After reading Show $\sum_{n=0}^\infty\frac{1}{a^2+n^2}=\frac{1+a\pi\coth a\pi}{2a^2}$ and noodling around on wolfram alpha, I discovered

$$ \begin{align} &\coth(x \pi)=\frac{x}{\pi}\sum_{n=-\infty}^\infty\frac{1}{x^2+n^2} & \cot(x \pi)= \frac{x}{\pi}\sum_{n=-\infty}^\infty\frac{1}{x^2-n^2} \\ & \text{csch}(x \pi )=\frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{(-1)^n }{x^2+n^2}} &\csc(x \pi)= \frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{(-1)^n }{x^2-n^2}} \\ & \tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}} &\tan(x \pi) = \frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2-4x^2}} \end{align} $$

I suspect (but don't know for sure) that these can all be justified by milking the techniques from the link above.

Go Go Gadget Calculus

We should be able to derive a few more identities. For example,

$$ \Big(\cot(x \pi) \Big)'= \Big(\frac{x}{\pi} \sum_{n=-\infty}^\infty{\frac{1}{x^2-n^2}} \Big )'$$

$$ \Big(\pi csc(\pi x)\Big)^2= \sum_{n=-\infty}^\infty\frac{x^2+n^2}{(x^2-n^2)^2}$$

And after some manipulations we can find

$$ \frac{\pi^2\csc^2(\pi/x)}{x^2}= \sum_{n=-\infty}^\infty\frac{1+(xn)^2}{(1-(xn)^2)^2} $$ Lovely. We can write then $$\frac{\pi^2}{9}=\sum_{n=-\infty}^\infty {\frac{1+(6n)^2}{(1-(6n)^2)^2}}$$ I have a feeling at this point that this must be a well-studied subject and I wonder where I can find some more identities of this class. Does anyone have a link/resource where I can read more on these. I don't really need their derivations if they are just the techniques of the link above + elementary calculus techniques. I am just looking for a well organized list that I can refer to.

  • $\begingroup$ $+1$ just for the "go go gadget calculus," because I get the reference. XD $\endgroup$ – Frpzzd Sep 15 '18 at 23:54
  • $\begingroup$ Also... How do you render $\csch$? $\endgroup$ – Mason Sep 15 '18 at 23:58
  • 1
    $\begingroup$ You can always use \operatorname{csch} $\endgroup$ – MPW Sep 16 '18 at 0:09
  • $\begingroup$ Might want to newcommand it if you're gonna use it a lot. Just type \newcommand{\csch}{\operatorname{csch}} at the top of your question and everytime you type \csch it turns it to \operatorname{csch}. It's quicker and easier that way. $\endgroup$ – Frank W. Sep 16 '18 at 4:31
  • $\begingroup$ This and This seem like they could be good resources. $\endgroup$ – Mason Sep 21 '18 at 23:27

I don’t have a list, but I can present the method to make a list systematically.

Let $P(x,n),Q(x,n)$ be two polynomials of $x,n$.

Suppose $$Q(x,r_1(x))=Q(x,r_2(x))=\cdots=Q(x,r_k(x))=0$$ for all $x$.

Suppose the sum $$S(x)=\sum^\infty_{n=-\infty}\frac{P(x,n)}{Q(x,n)}$$ converges whenever $r_{(\cdot)}(x)\not\in\mathbb Z$.

Then, by residue theorem, $$S(x)=-\pi\sum^k_{n=1}\operatorname*{Res}_{z=r_n(x)}\frac{\cot(\pi z)P(x,z)}{Q(x,z)}$$

(It is an issue that for some $x$, we might have $r_p(x)=r_q(x)$. When two or more $r$ functions take the same value, the residue should only be evaluated once.)


There is some nice things to see here:

$$\tanh(x \pi)=\frac{4x}{\pi}\sum_{n=-\infty}^\infty{\frac{1}{(2n+1)^2+4x^2}}$$

$$\Big(\tanh(x \pi) \Big)'= \Big( \frac{1}{\pi}\sum_{n=-\infty}^\infty{\frac{4x}{(2n+1)^2+4x^2}} \Big)'$$

$$ \pi\operatorname{sech}^2(\pi x)=\frac{1}{\pi}\sum_{n=-\infty}^{\infty}\frac{4\left(4n^2+4n-4x^2+1\right)}{\left(4n^2+4n+4x^2+1\right)^2}$$

Which is nice: taking $x=0$ we have


Interesting. I wonder if we can use this to demonstrate Takebe Kenko's somewhat similar looking (this can be found- with a tiny typo in denominator- on the last page of this):

$$\pi^2=4\Big(1+\sum_{n=1}^{\infty} \frac{2^{2n+1}(n!)^2}{(2n+2) !} \Big) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.