Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).


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

share|improve this question
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

1 Answer 1

up vote 5 down vote accepted

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 $\psi$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.