Previously posted(not by me) here without any context and received no response
I’m trying to prove the following neat identity:
$$ \sum_{j=1}^{\infty} \alpha_j^{-6} \left[ \frac{\sin \alpha_j - \sinh \alpha_j}{\cos \alpha_j + \cosh \alpha_j} \right]^2 = \frac{1}{80} $$
where the $\alpha_j$ are the positive solutions to the equation
$$ (\cos \alpha)(\cosh \alpha) + 1 = 0 $$
This identity follows from a problem in statistics, that of finding the distribution of a certain functional of a Gaussian process $\eta(t)$ with covariance kernel
$$ E[\eta(t), \eta(u)] = K(t,u) = \begin{cases} \frac{2}{3}(3t^2u - t^3) & 0 \le t \le u \le 1 \\ \frac{2}{3}(3u^2t - u^3) & 0 \le u \le t \le 1 \end{cases} $$
My initial approach was to observe that the transcendental equation
$$ \cos\alpha\cosh\alpha + 1 = 0 $$
can be written as the zeros of the entire function
$$ f(z) = \cos z\;\cosh z + 1 \;=\;\tfrac12\bigl(e^{iz}+e^{-iz}\bigr)\,\tfrac12\bigl(e^z+e^{-z}\bigr)+1, $$
so one might try to form a Weierstrass product for $f(z)$ and then exploit logarithmic differentiation to extract sums of powers of $\alpha_j$. In particular, if
$$ f(z)=C\,\prod_{j=1}^\infty\Bigl(1-\frac z{\alpha_j}\Bigr)\Bigl(1+\frac z{\alpha_j}\Bigr), $$
then
$$ \frac{f'(z)}{f(z)} =\sum_{j=1}^\infty\Bigl(\frac1{z-\alpha_j}+\frac1{z+\alpha_j}\Bigr). $$
However, carrying through this program and matching residues to pick out the specific combination $\bigl[\frac{\sin\alpha_j-\sinh\alpha_j}{\cos\alpha_j+\cosh\alpha_j}\bigr]^2$ has become algebraically quite heavy, and I haven’t been able to isolate the $1/80$ yet.
I also tried expanding
$$ \sin z-\sinh z,\quad \cos z+\cosh z $$
around each root $\alpha_j$ to understand their behaviour, but the expansions give rise to nested hyperbolic trigonometri series that I can’t seem to re‐sum neatly. Any pointers toward a cleaner contour or orthogonality argument (perhaps treating the $\alpha_j$ as eigenvalues of some boundary value problem) would be greatly appreciated!
I would like to see a proof using contour integration.
