I'm looking at proving the following sum:
$$\sum_{k=-\infty}^{\infty}\frac{1}{64k^4+1}=\frac{\pi}{4}\frac{1+\mathrm{sinh}(\pi/2)}{\mathrm{cosh}(\pi/2)}$$
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.
Thanks
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
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!!
