# 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.$$## 3 Answers The following identities hold for all real x>0. Your equality is the second identity at x=1.$$ -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, $$which is itself a corollary of the functional equation \Gamma(z+1)=z\Gamma(z). Taking more antiderivatives will give us identities involving \Re\psi^{(-3)},\Im\psi^{(-4)}, etc., but this isn't particularly interesting now that we know where they come from. All of joro's identities are now accounted for, I believe. The real part \Re\psi^{(-2)}(ix) or \Re\psi^{(-2)}(1+ix) picks out the odd, rather than even, zeta values in my series expansion. Because these numbers, \gamma,\zeta(3), \zeta(5), \zeta(7), \ldots, remain so mysterious to number theorists, I expect that \Re\psi^{(-2)}(1+i)=1.13063... does not have an closed-form expression in terms of "simpler" elementary/special function values, appropriately defined. I also expect that a proof of any such statement, even irrationality, is beyond the scope of current knowledge---but I would be glad to be corrected on either point. • I like your answer. I have one, similar but a litte more complicate, but you posted your better one, one or two hours before I have mine ready. – juan Jan 15 '15 at 19:42 Edited Maple's \psi disagrees with Wolfram Alpha and your integral, so here are some conjectures with both: According to Maple -- your equality fails with this definition of psi.$$ 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$$These were found using linear dependencies in real numbers (pari's lindep). Wolfram Alpha finds another expression for \psi. • BTW, Maple has different definitions for \psi^{(-2)}(z) and \psi^{(-1)}(z) than Mathematica (and Wolfram Alpha). – Vladimir Reshetnikov Jan 6 '15 at 18:12 • Maple's and Mathematica's functions \psi^{(-2)} both have second derivative \Gamma'(z)/\Gamma(z), so they differ only by some az+b function. – Gerald Edgar Jan 6 '15 at 19:00 I worked out the difference of Maple and Mathematica... Let \mathrm{Mple}(z) = \psi^{(-2)}(z) according to Maple's definition and \mathrm{Math}(z) = \psi^{(-2)}(z) according to Mathematica's definition. Then$$ \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$\$

• FWIW, here is actual Maple code for a slightly massaged version: (Zeta(1,-1,I)-Zeta(1,-1,-I))/(2*I)+(5/24)*Pi+polylog(2,exp(-2*Pi))/(4*Pi) – Neil Strickland Jan 6 '15 at 21:26
• Do you think the conjecture is within current knowledge or "hopeless"? – joro Jan 9 '15 at 13:23
• I believe it is within the current knowledge. Actually, I just found a proof (sort of) with some steps justified by symbolic computations in Mathematica, but I think it could be made simpler and completely manual. – Vladimir Reshetnikov Jan 11 '15 at 18:20