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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: $$\psi^{(-2)}(z)=\int_0^z\log\Gamma(x)\mathrm dx.$$

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Wolfram Alpha gives alternative representation for psi, check: wolframalpha.com/input/?i=polygammma%28-2%2C1%2BI%29 –  joro May 19 '13 at 6:43
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1 Answer

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 $\psi$.

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