How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right)  \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma  \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$

This is a partial answer which leaves one integral unevaluated. I'm not sure if it really helps or if it is just a reformulation replacing an unevaluated sum by an unevaluated integral.
We are looking for the sum
$$s = \sum_{k=3}^{\infty} \frac{\log(k)}{k^2-4}$$
Inserting
$$\log(k) = \int_0^1 \frac{x^{k-1}-1}{\log (x)} \, dx\tag{1}$$
and interchanging the order of summation and integration the integrand is given by the sum
$$i(x):=\frac{1}{\log (x)}\sum _{k=3}^{\infty } \frac{x^{k-1}-1}{k^2-4}\tag{2}$$
for which Mathematica gives
$$\begin{align}i(x)=\frac{4 x+2 x^2 -7 x^3 +x^4+4 \left(1-x^4\right) \log (1-x)}{16 x^3 \log (x)}\end{align}\tag{3}$$
now integrating gives the integral representation
$$s = s_i := \int_{0}^{1} i(x)\,dx= s(x)|_{x\to 1}-s(x)|_{x \to 0}\tag{4}$$
where antiderivative is given by
$$s(x) = \int i(x)\,dx = s_1(x) + s_2(x)\tag{5}$$
where
$$\begin{align}s_1(x) & = \int \frac{x^4-7 x^3+2 x^2+4 x}{16 x^3 \log (x)} \, dx\\
= & \frac{\text{Ei}(-\log (x))}{4}+\frac{1}{16} \text{Ei}(2 \log (x))-\frac{7 \text{li}(x)}{16}\\
& +\frac{1}{8} \log (\log (x))\end{align}\tag{6}$$
and
$$s_2(x) = \frac{1}{4}\int \frac{ \left(1-x^4\right) \log (1-x)}{ x^3 \log (x)} \, dx\tag{7}$$
Here $\text{Ei}(z)$ is the exponential integral. The integral $s_2$ remains unevaluated. And I am stuck here.
That the integral $s_{i}$ is convergent can be seen from the expansions at the endpoints of the integration interval.
We have
$$\begin{align}i(x \to 0)& \simeq -\frac{25}{48 \log (x)}+\frac{x^2}{5 \log (x)}\\
& +\frac{x^3}{12 \log (x)}+O\left(x^4\right)\end{align}\tag{8}$$
and
$$\begin{align}i(x \to 1) & \simeq -\frac{9}{16}+\frac{19 (x-1)}{32}\\
& +(x-2) \log (1-x) + O((1-x)^2)\end{align}\tag{9}$$
