I'm looking at proving the following sum:


I know this can be done by finding the residues and taking the sum of their negatives. Curious if anyone knows how to prove it in another way - maybe with Fourier transform? I thought about Parseval's theorem except the denominator isn't a perfect square.


  • $\begingroup$ Just a curiosity, but taking the residues of what ? $\endgroup$
    – Thomas
    Mar 26, 2021 at 16:23
  • $\begingroup$ $\pi\mathrm{cot}(\pi k)(1/(64k^4+1))$ $\endgroup$ Mar 26, 2021 at 16:26
  • $\begingroup$ That is not a complex function, I do not understand $\endgroup$
    – Thomas
    Mar 26, 2021 at 16:27
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    $\begingroup$ @Thomas There are then simple poles at the integers with the residues that make up the infinite sum, and also 4 simple poles at $\pm 2\pm 2i$ whose residues give the RHS. The integral round a big square of side $2N+1$ centered at $0$ tends to $0$. $\endgroup$ Mar 26, 2021 at 16:37
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    $\begingroup$ @ancientmathematician You may be interested in my answer; it's a very non-advanced method :) $\endgroup$ Mar 31, 2021 at 21:43

1 Answer 1


I came up with this method by myself, so there may be other easier or shorter methods. Enjoy (and don't give up-it does end eventually!).

We begin first with the following $2$ formulae: $$\frac{\sin x}{x}=\prod_{r=1}^{\infty}\left(1-\frac{x^2}{r^2\pi^2}\right)\tag{1.1}$$ $$\frac{\sinh x}{x}=\prod_{r=1}^{\infty}\left(1+\frac{x^2}{r^2\pi^2}\right).\tag{1.2}$$ Multiplying the $2$ formulae together, we obtain $$\frac{\sin x \sinh x}{x^2}=\prod_{r=1}^{\infty}\left(1-\frac{x^4}{r^4\pi^4}\right).\tag{1.3}$$ Replace $x$ with $\pi x$. $$\frac{\sin(\pi x)\sinh(\pi x)}{\pi^2x^2}=\prod_{r=1}^{\infty}\left(1-\frac{x^4}{r^4}\right)=\prod_{r=1}^{\infty}\left(\frac{r^4-x^4}{r^4}\right).\tag{1.4}$$

Take the natural logarithm of both sides and differentiate with respect to $x$. This leads us to $$\pi\cot(\pi x)+\pi\coth(\pi x)-\frac{2}{x}=\sum_{r=1}^\infty\frac{-4x^3}{r^4-x^4}$$ and therefore $$\frac{2-\pi x(\cot(\pi x)+\coth(\pi x))}{4x^4}=\sum_{r=1}^{\infty}\frac{1}{r^4-x^4}.\tag{1.5}$$ Before continuing, we need to know the following formula which is obtained in this video: $$\cot(x+iy)=\frac{\sin2x}{\cosh 2y-\cos2x}-i\frac{\sinh 2y}{\cosh2y-\cos2x}.\tag{1.6}$$ Now we let $x=ze^{i\pi/4}$, so that $x^4=-z^4$: $$\frac{2-\pi ze^{i\pi/4} (\cot(\pi ze^{i\pi/4})+\coth(\pi ze^{i\pi/4} ))}{-4z^4}=\sum_{r=1}^{\infty}\frac{1}{r^4+z^4}.\tag{1.7}$$ And now let $z=\frac{1}{64^{1/4}}=\frac{\sqrt{2}}{4}$: $$\frac{\pi \frac{\sqrt2}{4}e^{i\pi/4}(\cot(\pi \frac{\sqrt2}{4}e^{i\pi/4})+\coth(\pi \frac{\sqrt2}{4}e^{i\pi/4} ))-2}{4/64}=\sum_{r=1}^{\infty}\frac{1}{r^4+1/64}=64\sum_{r=1}^{\infty}\frac{1}{64r^4+1}.\tag{1.8}$$ Let's deal with the rather nasty looking expression inside the $\cot$ and $\coth$: $$\pi \frac{\sqrt2}{4}e^{i\pi/4}=\pi \frac{\sqrt2}{4}\left(\frac{1}{\sqrt 2}+i\frac{1}{\sqrt2}\right)=\frac{\pi}{4}+i\frac{\pi}{4}.$$ So we have $$\frac{\frac{\pi}{4}(1+i)(\cot(\frac{\pi}{4}+i\frac{\pi}{4})+\coth(\frac{\pi}{4}+i\frac{\pi}{4} ))-2}{4}=\sum_{r=1}^{\infty}\frac{1}{64r^4+1}.\tag{1.9}$$ Now, using equation $(1.6)$ we can see that $$\cot\left(\frac{\pi}{4}+i\frac{\pi}{4}\right)=\frac{1}{\cosh\frac{\pi}{2}}-i\frac{\sinh \frac{\pi}{2}}{\cosh\frac{\pi}{2}}$$ and using the identity $\coth z=i\cot iz$ we can also see that $$\coth\left(\frac{\pi}{4}+i\frac{\pi}{4}\right)=\frac{\sinh\frac{\pi}{2}}{\cosh\frac{\pi}{2}}-i\frac{1}{\cosh\frac{\pi}{2}}$$ and therefore $$\cot\left(\frac{\pi}{4}+i\frac{\pi}{4}\right)+\coth\left(\frac{\pi}{4}+i\frac{\pi}{4}\right)=\frac{1+\sinh\frac{\pi}{2}}{\cosh\frac{\pi}{2}}(1-i).\tag{1.10}$$ Substituting this values into equation $(1.9)$ we can see that $$\sum_{r=1}^{\infty}\frac{1}{64r^4+1}=\frac{\frac{\pi}{4}\frac{1+\sinh\frac{\pi}{2}}{\cosh\frac{\pi}{2}}(1-i)(1+i)-2}{4}\tag{1.11}$$ which simplifies to $$\sum_{r=1}^{\infty}\frac{1}{64r^4+1}=\frac{\frac{\pi}{2}\frac{1+\sinh\frac{\pi}{2}}{\cosh\frac{\pi}{2}}-2}{4}.\tag{1.12}$$

Now for the final part. Firstly, note that $$\sum_{r=-\infty}^{\infty}\frac{1}{64r^4+1}=1+2\sum_{r=1}^{\infty}\frac{1}{64r^4+1}.$$ Hence,

$$\sum_{r=-\infty}^{\infty}\frac{1}{64r^4+1}=1+\frac{\frac{\pi}{2}\frac{1+\sinh\frac{\pi}{2}}{\cosh\frac{\pi}{2}}-2}{2}\tag{2.1}$$ and finally

$$\mathbf{\sum_{r=-\infty}^{\infty}\frac{1}{64r^4+1}=\frac{\pi}{4}\frac{1+\sinh(\pi/2)}{\cosh(\pi/2)}}$$ as required!

I hope that was useful. If you have any questions please don't hesitate to ask :) I have enjoyed working this out enormously, thanks for the challenge!!

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    $\begingroup$ Nice work!..... $\endgroup$
    – Ron Gordon
    Mar 31, 2021 at 22:39
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    $\begingroup$ I like it! Very interesting (but I suppose not surprising) how it hovers around $\pi\cot\pi z$ and the $8$-th roots of unity, just like the residue calculus solution. $\endgroup$ Apr 1, 2021 at 6:39
  • $\begingroup$ @ancientmathematician Thanks! I'm still in high school so I don't know anything about residue calculus. Do you know anywhere where I could learn it? $\endgroup$ Apr 1, 2021 at 9:26
  • $\begingroup$ @RonGordon thank you! It means a lot, coming from someone as knowledgeable as you. $\endgroup$ Apr 1, 2021 at 10:03
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    $\begingroup$ Very nice! Sorry, I just got the email that I had answers to my question. Thanks for the approach. $\endgroup$ Apr 1, 2021 at 22:05

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