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For $x,y\in \mathbb{R}\backslash \mathbb{Z}$ let $$ f(x,y)=\sum_{n\in\mathbb{Z}} \frac{1}{(n-x)(n-y)} $$ Is there a closed formula for $f(x,y)$?

What is known:

We have $$ f(x,x)=\left(\frac{\pi}{\sin(\pi x)} \right)^2, $$ see, e.g. http://www.math.chalmers.se/~wastlund/Cosmic.pdf or

https://en.wikipedia.org/wiki/Basel_problem

Furthermore we have $$ f(x,y)\cdot (x-y)=\sum_{n\in\mathbb{Z}} \frac{x-y}{(n-x)(n-y)}=\sum_{n\in\mathbb{Z}} \left(\frac{1}{n-x}-\frac{1}{n-y} \right). $$ Hence for $m,k\in \mathbb{Z}$ we have $$ f(x,y)\cdot (x-y)=f(x+m,y+k)\cdot (x+m-(y+k)) $$ If $x-y\in \mathbb{Z}$ we get $f(x,y)=0$

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    $\begingroup$ According to Maple, $$f(x,y) = \frac{\pi (\cot(\pi x) - \cot(\pi y))}{y - x} $$ $\endgroup$
    – Robert Israel
    Commented Nov 20, 2020 at 19:05
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    $\begingroup$ You can easily find a proof of this formula for $f(x,y)$ using Euler's formula for the cotangent, see Chapter 23 (Cotangent and the Herglotz trick) in Proofs from the Book. $\endgroup$
    – François Brunault
    Commented Nov 20, 2020 at 20:40
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    $\begingroup$ This is a special case of the standard undergraduate complex variable exercise: $$\sum_{n\in Z}R(n)=-\pi\sum_a res_aR(z)\cot\pi z,$$ where $R$ is a rational function without poles at integers, $R(\infty)=0$ of order at least $2$, and summation in the right hand side is over poles of $R$. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 20, 2020 at 21:55

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