I have this issue that I'm kind of clueless about, it is peripheral to what I typically do. I will state all the assumptions meticulously, even though I suspect they are not all needed. It is problem 3.1.1 in Stengers 1993 "Numerical Methods Based on Sinc and Analytic Functions".

Let $d>0$ and define $$D_d= \{w\in \mathbb{C} : \lvert \Im w \rvert < d\}$$ and for $0<\epsilon<1$ $$D_d(\varepsilon)= \{w\in \mathbb{C} : \lvert \Re w \rvert < \varepsilon \lvert, \; \lvert \Im w \rvert < d (1-\varepsilon) \}$$ Let $1\leq p <\infty$ and define $$N_p(f,D_d) = \left ( \lim_{\varepsilon\to 0} \int_{\partial D_d(\varepsilon)} \lvert f(z) \rvert^p \lvert dz \rvert \right ) ^{1/p}$$

Define $H^p(D_d)$ the family of functions that are analytic and has $N_p(f,D_d)<\infty $.

Let $0<\delta < d$ and $h>0$. Define $$D_d(n,\delta)= \left\{w\in \mathbb{C} : \lvert \Re w \rvert < \left(n+\frac12\right) h , \; \lvert \Im w \rvert < \delta \right\}$$

Let $$ E(n,\delta,f)(z) = \frac{\sin(\pi z / h)}{2 \pi i} \int_{\partial D(n,\delta)} \frac{f(\zeta)}{(\zeta-z) \sin ( \pi \zeta /h)} d \zeta. $$

I would like to show that $$ E(n,\delta,f)(\zeta) = f(\zeta)- \sin(\pi \zeta/h) \sum_{k=-n}^n \frac{(-1)^k f(kh)}{\pi( \zeta-kh)/h} $$


We have $$ \frac{E(n,\delta,f)(\zeta)}{\sin(\pi\zeta/h)} = \frac{1}{2 \pi i} \int_{\partial D(n,\delta)} \frac{f(w)}{(w-\zeta) \sin ( \pi w /h)} \,\mathrm{d}w. $$ The RHS is almost the Cauchy integral formula for $\dfrac{f(\zeta)}{\sin(\pi\zeta/h)}$, except for the fact that $\sin(\pi w/h)$ has simple zeroes at $w=kh$, $k=-n,-n+1,\dots,n-1,n$ lying inside $D(n,\delta)$ contributing to the integral (and if $\zeta$ happens to be one of these points you need to work a bit more for this double pole, for example, take limit over different $\zeta$s).

So by the homology form of Cauchy's theorem we just need to prove $$ \frac{1}{2\pi i}\int_{\partial B_\epsilon(kh)} \frac{f(w)}{(w-\zeta) \sin ( \pi w /h)} \,\mathrm{d}w=\frac{(-1)^k f(kh)}{\pi( kh-\zeta)/h} $$ where $0<\epsilon<h/2$ is chosen so that $\zeta\notin\overline{B_\epsilon(kh)}$. But near $w=kh$, we have a holomorphic $g$ vanishing at $kh$ such that $$ \sin(\pi w/h)=(-1)^k\frac{\pi}{h}(w-kh)\cdot(1+g(w)). $$ So \begin{align*} &\frac1{2\pi i}\int_{\partial B_\epsilon(kh)} \frac{f(w)}{(w-\zeta) \sin ( \pi w /h)} \,\mathrm{d}w\\ &=\frac1{2\pi i} \int_{\partial B_\epsilon(kh)} \frac{(-1)^k \dfrac{f(w)}{(w-\zeta)(1+g(w))\pi/h}}{w-kh} \,\mathrm{d}w\\ &=\left.(-1)^k \frac{f(w)}{(w-\zeta)(1+g(w))\pi/h}\right\rvert_{w=kh} \end{align*} and the result follows.


  • The function $g(z)$ is easily constructed from the Taylor series expansion of $\sin(\pi z/h)$ at $z=kh$ \begin{align*} \sin(\pi z/h)&=(-1)^k\sin(\pi(z-kh)/h)\\ &=(-1)^k\sum_{j=0}^\infty\frac{(-1)^j}{(2j+1)!}\left[\frac{\pi(z-kh)}{h}\right]^{2j+1}\\ &=(-1)^k\frac{\pi(z-kh)}{h}\times\\ &\quad\left\{1+\sum_{j=1}^\infty\frac{(-1)^j}{(2j+1)!}\left[\frac{\pi(z-kh)}{h}\right]^{2j}\right\} \end{align*} so $g(z)=\sum_{j=1}^\infty\frac{(-1)^j}{(2j+1)!}\left[\frac{\pi(z-kh)}{h}\right]^{2j}$.

  • When $\zeta=\ell h$, we can note that $$\int_{\partial D(n,\delta)} \frac{f(w)}{(w-\zeta) \sin ( \pi w /h)}\,\mathrm{d}w$$ is finite, multiplying by $\sin(\pi\zeta/h)/(2\pi i)=\sin(\ell\pi)/(2\pi i)=0$ thus gives you $0$. This is also the limit as $\zeta'\to\zeta$ of \begin{align*} E(n,\delta,f)(\zeta')&=f(\zeta')- \sum_{k=-n}^n \frac{(-1)^k f(kh)\sin(\pi\zeta'/h)}{\pi( \zeta'-kh)/h}\\ &\to f(\ell h)- \sum_{k=-n}^n (-1)^k f(kh)\color{red}{\underbrace{\lim_{\zeta'\to\ell h}\frac{\sin(\pi \zeta'/h)}{\pi( \zeta'-kh)/h}}_{\begin{cases}(-1)^\ell&\text{ if }k=\ell\\0&\text{otherwise}\end{cases}}}\\ &= 0, \end{align*} which is morally how we should make sense of the formula.

  • $\begingroup$ I did not know any of the things you mentioned, but I was able with some googling to make your answer make sense. Would you mind elaborating on how $g$ is found? And/or how double poles are handled? $\endgroup$ – Henrik Dec 18 '18 at 21:23
  • 1
    $\begingroup$ I've added that in the answer. Is there any more you want me to clarify? $\endgroup$ – user10354138 Dec 19 '18 at 12:25
  • $\begingroup$ Nope! Thank you very much! $\endgroup$ – Henrik Dec 19 '18 at 18:10

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