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{eqnarray*} &&\sum_{0\f a\f b}\frac1{a^2b2^b}\\ &=&\sum_{0\f a,b}\frac1{a^2(a+b)2^{a+b}}\\ &=&\sum_{0\f a,b}\frac1{a^22^{a+b}}\int_0^1 x^{a+b-1}dx\\ &=&\int_0^1\frac1x\sum_{0\f a,b}\frac{\l\frac x2\r^a\l\frac x2\r^b}{a^2}dx\\ &=&\int_0^1 \frac{\Li_2\l\frac x2\r}{2-x}dx\\ &=&\int_0^{\frac12}\frac{\Li_2(t)}{1-t}dt~~~~~~~~~~(x=2t)\\ &=&-\left[\Li_2(t)\log(1-t) \right]_0^{\frac12}-\int_0^{\frac12}\frac{\log^2(1-t)}tdt\\ &=&-\left[\log t\log^2(1-t) \right]_0^{\frac12}-2\int_0^{\frac12}\frac{\log t\log(1-t)}{1-t}dt-\Li_2\l\frac12\r\log\frac12\\ &=&-2\left[\Li_2(1-t)\log(1-t) \right]_0^{\frac12}-2\int_0^{\frac12}\frac{\Li_2(1-t)}{1-t}dt-\log\frac12\log^2\frac12+\Li_2\l\frac12\r\log2\\ &=&2\left[\Li_3(1-t)\right]_0^{\frac12}+\log^32-2\Li_2\l\frac12\r\log\frac12+\Li_2\l\frac12\r\log2\\ &=&2\Li_3\l\frac12\r-2\z(3)+3\Li_2\l\frac12\r\log2+\log^32\\ &=&2\Li_3\l\frac12\r+3\Li_2\l\frac12\r\log2+\log^32-2\z(3)\\ &=&\frac74\z(3)+\frac13\log^32-\frac{\pi^2}6\log2+\frac{\pi^2}4\log2-\frac32\log^32+\log^32-2\z(3)\\ &=&\frac{\pi^2}{12}\log2-\frac16\log^32-\frac14\z(3) \end{eqnarray*} $

よって、この問題の解答は$\displaystyle\frac{\pi^2}{12}\log2-\frac16\log^32-\frac14\z(3)$となります。