正規化によって得られた発散級数の関係式

\begin{align} \sum_{n=1}^\infty n^2&=\lim_{s\to 0}\left(\sum_{n=1}^\infty\frac{n^2}{n^s}\right)\\&=\zeta(-2)\\&=0 \end{align}

\begin{align} \sum_{n=1}^\infty(-1)^{n-1}n&= \lim_{x\to1^-}\left(\sum_{n=0}^\infty (-1)^n(n+1)x^n\right)\\&=\lim_{x\to1^-}\left(\frac{1}{(1+x)^2}\right)\\&=\frac{1}{4} \end{align}

$\displaystyle \int_0^1\frac{x}{1+(xt)}dt$を考える
$\displaystyle \int_0^1\frac{x}{1+(xt)}dt=\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}$
$\displaystyle \frac{d^2}{dx^2}\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}=\sum_{n=0}^\infty n(-1)^nx^n$
$\displaystyle \frac{d^2}{dx^2} \int_0^1\frac{x}{1+(xt)}dt=-\int_0^1\frac{2t}{(1+xt)^3}dt$
$x=1$とすれば$\displaystyle \sum_{n=1}^\infty n(-1)^{n+1}=\frac{1}{4}$を得る

$\displaystyle \zeta(\overline{-1})^2$を考える
調和積とDirichlet正規化より
\begin{align} \zeta(\overline{-1})^2&=\left(\sum_{n=1}^\infty(-1)^nn\right)^2\\&=\sum_{n=1}^\infty n^2+2\sum_{0< n_1< n_2}(-1)^{n_1+n_2}n_1n_2\\&=\zeta(-2)+2\zeta(\overline{-1},\overline{-1})\\&=2\zeta(\overline{-1},\overline{-1}) \end{align}
$\displaystyle \int_0^1\frac{2t}{(1+t)^3}dt=\eta(-1)$だった
シャッフル積より
\begin{align} \left(\int_0^1\frac{2t}{(1+t)^3}dt\right)^2&=8\int_{0< t_1< t_2<1}\frac{t_1t_2}{(1+t_1)^3(1+t_2)^3}dt_1dt_2\\&=\frac{1}{6}\sum_{n=0}^\infty(-1)^n(n+1)(n+2)(n+3)\\&=-\frac{1}{3}\sum_{n=1}^\infty(-1)^nn(n+1)(n+2)\\&=-\frac{1}{3}\sum_{n=1}^\infty(-1)^n\left(\sum_{k=1}^nk\right)(n+2)\\&=-\frac{1}{3}\sum_{0< n_1\le n_2}(-1)^{n_2}n_1n_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,\overline{0}) \end{align}
よって、示せた

\begin{align} \left(\int_0^1\frac{dt}{1-t}\right)\left(\int_0^1(1-t)^mdt\right)&=\frac{1}{m+1}\sum_{n=1}^\infty\frac{1}{n}…(1)\\&=\int_0^1(1-t_2)^mdt_2\int_0^{t_2}\frac{dt_1}{1-t_1}+\int_0^1\frac{dt_2}{1-t_2}\int_0^{t_2}(1-t_2)^mdt_2\\&=\sum_{n=1}^\infty\frac{1}{n}\frac{n!m!}{(n+m+1)!}+\frac{1}{m+1}\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{(m+1)^2}…(2) \end{align}
(1)=(2)より得る