Sum of the squares of the reciprocals of the fixed points of the tangent function The sum of the squares of the reciprocals of the positive fixed points of the tangent function is $1/10$.
I've seen this proved by means of residues, but I don't remember the details.
I've also heard it asserted that it that it can be done by means of Green's functions.
What proofs of this fact are published or otherwise known?
PS: Maybe it is of interest to note that the positive fixed points of the tangent function are also the abscissas of the extreme values of the function $x\mapsto\dfrac{\sin x} x.$

Here is a very different proof :
Let $0 < x_1 < x_2 < x_3 < \cdots $ be the positive roots of equation 
$$\tan(x)=x \tag 0$$
These roots are such that 
$$\sin(x_k)= x_k \cos(x_k) \tag 1 $$
(this relationship will be used later on).
We have to show that :
$$\sum_{k=1}^\infty \frac{1}{x_k^2}=\frac{1}{10} \tag 2$$
The framework is Hilbert space $L^2[0,1]$ of square integrable functions on $[0,1]$ with bilinear form $\langle f|g\rangle:=\int_0^1fg$. 
Let us consider its closed subspace defined by 
$$E=\{f\mid f(0)=0 \ \ \& \ \ f(1)=f'(1) \ \ \& f \in C^2\}, \tag 3 $$
itself a Hilbert space.
Operator $\Delta : f \mapsto f''$ is a self-adjoint operator on $E$, due to the fact that a double integration by parts gives :
$$\int f g'' = \int f'' g \ \ \iff \ \ \langle f|\Delta g \rangle=\langle \Delta f| g \rangle \tag 4 $$
The kernel of $\Delta$ is one-dimensional : it is the set of linear functions $f(x)=ax$. Let $\varphi_0(x)=x$ be the basis of this kernel. 
Functions defined by $\varphi_k(x)=\sin(x_kx)$ (with $x_k$ as defined in (2)), due to property (1), belong to space $E$ and moreover are such that :
$$\Delta \varphi_k = x_k^2 \varphi_k \tag 5 $$
i.e., they are eigenvectors of operator $\Delta$ with associated eigenvalues $x_k^2$.
Their norms are found to be 
$$\|\varphi_0\|^2=\frac{1}{3} \ \ \text{and} \ \ \|\varphi_k\|^2=\frac{1}{2}\sin^2(x_k)\tag{6}$$
Being eigenvectors of a self-adjoint operator, functions $\varphi_k$ are mutually orthogonal. We admit that they constitute an Hilbert basis of $E$, allowing us to write, for any $f \in E$ :
$$f=\sum_{k=0}^\infty \frac{1}{\|\varphi_k\|^2} \langle f
\mid \varphi_k \rangle\tag{7}$$
Now consider the particular function $f(x)=x^3-2x^2 \in E$.
We have the following chain of equalities :


*

*in the case [$k>0$] : 


$$\langle f\mid \varphi_k \rangle=\langle f\mid \frac{1}{x_k^2}\Delta\varphi_k \rangle=\frac{1}{x_k^2}\langle \Delta f\mid\varphi_k \rangle=\frac{1}{x_k^2}(-6\langle \varphi_0 \mid \varphi_k \rangle+  \langle 4\mid\varphi_k \rangle) = \frac{4}{x_k^2}(1-\cos(x_k))\tag{8}$$


*

*in the case [$k=0$] : 


$$\langle f\mid \varphi_0 \rangle=\int_0^1 x(x^3-2x^2) \, dx=-\tfrac{3}{10}\tag{9}$$
Now, using (7)
$$f(x):=x^3-2x^2=-\frac{9}{10}x+\sum_{k=1}^\infty \frac{8}{x_k^2(1+\cos(x_k))}\sin(x_kx)\tag{10}$$
Let us terminate by computing, thanks to (10), the expression  $f'(0)+f'(1)$ in two ways (with the LHS and the RHS), from which we deduce (2).
