Beautiful sum of trigonometric roots I was studying the sinc function, it led me to the study the following equation on $\mathbb{R}$
$$
x=\tan\left(x\right) \ \ \ \ \left(\star\right)
$$
The equation $\left(\star\right)$ has a unique solution $x_n$ on $\displaystyle I_n=\left]\frac{\pi}{2}+n\pi, \frac{\pi}{2}+\left(n+1\right)\pi\right[$ for $n \in \mathbb{Z}$ with $x_{n}=-x_{-n}$ and $x_0=0$, which allows us to define a sequence on $\mathbb{N}$ only.
I've read somewhere the following ( astonishing ) equality

$$
\sum_{n=0}^{+\infty}\frac{1}{\left(x_n\right)^2}=\frac{1}{10}
$$

This sum does exist, because I've shown that
$$
\frac{1}{\left(x_n\right)^2} \underset{(+\infty)}{\sim}\frac{1}{\pi^2 n^2}
$$
But I dont know how to compute it. I thought about residue theorem making $x_n$ appears in a pôle but cannot find a way to prove it. Any help would be great.
 A: If the roots of an analytic function $f(z)$ are complex numbers $\alpha_n$ for integer $n$, then we can attempt to compute summations of the form:
$$S_r = \sum_{n=0}^{\infty}\frac{1}{\alpha_n^r}$$
for integer $r$, by considering the contour integral:
$$\oint_{C(R)}\frac{1}{z^r}\frac{f'(z)}{f(z)}dz$$
where $c(R)$ is a circle with the origin at the center. The zeros $\alpha_n$ of $f(z)$ inside the contour are then simple poles with residue $\frac{1}{\alpha_n^r}$. If the contour integral tends to zero for $R\to\infty$, then $S_r$ plus the residue at zero must be zero. In that case the summation is thus given by minus the coefficient of $z^{r-1}$ in the series expansion of the logarithmic derivative of $f(z)$. We can then also say that if the condition for the contour integral to tend to zero is met, then:
$$S_r = - r \text{ coefficient of } z^r\text{ of} \log\left[f(z)\right]$$
In your case, you can take $f(z) = \sin(z) - z\cos(z)$
and then you'll easily find that the sum over all zeros of the squared reciprocals is $\dfrac{1}{5}$ and this is twice the summation of the zeroes on the positive real axis.
