級数解説03

2020/11/19に出題した問題です。

https://mathlog.info/articles/770

[解説]

$\begin{array}{rcl}& & \sum _{0<a<b}\frac{1}{a^{2}b{2}^b}\\ & =& \sum _{0<a,b}\frac{1}{a^2(a+b)2^{a+b}}\\ & =& \sum _{0<a,b}\frac{1}{a^2{2}^{a+b}}{\int }_0^1{x}^{a+b-1}dx\\ & =& {\int }_0^1\frac{1}{x}\sum _{0<a,b}\frac{{\left(\frac{x}{2}\right)}^a{\left(\frac{x}{2}\right)}^b}{a^2}dx\\ & =& {\int }_0^1\frac{{\text{Li}}_2\left(\frac{x}{2}\right)}{2-x}dx\\ & =& {\int }_0^{\frac{1}{2}}\frac{{\text{Li}}_2(t)}{1-t}dt(x=2t)\\ & =& -{[{\text{Li}}_2(t)\log (1-t)]}_0^{\frac{1}{2}}-{\int }_0^{\frac{1}{2}}\frac{{\log }^2(1-t)}{t}dt\\ & =& -{[\log t{\log }^2(1-t)]}_0^{\frac{1}{2}}-2{\int }_0^{\frac{1}{2}}\frac{\log t\log (1-t)}{1-t}dt-{\text{Li}}_2\left(\frac{1}{2}\right)\log \frac{1}{2}\\ & =& -2{[{\text{Li}}_2(1-t)\log (1-t)]}_0^{\frac{1}{2}}-2{\int }_0^{\frac{1}{2}}\frac{{\text{Li}}_2(1-t)}{1-t}dt-\log \frac{1}{2}{\log }^2\frac{1}{2}+{\text{Li}}_2\left(\frac{1}{2}\right)\log 2\\ & =& 2{[{\text{Li}}_3(1-t)]}_0^{\frac{1}{2}}+{\log }^{3}2-2{\text{Li}}_2\left(\frac{1}{2}\right)\log \frac{1}{2}+{\text{Li}}_2\left(\frac{1}{2}\right)\log 2\\ & =& 2{\text{Li}}_3\left(\frac{1}{2}\right)-2\zeta (3)+3{\text{Li}}_2\left(\frac{1}{2}\right)\log 2+{\log }^{3}2\\ & =& 2{\text{Li}}_3\left(\frac{1}{2}\right)+3{\text{Li}}_2\left(\frac{1}{2}\right)\log 2+{\log }^{3}2-2\zeta (3)\\ & =& \frac{7}{4}\zeta (3)+\frac{1}{3}{\log }^{3}2-\frac{{\pi }^2}{6}\log 2+\frac{{\pi }^2}{4}\log 2-\frac{3}{2}{\log }^{3}2+{\log }^{3}2-2\zeta (3)\\ & =& \frac{{\pi }^2}{12}\log 2-\frac{1}{6}{\log }^{3}2-\frac{1}{4}\zeta (3)\end{array}$

よって、この問題の解答は${\displaystyle \frac{{\pi }^2}{12}\log 2-\frac{1}{6}{\log }^{3}2-\frac{1}{4}\zeta (3)}$となります。