Is this combination of generalized polygamma and dilogarithm actually zero? $\Im\;\psi^{(-2)}(1+i)+\frac1{4\pi}\text{Li}_2(e^{-2\pi})-\log\sqrt{2\pi}+\frac{5\pi}{24}+\frac12$ 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:
$$\psi^{(-2)}(z)=\int_0^z\log\Gamma(x)\mathrm dx.$$
 A: Edited Maple's $\psi$ disagrees with Wolfram Alpha and your integral,
so here are some conjectures with both:
According to Maple -- your equality fails with this definition of psi.
$$ 24 \Im{\psi^{(-2)}}(i)+6 Li_2(e^{-2 \pi}) / \pi + 5 \pi - 12= 0$$
$$ 24 \Im{\psi^{(-2)}}(1+i)+6 Li_2(e^{-2 \pi}) / \pi + 5 \pi + 12= 0$$
Simlarly for $Li_4$,
$$ -1440 \Im{\psi^{(-4)}}(1+i)+ 90 Li_4(e^{-2 \pi}) / \pi^3 - \pi + 220 = 0$$
Checked with precision 1000 decimal digits.
Using your integral and mpmath, these appear to hold for $\psi^{(-2)}(i)$
and $\psi^{(-2)}(2+i)$
$$ -24 \Im{\psi^{(-2)}}(i)-6 Li_2(e^{-2 \pi}) / \pi - 5 \pi + 12 +24 \log{\sqrt{\pi}} + 24 \log{\sqrt{2}} = 0$$
$$ -24 \Im{\psi^{(-2)}}(2+i)-6 Li_2(e^{-2 \pi}) / \pi + \pi - 36 +24 \log{\sqrt{\pi}} + 48 \log{\sqrt{2}} = 0$$
These were found using linear dependencies in real numbers (pari's lindep).
Wolfram Alpha finds
another expression for $\psi$.
A: I worked out the difference of Maple and Mathematica...  Let
$\mathrm{Mple}(z) = \psi^{(-2)}(z)$ according to Maple's definition and
$\mathrm{Math}(z) = \psi^{(-2)}(z)$ according to Mathematica's definition.  Then
$$
\mathrm{Mple}(z) = \mathrm{Math}(z) + az+b
$$
where
$$
a = \frac{-\log(2\pi)}{2} \approx −0.9189385
\\
b = \zeta'(-1)-\frac{1}{12} \approx −0.248754
$$
The conjecture
$$
\mathrm{Im}\big(\mathrm{Math}(1+i)\big)
+\frac{\mathrm{Li}_2(e^{-2\pi})}{4\pi}-\log\sqrt{2\pi}+\frac{5\pi}{24}+\frac12\stackrel{?}{=}0
$$
becomes
$$
\mathrm{Im}\big(\mathrm{Mple}(1+i)\big)
+\frac{\mathrm{Li}_2(e^{-2\pi})}{4\pi}+\frac{5\pi}{24}+\frac12\stackrel{?}{=}0
$$
(Whatever $\zeta'(-1)$ is, its imaginary part is zero.)  
Maple writes its thing in terms of the Hurwitz Zeta function, so the conjecture is
$$
\mathrm{Im}\left(\zeta^{(1)}(-1,1+i)\right)
+\frac{\mathrm{Li}_2(e^{-2\pi})}{4\pi}+\frac{5\pi}{24}\stackrel{?}{=}0
$$
A: The following identities hold for all real $x>0$.  Your equality is the second identity at $x=1$.
$$
-x + \frac{\pi}{12}(6x^2-1) + x\log{x} - x \log(2\pi) + 2\Im{\psi^{(-2)}(ix)} + \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = 0,\\
x + \frac{\pi}{12}(6x^2-1) - x\log{x} - x \log(2\pi) + 2\Im{\psi^{(-2)}(1+ix)} + \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = 0.
$$
My proofs are series manipulations, which I'll now sketch. To prove the first identity, consider the product expansion for the gamma function:
$$
\Gamma(z) = \frac{e^{-\gamma z}}{z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)^{-1}e^{z/n}.
$$
Take the logarithm of both sides; the RHS is now an infinite sum.  Expand the $\log\left(1+\frac{z}{n}\right)$ term as power series about $z=0$, and switch the order of summation.  Now we have:
$$
\log{\Gamma(z)} = -\gamma z - \log{z} + \sum_{k=2}^\infty \frac{(-1)^k}{k}\zeta(k)z^{k}
$$
The polygamma $\psi^{(-2)}(z)$ (by which I mean the Mathematica function $\texttt{PolyGamma[-2,z]}$) is the antiderivative of $\log{\Gamma(z)}$ with $\psi^{(-2)}(0)=0$.  Integrating term-by-term, plugging in $z=ix$, and taking just the imaginary terms, we obtain
$$
\Im{\psi^{(-2)}(ix)} = -x\log{x} + x + \sum_{\ell=1}^\infty \frac{(-1)^\ell}{2\ell(2\ell+1)}\zeta(2\ell) x^{2\ell+1}.
$$
To obtain the appropriate series expansion for $\text{Li}_2(e^{-2\pi x})$, start with its second derivative
$$
\frac{d^2}{dx^2} \frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{2\pi}{e^{2\pi x}-1} = x^{-1} + \sum_{k=1}^\infty \frac{B_k}{k!}(2\pi)^k x^{k-1}.
$$
Here, $B_k$ is the $k^{\rm th}$ Bernoulli number.  After integrating twice, the series we get for $\text{Li}_2(e^{-2\pi x})$ is
$$
\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{\pi}{12} + x\log{x} + (\log(2\pi)-1)x + \sum_{k=1}^\infty \frac{B_k(2\pi)^k}{k(k+1)k!} x^{k+1}.
$$
The $\frac{\pi}{12}$ term comes from the fact that $\text{Li}_2(1) = \zeta(2) = \frac{\pi^2}{6}$.  Using the relation between Bernoulli numbers and even zeta values, this series may be rewritten
$$
\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x}) = \frac{\pi}{12} + x\log{x} + (\log(2\pi)-1)x - \frac{\pi}{2}x^2 - \sum_{\ell=1}^\infty \frac{(-1)^\ell}{\ell(2\ell+1)} \zeta(2\ell) x^{2\ell+1}.
$$
To obtain my first identity, just add the series for $2\Im{\psi^{(-2)}(ix)}$ and $\frac{1}{2\pi}\text{Li}_2(e^{-2\pi x})$; everything cancels beyond the quadratic term.  The second identity---and identities for every $m + ix$, $m \in \mathbb{Z}$---follow from the first via the formula
$$
\psi^{(-2)}(z+1) = \psi^{(-2)}(z) + \frac{1}{2}\log(2\pi) + z\log{z} - z,
$$
which is itself a corollary of the functional equation $\Gamma(z+1)=z\Gamma(z)$.
Taking more antiderivatives will give us identities involving $\Re\psi^{(-3)},\Im\psi^{(-4)},$ etc., but this isn't particularly interesting now that we know where they come from.  All of joro's identities are now accounted for, I believe.
The real part $\Re\psi^{(-2)}(ix)$ or $\Re\psi^{(-2)}(1+ix)$ picks out the odd, rather than even, zeta values in my series expansion.  Because these numbers, $\gamma,\zeta(3), \zeta(5), \zeta(7), \ldots$, remain so mysterious to number theorists, I expect that $\Re\psi^{(-2)}(1+i)=1.13063...$ does not have an closed-form expression in terms of "simpler" elementary/special function values, appropriately defined. I also expect that a proof of any such statement, even irrationality, is beyond the scope of current knowledge---but I would be glad to be corrected on either point.
