I encountered this quantity in my calculations and tried to simplify it. Approximate numeric calculations suggested it could be zero (more precisely, it is certainly less than $10^{-4\times10^3}$ in absolute value).

$\hspace{1in}$$\Im\;$$\psi^{(-2)}$$(1+\;$$i$$)+\frac1{4\pi}$$\text{Li}_2$$(e^{-2\pi})-\log\sqrt{2\pi}+\frac{5\pi}{24}+\frac12\stackrel{?}{=}0$.

But I am stuck finding the proof. Could you please help me? I am also curious if this formula could be generalized for other arguments of $\psi^{(-2)}(z)$, and what is the value of the real part $\Re\;\psi^{(-2)}(1+i)$ in terms of simpler functions.

The polygamma function of the negative order $-2$ can be defined as: $$\psi^{(-2)}(z)=\int_0^z\log\Gamma(x)\mathrm dx.$$