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How do I calculate the following sum $$\sum_{n\geq1}\frac{1}{n^4+1}$$

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    $\begingroup$ To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. Finally, many would consider your post rude because it is a command ("Calculate..."), not a request for help, so please consider rewriting it. $\endgroup$ – Zev Chonoles Feb 16 '13 at 22:26
  • $\begingroup$ ok :) I have calculated the sum partial. Right? $\endgroup$ – Sophie Germain Feb 16 '13 at 22:26
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    $\begingroup$ $$\sum_{n=1}^\infty \frac{1}{n^4+1}=-\frac{1}{4}(-2+(-1)^{1/4}\pi \cot((-1)^{1/4}\pi)+(-1)^{3/4}\pi \cot((-1)^{3/4}\pi))$$ $\endgroup$ – Ethan Feb 16 '13 at 22:29
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    $\begingroup$ @julien a root of unity $\endgroup$ – Ethan Feb 16 '13 at 22:41
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    $\begingroup$ @Ethan Which one? Does it mean your forumla is the same no matter what fourth root of $-1$ you choose? $\endgroup$ – Julien Feb 16 '13 at 22:43
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Hint Consider the function $f(z) = \dfrac{\pi \cot \pi z}{z^4+1}$. I assume you have seen other examples of computing series with residue calculus (since you marked the question complex-analysis).


A few more details. Let $C(z) = \pi\cot \pi z$. Then $C$ is holomorphic everywhere except at the integers, and $\newcommand{\Res}{\operatorname{Res}}\Res(C;z=n) = 1$ for every $n \in \mathbb{Z}$. The function $f$ has additional simple poles at points where $z^4+1=0$.

Let $\gamma_N$ be the positively oriented square with corners at $\pm(N+\frac12) \pm(N+\frac12)i$. A tedious computation will show that $|C(z)|$ is bounded on $\gamma_N$ by a constant, not depending on $N$. (It you haven't seen this, try to work out the details yourself. Check a textbook if you can't get it.)

The residue theorem gives

$$\int_{\gamma_N} f(z)\,dz = \sum_{k=-N}^N \Res(f;z=k) + \sum_{\alpha^4=-1} \Res(f;z=\alpha).$$

Letting $N\to\infty$, we end up with (since $|f|\to0$ quickly enough at $\infty$):

$$0 = \sum_{k=-\infty}^\infty \frac{1}{k^4+1} + \sum_{\alpha^4=-1} \Res(f;z=\alpha).$$

What remains is to compute the four extra residues, and manipulate the doubly infinite series a little, but since it's homework, I'm not going to finish things off for you.

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  • $\begingroup$ thanx ;) I will do it $\endgroup$ – Sophie Germain Feb 16 '13 at 22:47
  • $\begingroup$ I can't with this exercise, please help me. $\endgroup$ – Sophie Germain Feb 17 '13 at 1:50
  • $\begingroup$ ok, thanx :) it's very easy. $\endgroup$ – Sophie Germain Feb 17 '13 at 17:13
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First, we write the sum in the following form

$$ \sum_{n=0}^{\infty}\frac{1}{n^4+1} = \frac{i}{2}\sum_{n=0}^{\infty}\frac{1}{n^2+i} -\frac{i}{2} \sum_{n=0}^{\infty}\frac{1}{n^2-i}\quad i=\sqrt{-1}, $$

Then, we use the result

$$\sum_{n=0}^{\infty}\frac{1}{n^2+a^2} =\frac{\pi}{a} \frac{e^{2a\pi}}{e^{2a\pi}-1} \,. $$

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$$\frac{1}{x+1}=\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}+\frac{1}{x^5}..$$ $$\sum_{n=1}^\infty \frac{1}{n^4+1}=\zeta(4)-\zeta(8)+\zeta(12)-\zeta(16)+\zeta(20)...$$ $$\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2} = \zeta(2) t^2 - \zeta(4) t^4 + \zeta(6) t^6 - \zeta(8)t^8+\zeta(10)t^{10}\cdots.$$ $$\frac{\pi ti}{e^{2\pi t i}-1}-\frac{1}{2}+\frac{\pi t i}{2} = -\zeta(2) t^2 - \zeta(4) t^4 -\zeta(6) t^6 -\zeta(8)t^8-\zeta(10)t^{10} \cdots.$$ $$\frac{\pi ti}{e^{2\pi t i}-1}-\frac{1}{2}+\frac{\pi t i}{2} +\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2}=-2( \zeta(4)t^4+\zeta(8)t^8+\zeta(12)t^{12}+\zeta(16)t^{16}..)$$

I think you can figure out the rest...

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    $\begingroup$ i loved wolfram alpha too but I need the development of the exercise. $\endgroup$ – Sophie Germain Feb 17 '13 at 1:48
  • $\begingroup$ @SophieGermain Are you referring to my previous comment? $\endgroup$ – Ethan Feb 17 '13 at 3:58
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    $\begingroup$ @SophieGermain: Ethan's answer here can be turned into a full solution. Let $$f(t)=\frac{\pi t}{e^{2\pi t}-1}-\frac{1}{2}+\frac{\pi t}{2},$$ and then consider $$f(\zeta_8 t)+f(\zeta_8^3t)$$ where $\zeta_8=e^{\pi i/8}$ is an eight root of unity. Now, to get the series $\sum_{n=1}^\infty \frac{1}{n^4+1}$, you need to take the limit as $t\rightarrow 1$, and use Abel's limit theorem to justify the switching of the orders. $\endgroup$ – Eric Naslund Feb 17 '13 at 16:43
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I think that:

If we know $$\sum_{n\geq1}\frac{1}{n^4-a^4} = \frac{1}{2a^2}-\frac{\pi}{4 a^3}(\cot \pi a+\coth \pi a)$$ Then, with $a = \sqrt[4]{-1}$: $$\sum_{n\geq1}\frac{1}{n^4+1} \approx 0.57847757966713683831802219...$$ Right?

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It is enough to apply the Poisson summation formula. Since the Cauchy distribution and the Laplace distribution are conjugated via the Fourier transform, a simple partial fraction decomposition ensures

$$ \mathscr{F}\left(\frac{1}{1+x^4}\right)(s) =\frac{\pi}{\sqrt{2}}e^{-\pi\sqrt{2}|s|}(\cos+\sin)(\pi\sqrt{2}|s|)$$ hence by summing the real and imaginary part of a geometric series $$ \sum_{n\in\mathbb{Z}}\frac{1}{1+n^4}=\frac{\pi}{\sqrt{2}}\cdot\frac{(b-a)+ab(a+b)}{a^2+b^2} $$ where $a=\cot\frac{\pi}{\sqrt{2}}$ and $b=\coth\frac{\pi}{\sqrt{2}}$.

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