closed form for $\psi^{(-2)}\left(\frac{1}{4} \right)$

Question

the polygamma function of negative order exist in a lot of complicated integrals like this integral: $$\int_0^{\frac{1}{4}} x \psi(x) dx=\left(x \psi^{(-1)}(x) \right)^{\frac{1}{4}}_0-\int_0^{\frac{1}{4}} \psi^{(-1)}(x)dx=\frac{1}{4} \ln \Gamma{\left(\frac{1}{4}\right)}-\psi^{(-2)}\left(\frac{1}{4} \right) $$ now how can we prove the closed form $$\psi^{(-2)}\left(\frac{1}{4} \right)=\frac{1}{8}\ln \left(2\pi A^9 \right)+\frac{C}{4\pi}$$ Where $A$ is Glaisher-Knkelin constant $C$ is Catalan's constant

Answer 1

From here we get that: $$\psi^{(-2)}(z)=\frac{(1-z)z}{2}+\frac{z}{2}\log(2\pi)-\zeta'(-1)+\zeta'(-1,z)$$ Additionally we can note that: $$\zeta'(-1)=\frac1{12}-\ln(A)$$ As for the derivative of the incomplete zeta function, I have not found any "nice forms" for it yet but will keep searching