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In complex analysis, there is a formula involving the residues of complex functions that one can employ to find the value of certain infinite series.

If the function $f: \mathbb{C} \to \mathbb{C} $ satisfies certain conditions$\raise0.9ex\small \hbox{a}$, sums can be evaluated by means of this theorem according to the following result: $$\lim_{N \to +\infty} \sum_{k=-N}^{N} f(k) = - \ \{ \text{sum of the residues of } \pi f(z) \cot(\pi z) \text{ at the poles of }f(z) \} . \tag{*}\label{star}$$

It appears to be the case that residues also exist and can be computed in quaternionic analysis:

I am curious$\raise0.9ex\small\hbox{b}$ as to whether it may be possible to formulate an analogue of \eqref{star} for a function $g$ of a quaternionic variable.

Questions:

  1. Does an analogue of \eqref{star} exist for functions $g$ of a quaternionic variable? So functions like $g: \mathbb{H} \to \mathbb{H}$?
  2. If so, what conditions must $g$ satisfy?
  3. Can this quaternionic analogue of the residue formula for series be applied to evaluate infinite series that cannot be evaluated by means of the residue sum formula from complex analysis?

Notes:

$\raise0.9ex\small\hbox{a}$ The complex function must satisfy the condition that $$\lvert f(z)\rvert < \frac{M}{z^{k}} $$ — where $k>1$ and $M$ are constants independent of $N$ — along the path $C_{N}$. This path goes counterclockwise along the rectangle $(N + (1/2))(1-i)$, $(N+(1/2))(1+i)$, $(N+(1/2))(-1-i)$ and $(N+(1/2))(-1-i)$ and it encloses the poles of $f$. See pp. 3–5 of Hughes - Infinite Series and the Residue Theorem for more details.

$\raise0.9ex\small\hbox{b}$ I've asked a similar question Is there a residue sum formula in quaternionic analysis? on MSE.

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  • $\begingroup$ @Wojowu No, I indeed meant what is written at the moment: at $f$'s poles. See for instance p. 5 of the Hughes document on infinite series and the residue theorem $\endgroup$
    – Max Muller
    May 1, 2022 at 18:10
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    $\begingroup$ I see, I completely misinterpreted what is going on here. Apologies. $\endgroup$
    – Wojowu
    May 1, 2022 at 18:17
  • $\begingroup$ The inequality on $f$ seems to be equivalent to saying that $f$ is homomorphic at $\infty$ with vanishing order $\ge k$, by applying the proof of Riemann's existence theorem, using the inverses of $C_N$ as contours, to the function $u\mapsto u^{-k}f(1/u)$ around $0$. $\endgroup$
    – Z. M
    May 1, 2022 at 19:26
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    $\begingroup$ @LSpice Do you know whether there is some documentation on hyperlink best practices for MO and SE websites in general? $\endgroup$
    – Max Muller
    May 2, 2022 at 10:15
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    $\begingroup$ Re, my personal guidance is to link to abstract pages instead of directly to PDFs, and to include the name of the target of any link in case of link rot. As far as I know, there is no documentation insisting that this is best practice on MO or elsewhere. I have hoped that it is a net benefit, but, if you find it a negative, then please feel free to revert or rollback, and please accept my apologies for an unwelcome edit. $\endgroup$
    – LSpice
    May 2, 2022 at 14:39

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