I was wondering if there is a closed form expression for the zeros of the following equation:
$$\sum\limits_{n=1}^\infty\frac{1}{n^4 + x^2} \text{ where } x \in \rm I\!R$$
If it not exists, could you suggest a numerical method for calculate the approximate values of these zeros.
Thanks you in advance for every response!
See this answer in which we determine that, for $a \gt 0$,
$$\sum_{n=-\infty}^{\infty} \frac{1}{n^4+a^4} = \frac{\pi}{\sqrt{2} \, a^3} \frac{\sinh{(\sqrt{2} \pi a)}+\sin{(\sqrt{2} \pi a)}}{\cosh{(\sqrt{2} \pi a)} -\cos{(\sqrt{2} \pi a)}}$$
Therefore,
$$\sum_{n=1}^{\infty} \frac{1}{n^4+x^2} = \frac12 \left [ \frac{\pi}{\sqrt{2 |x|^3} } \frac{\sinh{(\pi \sqrt{2 |x|} )}+\sin{((\pi \sqrt{2 |x|})}}{\cosh{((\pi \sqrt{2 |x|})} -\cos{((\pi \sqrt{2 |x|})}}- \frac1{x^2} \right ] $$
Not sure how you expect to derive zeroes from the sum.