Compute polylog of order $3$ at $\frac{1}{2}$ 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.
 A: $$\begin{aligned}I_1&=\int_0^{1/2}\frac{\ln^2(1-x)}x\mathrm {d}x\\
I_2&=\int_0^{1/2}\frac{\ln x\ln(1-x)}x\mathrm {d}x\\
I_3&=\int_0^{1/2}\frac{\ln^2 x}{1-x}\mathrm {d}x\\
I_4&=\int_0^{1/2}\frac{\ln x\ln(1-x)}{1-x}\mathrm {d}x\end{aligned}$$
$$I_1+I_3=\int_0^1\frac{\ln^2 x}{1-x}\mathrm {d}x=2\zeta(3)$$
$$I_2+I_4=\int_0^1\frac{\ln x\ln(1-x)}x\mathrm {d}x=\zeta(3)$$
$$I_1=\ln^2(1-x)\ln x\big|_0^{1/2}+2\int_0^{1/2}\frac{\ln(1-x)\ln x}{1-x}\mathrm{d}x=-\ln^32+2I_4$$
$$I_2-I_1=\int_0^{1/2}\ln\frac x{1-x}\frac{\ln(1-x)}x\mathrm{d}x=-\int_0^1\frac{\ln u\ln(1+u)}{u(1+u)}\mathrm{d}u=\frac58\zeta(3)$$(using substitution $u=\frac{x}{1-x}$) 
Hence,$$I_1=\frac14\zeta(3)-\frac13\ln^32,\ I_2=\frac78\zeta(3)-\frac13\ln^32,$$
$$I_3=\frac74\zeta(3)+\frac13\ln^32,\ I_4=\frac18\zeta(3)+\frac13\ln^32.$$
$$\operatorname{Li}_3\left(\frac12\right)=\int_0^{1/2}\frac{\operatorname{Li}_2(x)}x\mathrm{d}x\\
=\operatorname{Li}_2(x)\ln x\Big|_0^{1/2}+\int_0^{1/2}\frac{\ln(1-x)\ln x}x\mathrm{d}x\\
=-\frac1{12}\pi^2\ln2+\frac12\ln^32+\frac78\zeta(3)-\frac13\ln^32$$
