Sum of $\sum_0^\infty \frac1{n^4+a^4}$

Question

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!

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

Answer 1

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.

Answer 2

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.

Answer 3

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 $p$ $$S_p=\sum_{n=0}^p \frac 1{n^4+a^4}=\frac{1}{2 a^4}+\frac{\pi }{2 \sqrt{2} a^3} \frac{\sinh \left(\sqrt{2} \pi a\right)+\sin \left(\sqrt{2} \pi a\right)}{\cosh \left(\sqrt{2} \pi a\right)-\cos \left(\sqrt{2} \pi a\right)}-\frac{1}{3 p^3}+O\left(\frac{1}{p^4}\right)$$

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