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 and not very rigorous (I don't know how to make it rigorous) but i think this way might be able to produce some results.
$$ \sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} - \sum_{k=3}^{\infty} \frac{\ln(k)}{k^2} = 4\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2(k^2-4)} $$
Now Consider the function
$$f(a) = \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)}$$
such that :
$$\lim_{a \to 2}\, \frac{d}{da}\, f(a) = -\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2(k^2-4)} $$
Then using
$$ \int_0^1 x^{k-1} \, dx = \frac{1}{k} $$
One can show that if $a$ is an EVEN INTEGER greater than $0$
$$ \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)} = \left.\left(\frac{1}{2^{a+2}x}+\frac{(4^{\frac{a}{2}+1}-3)x}{3*2^{a+2}}+\frac{(2a+1)x^2}{4*2^{a+2}} +\frac{\ln(1-x)}{2^{a+2}x^2}-\frac{x^2\ln(1-x)}{2^{a+2}}-\sum_{n=1}^{\frac{a}{2}}\frac{\operatorname{Li}_{(a-2n+2)}(x)}{2^{2n}} \right)\right\vert_{x=0}^{x=1} $$
Hopefully someone is able to produce some results from this , cheers.
Edit #1
$$ \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)} = \frac{1}{3} +\frac{1}{2^{a+3}} + \frac{1}{2^{a+4}}+\frac{a}{2^{a+3}} - \sum_{n=1}^{\frac{a}{2}}\frac{\zeta{(a-2n+2)}}{2^{2n}} $$
Edit #2 (27/05/2021)
One last remark before i stop trying the problem. Once Again , not rigorous.
Consider the following identity for EVEN INTEGERS and $ x\ge 2$ which i will not prove but you can find in a similar manner here in result #20 :
https://mathworld.wolfram.com/InfiniteProduct.html
$$\prod_{n=3}^{\infty} \left(1-\frac{4}{n^x}\right) = \frac{1}{6\, \pi^{x/2} \left(1-\frac{1}{2^{x-2}}\right) i^{x/2-1}} \prod_{n=1}^{x/2} \sin\left(\pi \,2^{2/x} \,(-1)^{2n/x}\right) $$
Taking Natural Logs of both sides and differentiating with respect to x and some switching around we can obtain a neat little identity :
$$ -\frac{i \pi}{4} - \frac{4 \ln{(2)}}{2^x-4} - \frac{\ln(\pi)}{2} = 4 \sum_{n=3}^{\infty}\frac{\ln(n)}{n^x-4} - \frac{d}{dx} \sum_{n=1}^{x/2} \ln\left(\sin\left(\pi \,2^{2/x} \,(-1)^{2n/x}\right)\right) $$
Note : One can encounter similar sums of these forms the one with the $**\ln(\sin)**$ inside and these arise when dealing with Catalan constant ,$\zeta(3)$ and PolyGamma functions as far as i know, we simply don't understand their behavior yet. Providing further evidence that at least right now , the sum you asked for may not have a closed form.
