples above. – Anixx Nov 30 '20 at 18:18
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Except in rare instances, the series from this operator do not converge. This is why the Euler-Maclaurin Sum Formula usually provides asymptotic formulas rather than convergent ones. – robjohn Nov 30 '20 at 18:20
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The examples in your question seem to be $f(x)\to g(x)$ where $f'(x)=g(x+1)-g(x)$. However, note that $g(x)+C$ works just as well on the right hand side, so I don't think there will be an unambiguous operator that will not involve some constant that can be adjusted. – robjohn Nov 30 '20 at 18:47
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@robjohn I answered myself – Anixx Nov 30 '20 at 19:03
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Seems, something like this:
$$\frac D{e^D-1}f(x)=\frac1{2 \pi }\int_{-\infty }^{+\infty } \frac{e^{-iwx}}{e^{-i w}-1}\int_{-\infty }^{+\infty } e^{i t w} f'(t) \, dt \, dw$$
Anixx
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