• It is known that: \begin{align} &\int _{-\pi }^{\pi }\log\left(\,{\left\vert \,{ 1 + \mathrm{e}^{\mathrm{i}x}}\,\right\vert} \,\right)\,\mathrm{d}x = 0 \\[5mm] &\ \int _{-\pi }^{\pi }\int_{-\pi }^{\pi } \log\left(\,{\left\vert\,{1 + \mathrm{e}^{\mathrm{i} x} + \mathrm{e}^{\mathrm{i}y}} \,\right\vert}\,\right) \,\mathrm{d}x\,\mathrm{d}y \\[2mm] = &\ \frac{\pi\left[% \psi ^{\left(1\right)}\left(1/3\right) - \psi^{\left(1\right)}\left(2/3\right)\right]}{\,\sqrt{\,3\,}\,} \end{align} which are direct consequences of
    • Cauchy integral,
    • Poisson integral formula$\,+\,$Fourier expansion, respectively.
  • However, I have no idea for the $3$-dimensional case: \begin{align} &\int_{-\pi}^{\pi}\int_{-\pi}^{\pi} \int _{-\pi}^{\pi} \log\left(\,{\left\vert\,{1 + \mathrm{e}^{\mathrm{i}x} + \mathrm{e}^{\mathrm{i}y} + \mathrm{e}^{\mathrm{i} z}} \,\right\vert}\,\right) \,\mathrm{d}x\,\mathrm{d} y\,\mathrm{d}z \\[2mm] = &\ 28\pi\,\zeta\left(\,{3}\,\right) \end{align}

Any kind of help is appreciated.

  • $\begingroup$ Totally not a solution but this is something that could be 'checked' numerically. $\endgroup$
    – quarague
    Mar 12, 2020 at 10:09
  • $\begingroup$ V. beautiful... but, belatedly, and v. ignorantly, what is $\psi^{(1)}$? Could you update your question to include the definition? Thanks. $\endgroup$
    – peter a g
    Jun 29, 2022 at 13:38

1 Answer 1


J. M. Borwein's article Mahler measures, short walks and log-sine integrals offers an elegant proof. Denote $W_n(s)=\int_{(0,1)^n}\left|\sum_{k=1}^n e^{2\pi i x_k}\right|^s dx_1\cdots dx_n$, then according to R. Crandall's Analytic representations for circle-jump moments, $W_4(s)$ enjoys a Meijer-G representation reducible to hypergeometric functions via functional identities: $$\scriptsize W_4(s)=\binom{s}{\frac{s}{2}} \, _4F_3\left(\frac{1}{2},-\frac{s}{2},-\frac{s}{2},-\frac{s}{2};1,1,\frac{1-s}{2};1\right)+\frac{1}{4^s}{\binom{s}{\frac{s-1}{2}}^3 \tan \left(\frac{\pi s}{2}\right) \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{s}{2}+1;\frac{s+3}{2},\frac{s+3}{2},\frac{s+3}{2};1\right)}$$ Differentiating both sides w.r.t $s$, let $s\rightarrow 0$ one have (the final $_pF_q$ series is rational thus trivial) $$\small W_4^{'}(0)=\frac{4}{\pi ^2}{_4F_3\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)}=\frac{7 \zeta (3)}{2 \pi ^2}$$ On the other hand, by scaling, using symmetry and periodicity of the original integrand and noticing $\log\left|e^{i \phi}\right|=0$ (thus the integral is invariant after extracting $e^{\text{one variable}}$), it's clear that: $$2\pi I=\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\int _{-\pi }^{\pi }\log \left|e^{i w}+e^{i x}+e^{i y}+e^{i z}\right|dwdxdydz=16\pi^4 W_4^{'}(0)$$ From which we get the desired result $I=28\pi \zeta(3)$.


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