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How to find the following series as pretty closed form by $a$? $$S=\sum_0^\infty \frac1{n^4+a^4}$$ I first considered applying Herglotz trick, simply because the expressions are similar. So I changed it like this... $$2S-a^{-4}=\sum_{-\infty}^\infty \frac1{n^4+a^4}$$ However, the attempt failed to find such an appropriate function like $\pi\cot\pi z$ in this post.

Next I found this post and used Fourier transform in a similar way, and the result was a nightmare!

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How on earth can I calculate the value of this series?

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  • $\begingroup$ According to Wolfram Alpha we have that $\sum_{n=-\infty}^\infty\frac{1}{n^4+a^4}=-\frac{\pi\left(\sin(\sqrt{2}\pi a)+\sinh(\sqrt{2}\pi a)\right)}{\sqrt{2}a^3\left(\cos(\sqrt{2}\pi a)-\cosh(\sqrt{2}\pi a)\right)}$, so I'm suspecting there's no short solution $\endgroup$
    – Lorago
    Commented Sep 3, 2022 at 20:23
  • $\begingroup$ I must have misjudged that that would be simpler than I thought. Thanks for letting me know, @Lorago . Nevertheless, I want to know how that is induced. $\endgroup$
    – MH.Lee
    Commented Sep 3, 2022 at 20:26
  • $\begingroup$ several solutions are here - math.stackexchange.com/questions/384780/… IMHO, the most regular way is the integration in the complex plane with the functions $\pi\cot\pi z$ or $\frac{\pi}{\sin\pi z}$- what allows to get closed forms for different sums of types $\frac{1}{n^{2k}+a}, \frac{(-1)^n}{n^{2k}+a}$, etc. $\endgroup$
    – Svyatoslav
    Commented Sep 3, 2022 at 21:09

3 Answers 3

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We start with partial fractions, $$\frac{1}{n^4+a^4} = \frac{i}{2a^2}\left(\frac{1}{n^2 + i a^2} - \frac{1}{n^2 - i a^2}\right) \tag 1$$ And then note that (Series expansion of $\coth x$ using the Fourier transform) $$\coth x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{x}{x^2+\pi^2n^2}$$ which we rearrange to give: $$\sum_{n=1}^\infty \frac{1}{\left(\frac{x}{\pi}\right)^2+n^2} = \frac{\pi^2}{2x}\left(\coth x - \frac{1}{x} \right) $$ $$\sum_{n=0}^\infty \frac{1}{\left(\frac{x}{\pi}\right)^2+n^2} = \frac{\pi^2}{x^2}+\frac{\pi^2}{2x}\left(\coth x - \frac{1}{x} \right) =\frac{\pi^2}{2x^2}\left(x \coth x + 1 \right) \tag 2$$ and so $$\sum_{n=0}^\infty \frac{1}{n^4+a^4} = \frac{i}{2a^2}\sum_{n=0}^\infty \left(\frac{1}{n^2 + i a^2} - \frac{1}{n^2 - i a^2}\right) $$ $$ = \frac{i}{2a^2}\frac{\pi^2}{2ia^2\pi^2}\left(\sqrt{i}a \pi \coth \sqrt{i}a - \sqrt{-i}a \pi \coth \sqrt{-i}a\right) $$ $$ = \frac{\pi}{4a^3}\left(e^{i\pi/4} \coth \sqrt{i}a - e^{3i\pi/4} \coth \sqrt{-i}a\right) $$ which I'm sure can be further simplified but I will leave there for now.

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    $\begingroup$ Holy god! I've never thought of that before!!!! Thanks! It seems to have been resolved perfectly. $\endgroup$
    – MH.Lee
    Commented Sep 3, 2022 at 21:00
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Excuse my short answer but the result is

$\sum _{n=0}^{\infty } \frac{1}{a^4+n^4}=\frac{1}{2 a^4}+\frac{\pi \left(\sin \left(\sqrt{2} \pi a\right)+\sinh \left(\sqrt{2} \pi a\right)\right)}{2 \sqrt{2} a^3 \left(\cosh \left(\sqrt{2} \pi a\right)-\cos \left(\sqrt{2} \pi a\right)\right)} $

No complex numbers are needed! For the calculation the path of Blitzer is appropriate but the calculations are seemingly wrong to me. My result has both sites are real as required. My solution can be verified on wolframcloud are no extra cost. Simply type in the infinite sum as given. Hope that helps so far. Here is a reference of the formula under consideration: Coth with constraints. Here some formula how to deal with the imaginary unit: Coth.

Hope that helps.

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If you write $$n^4+a^4=(n-\alpha)(n-\beta)(n-\gamma)(n-\delta)$$ $$\alpha=-\frac{(1+i) a}{\sqrt{2}}\qquad \beta=\frac{(1+i) a}{\sqrt{2}}\qquad \gamma=-\frac{(1-i) a}{\sqrt{2}} \qquad \delta=\frac{(1-i) a}{\sqrt{2}}$$ Using partial fraction $$\frac 1{n^4+a^4}=\frac{1}{(\alpha-\beta) (\alpha-\gamma) (\alpha-\delta) (x-\alpha)}+\frac{1}{(\beta-\alpha) (\beta-\gamma) (\beta-\delta) (x-\beta)}+$$ $$\frac{1}{(\gamma-\alpha) (\gamma-\beta) (\gamma-\delta) (x-\gamma)}+\frac{1}{(\delta-\alpha) (\delta-\beta) (\delta-\gamma) (x-\delta)}$$

Now,consider the partial sum $$S_p(\epsilon)=\sum_{n=0}^p \frac 1{n-\epsilon}=\psi (p+1-\epsilon )-\psi (-\epsilon )$$ Compute all sums and use the asymptotics of the digamma function for large $p$ to obtain for large $p$ $$S_p=\sum_{n=0}^p \frac 1{n^4+a^4}=\frac{1}{2 a^4}+\frac{\pi }{2 \sqrt{2} a^3} \frac{\sinh \left(\sqrt{2} \pi a\right)+\sin \left(\sqrt{2} \pi a\right)}{\cosh \left(\sqrt{2} \pi a\right)-\cos \left(\sqrt{2} \pi a\right)}-\frac{1}{3 p^3}+O\left(\frac{1}{p^4}\right)$$

$$\color{red}{S_\infty=\sum_{n=0}^\infty \frac 1{n^4+a^4}=\frac{1}{2 a^4}+\frac{\pi }{2 \sqrt{2} a^3} \frac{\sinh \left(\sqrt{2} \pi a\right)+\sin \left(\sqrt{2} \pi a\right)}{\cosh \left(\sqrt{2} \pi a\right)-\cos \left(\sqrt{2} \pi a\right)}}$$

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