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How to compute the following series:

$$\sum_{n=1}^{\infty}\frac{1}{2^nn^3}$$

I am aware this equals polylog of order $3$ at $\frac{1}{2}$ or $\operatorname{Li}_3\left(\frac{1}{2}\right)$, but how to prove it using integral or Euler sum only (without using any polylog identities)? I know how to prove $$\sum_{n=1}^{\infty}\frac{1}{2^nn^2}$$ or dilogarithm at $\frac{1}{2}$ like answer provided by Tunk-Fey in my previous OP, but I do not know how to use that fact to compute the polylog of order $3$ at $\frac{1}{2}$. My instructor told me to use geometric series yet I can't figure it out that clue. Any idea how to compute it without using using polylog identity (integral or infinite sum only)? Any help would be appreciated. Thanks in advance.

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  • $\begingroup$ $\text{Li}_3\bigg(\dfrac12\bigg) = \dfrac78\zeta(3) - \dfrac{\zeta(2)\ln2}2 + \dfrac{\ln^32}6\qquad$ $\endgroup$ – Lucian Sep 23 '14 at 12:03
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    $\begingroup$ @Lucian I know that, Wiki told me so. But how to get it? $\endgroup$ – Venus Sep 23 '14 at 12:17
  • $\begingroup$ @user153012 No. The answer must be like this math.stackexchange.com/a/810669/146687 $\endgroup$ – Venus Sep 23 '14 at 13:08
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Note Landen's identity $$\color{red}{{\rm Li}_2(z)+{\rm Li}_2\left(\frac{z}{z-1}\right)=-\frac{1}{2}\ln^2(1-z)}\tag1$$ which can be easily derived by differentiating ${\rm Li}_2\left(\frac{z}{z-1}\right)$ then integrating back. Plucking in $z=.5$ yields $$\color{red}{{\rm Li}_2\left(\frac{1}{2}\right)=\frac{\pi^2}{12}-\frac{1}{2}\ln^2{2}}$$ since $\displaystyle{\rm Li}_2(-1)=-\eta(1)=\small{-\frac{\pi^2}{12}}$. We will use the same idea to evaluate ${\rm Li}_3\left(\frac{1}{2}\right)$.

Observe $D_z{\rm Li}_{s+1}\left(\frac{z}{z-1}\right)=\frac{1}{z(1-z)}{\rm Li}_{s}\left(\frac{z}{z-1}\right)$. So dividing $(1)$ throughout by $z(1-z)$ then integrating gives us $${\rm Li}_3(z)+\int\frac{{\rm Li}_2(z)}{1-z}{\rm d}z+{\rm Li}_3\left(\frac{z}{z-1}\right)=\frac{1}{6}\ln^3(1-z)-\frac{1}{2}\int\frac{\ln^2(1-z)}{z}{\rm d}z\tag2$$ The integral on the LHS evaluates to \begin{align} \int\frac{{\rm Li}_2(z)}{1-z}{\rm d}z &=-{\rm Li}_2(z)\ln(1-z)-\int\frac{\ln^2(1-z)}{z}{\rm d}z \end{align} so we can rewrite $(2)$ as \begin{align} {\rm Li}_3(z)+{\rm Li}_3\left(\frac{z}{z-1}\right)={\rm Li}_2(z)\ln(1-z)+\frac{1}{6}\ln^3(1-z)+\frac{1}{2}\int\frac{\ln^2(1-z)}{z}{\rm d}z\tag3 \end{align} and the integral on the RHS of $(3)$ is \begin{align} \frac{1}{2}\int\frac{\ln^2(1-z)}{z}{\rm d}z &=\frac{1}{2}\ln{z}\ln^2(1-z)+\int\frac{\ln{z}\ln(1-z)}{1-z}{\rm d}z\\ &=\frac{1}{2}\ln{z}\ln^2(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\int\frac{{\rm Li}_2(1-z)}{1-z}{\rm d}z\\ &=\frac{1}{2}\ln{z}\ln^2(1-z)+{\rm Li}_2(1-z)\ln(1-z)-{\rm Li}_3(1-z)+C\tag4\\ \end{align} Plucking $(4)$ into $(3)$ then moving all the trilog terms to one side gives us \begin{align} &{\rm Li}_3(z)+{\rm Li}_3(1-z)+{\rm Li}_3\left(\frac{z}{z-1}\right)\\ =&\frac{1}{6}\ln^3(1-z)+\Big{[}{\rm Li}_2(z)+{\rm Li}_2(1-z)\Big{]}\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+C\\ =&\frac{1}{6}\ln^3(1-z)-\frac{1}{2}\ln{z}\ln^2(1-z)+\frac{\pi^2}{6}\ln(1-z)+C\tag5 \end{align} where we applied the dilogarithm reflection formula $$\color{red}{{\rm Li}_2(z)+{\rm Li}_2(1-z)=\frac{\pi^2}{6}-\ln{z}\ln(1-z)}\tag6$$ to get to $(5)$. This formula can be easily derived by differentiating ${\rm Li}_2(1-z)$ then integrating back. Now, by letting $z=0$ in $(5)$, it is evident that $C={\rm Li}_3(1)=\zeta(3)$. Therefore, \begin{align}&\color{red}{{\rm Li}_3(z)+{\rm Li}_3(1-z)+{\rm Li}_3\left(\frac{z}{z-1}\right)}\\=&\color{red}{\frac{1}{6}\ln^3(1-z)-\frac{1}{2}\ln{z}\ln^2(1-z)+\frac{\pi^2}{6}\ln(1-z)+\zeta(3)}\tag7\end{align} Finally, letting $z=.5$ in $(7)$, we get $$2{\rm Li}_3\left(\frac{1}{2}\right)-\frac{3}{4}\zeta(3)=-\frac{1}{6}\ln^3{2}+\frac{1}{2}\ln^3{2}-\frac{\pi^2}{6}\ln{2}+\zeta(3)$$ which implies $$\color{blue}{{\rm Li}_3\left(\frac{1}{2}\right)=\frac{7}{8}\zeta(3)-\frac{\pi^2}{12}\ln{2}+\frac{1}{6}\ln^3{2}}\tag8$$

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    $\begingroup$ Thanks. Nice answer & format. If possible, I want the answer avoiding using polylog identity. +1 $\endgroup$ – Venus Sep 23 '14 at 13:43
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$$\begin{aligned}I_1&=\int_0^{1/2}\frac{\ln^2(1-x)}x\mathrm {d}x\\ I_2&=\int_0^{1/2}\frac{\ln x\ln(1-x)}x\mathrm {d}x\\ I_3&=\int_0^{1/2}\frac{\ln^2 x}{1-x}\mathrm {d}x\\ I_4&=\int_0^{1/2}\frac{\ln x\ln(1-x)}{1-x}\mathrm {d}x\end{aligned}$$ $$I_1+I_3=\int_0^1\frac{\ln^2 x}{1-x}\mathrm {d}x=2\zeta(3)$$ $$I_2+I_4=\int_0^1\frac{\ln x\ln(1-x)}x\mathrm {d}x=\zeta(3)$$ $$I_1=\ln^2(1-x)\ln x\big|_0^{1/2}+2\int_0^{1/2}\frac{\ln(1-x)\ln x}{1-x}\mathrm{d}x=-\ln^32+2I_4$$ $$I_2-I_1=\int_0^{1/2}\ln\frac x{1-x}\frac{\ln(1-x)}x\mathrm{d}x=-\int_0^1\frac{\ln u\ln(1+u)}{u(1+u)}\mathrm{d}u=\frac58\zeta(3)$$(using substitution $u=\frac{x}{1-x}$)

Hence,$$I_1=\frac14\zeta(3)-\frac13\ln^32,\ I_2=\frac78\zeta(3)-\frac13\ln^32,$$ $$I_3=\frac74\zeta(3)+\frac13\ln^32,\ I_4=\frac18\zeta(3)+\frac13\ln^32.$$

$$\operatorname{Li}_3\left(\frac12\right)=\int_0^{1/2}\frac{\operatorname{Li}_2(x)}x\mathrm{d}x\\ =\operatorname{Li}_2(x)\ln x\Big|_0^{1/2}+\int_0^{1/2}\frac{\ln(1-x)\ln x}x\mathrm{d}x\\ =-\frac1{12}\pi^2\ln2+\frac12\ln^32+\frac78\zeta(3)-\frac13\ln^32$$

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