Integrals of the kind $$\int_0^1 \frac{\log(1-z)\log(1-z^n)}{z^2}dz$$ where $n\geq 1$ is an integer, arises from a natural way when one apply Möbius inversion to get identities realated to $\zeta(2)$ (see my appendix if there are no mistakes), but how

Question. Can you state a closed form for $$\int_0^1 \frac{\log(1-z)\log(1-z^3)}{z^2}dz?$$

I am asking for a detailed explanationabout how get the definite integral. Many thanks.

I don't know if the general integral was in the literature but my trials with Wolfram Alpha, seems that is very difficult to get particular cases. See this

Example. Other case of the indefinite integral, for example $n=5$ also is provided by Wolfram Alpha online calculator when one type

integrate (log(1-x)log(1-x^5))/x^2 dx

My idea was multiply both formulas of the first paragraph in page 77 from:

Benito, Navas, Varona, Möbius inversion from the point of view of flows, Proceedings of the Segundas Jornadas de Teoría de Números, Biblioteca de la Revista Matemática Iberoamericana (2008),

then integrating one gets $$\zeta(2)=\int_0^1\sum_{n=1}^\infty\frac{z^{n-1}}{n}dz=\int_0^1 \left(\frac{\log(1-z)}{z^2}\sum_{n=1}^\infty\frac{\mu(n)}{n}\log(1-z^n)\right)dz.$$ Here as you see $\zeta(s)$ is the Riemann's Zeta function and $\mu(n)$ is the Möbius function. And now from there one more time by absolute convergence, I've asked myself what's about these coefficients

$$a_n=\int_0^1 \frac{\log(1-z)\log(1-z^n)}{z^2}dz,$$ that for integers $n\geq 1$ satisfy $$\zeta(2)=\sum_{n=1}^\infty\frac{\mu(n)}{n}a_n.$$

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    $\begingroup$ The required closed form is $$ \frac{1}{6} \left(\left(-3-3 i \sqrt{3}\right) \text{Li}_2\left(\frac{1}{1+\sqrt[3]{-1}}\right)+3 \left(\sqrt{3}+i\right) i \text{Li}_2\left(\frac{1}{6} \left(3+i \sqrt{3}\right)\right)+2 \pi ^2-3 \log ^2(3)\right) $$ $\endgroup$ – tired Nov 2 '16 at 15:03
  • $\begingroup$ Very nice @tired , but I don't know how get the indefinite integral, case $n=3$ and/or your definite integral. Many thanks for your attention and answer. $\endgroup$ – user243301 Nov 2 '16 at 15:12
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    $\begingroup$ the keyideas are the following $\log(1+z^3)=\sum_i\log(1-\omega_i z)$ where $\omega_i $ are definded through the condition $1-z^3=0$ then we are left with integrals of the type $$ \int_0^1\frac{\log(1-z)\log(1-\omega_i z)}{z^2}dz $$ now we integrate by parts: $$ \int_0^1\left(-\frac{\log (1-z)}{z}+\log (1-z)+\log (z)\right)\frac{1}{1-\omega_i z} $$ the resulting integrals can be now brought straightforwardly into a form which contains logartihms and dilogarithms $\endgroup$ – tired Nov 2 '16 at 15:42
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    $\begingroup$ this idea also works also for powers $z^n$ in the second log of the original integral $\endgroup$ – tired Nov 2 '16 at 15:43
  • $\begingroup$ As I said your formula and idea is very nice, thanks @tired $\endgroup$ – user243301 Nov 2 '16 at 16:31

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