正規化によって得られた発散級数の関係式

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}{n}^2& =\underset{s\to 0}{lim}\left(\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{n^s}\right)\\ & =\zeta (-2)\\ & =0\end{array}$$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}n& =\underset{x\to 1^{-}}{lim}(\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(n+1)x^n)\\ & =\underset{x\to 1^{-}}{lim}\left(\frac{1}{(1+x{)}^2}\right)\\ & =\frac{1}{4}\end{array}$$

${\displaystyle {\int }_0^1\frac{x}{1+(xt)}dt}$を考える
${\displaystyle {\int }_0^1\frac{x}{1+(xt)}dt=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{x^{n+1}}{n+1}}$
${\displaystyle \frac{d^2}{d{x}^2}\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{x^{n+1}}{n+1}=\sum _{n=0}^{\mathrm{\infty }}n(-1{)}^n{x}^n}$
${\displaystyle \frac{d^2}{d{x}^2}{\int }_0^1\frac{x}{1+(xt)}dt=-{\int }_0^1\frac{2t}{(1+xt{)}^3}dt}$
$x=1$とすれば${\displaystyle \sum _{n=1}^{\mathrm{\infty }}n(-1{)}^{n+1}=\frac{1}{4}}$を得る

${\displaystyle \zeta (\overline{-1}{)}^2}$を考える
調和積とDirichlet正規化より
$$\begin{array}{rl}\zeta (\overline{-1}{)}^2& ={(\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n}n)}^2\\ & =\sum _{n=1}^{\mathrm{\infty }}{n}^2+2\sum _{0<n_1<n_2}(-1{)}^{n_1+n_2}{n}_1{n}_2\\ & =\zeta (-2)+2\zeta (\overline{-1},\overline{-1})\\ & =2\zeta (\overline{-1},\overline{-1})\end{array}$$
${\displaystyle {\int }_0^1\frac{2t}{(1+t{)}^3}dt=\eta (-1)}$だった
シャッフル積より
$$\begin{array}{rl}{\left({\int }_0^1\frac{2t}{(1+t{)}^3}dt\right)}^2& =8{\int }_{0<t_1<t_2<1}\frac{t_1{t}_2}{(1+t_1{)}^3(1+t_2{)}^3}d{t}_{1}d{t}_2\\ & =\frac{1}{6}\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n(n+1)(n+2)(n+3)\\ & =-\frac{1}{3}\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n}n(n+1)(n+2)\\ & =-\frac{1}{3}\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\left(\sum _{k=1}^{n}k\right)(n+2)\\ & =-\frac{1}{3}\sum _{0<n_1\le n_2}(-1{)}^{n_2}{n}_1{n}_2-\frac{2}{3}\sum _{0<n_1\le n_2}(-1{)}^{n_2}{n}_1\\ & =-\frac{1}{3}{\zeta }^{\star }(-1,\overline{-1})-\frac{2}{3}{\zeta }^{\star }(-1,\bar{0})\end{array}$$
よって、示せた

$$\begin{array}{rl}\left({\int }_0^1\frac{dt}{1-t}\right)({\int }_0^1(1-t{)}^{m}dt)& =\frac{1}{m+1}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\dots (1)\\ & ={\int }_0^1(1-t_2{)}^{m}d{t}_2{\int }_0^{t_2}\frac{d{t}_1}{1-t_1}+{\int }_0^1\frac{d{t}_2}{1-t_2}{\int }_0^{t_2}(1-t_2{)}^{m}d{t}_2\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\frac{n!m!}{(n+m+1)!}+\frac{1}{m+1}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}-\frac{1}{(m+1{)}^2}\dots (2)\end{array}$$
(1)=(2)より得る