ζ(1,1,･･･,1,2)について

\begin{align} \Gamma(s)&=\int_{0}^{\infty}x^{s-1}e^{-x}dx\\ &=n^s\int_{0}^{\infty}t^{s-1}e^{-nt}dt \quad(x=nt)\\ \end{align}
$$\therefore\frac{1}{n^s}=\frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}e^{-nt}dt$$
両辺$1\leq n$の範囲で和をとると
\begin{align} \zeta(s)&=\sum_{n=1}^{\infty}\frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}e^{-nt}dt\\ &=\frac{1}{\Gamma(s)}\int_{0}^{\infty}t^{s-1}\sum_{n=1}^{\infty}e^{-nt}dt\\ &=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{t^{s-1}}{e^t-1}dt\quad \blacksquare \end{align}

シャッフル積を用いる
\begin{align} \frac{1}{r!}\Big(-\log{(1-x)}\Big)^{r} &=\frac{1}{r!}\Big(\int_{0}^{x}\frac{1}{1-t}dt\Big)^{r}\\ &=\int_{0\lt t_1 \lt t_2 \lt \cdots \lt t_r \lt x}\frac{dt_1}{1-t_1}\frac{dt_2}{1-t_2}\cdots\frac{dt_r}{1-t_r}\\ &=\int_{0}^{x}\frac{dt_r}{1-t_r}\cdots\int_{0}^{t_4}\frac{dt_3}{1-t_3}\int_{0}^{t_3}\frac{dt_2}{1-t_2}\int_{0}^{t_2}\frac{dt_1}{1-t_1}\\ &=\int_{0}^{x}\frac{dt_r}{1-t_r}\cdots\int_{0}^{t_4}\frac{dt_3}{1-t_3}\int_{0}^{t_3}\frac{dt_2}{1-t_2}\int_{0}^{t_2}\sum_{1\leq n_1}{t_1}^{n_1-1}\\ &=\sum_{1\leq n_1}\int_{0}^{x}\frac{dt_r}{1-t_r}\cdots\int_{0}^{t_4}\frac{dt_3}{1-t_3}\int_{0}^{t_3}\frac{dt_2}{1-t_2}\int_{0}^{t_2}{t_1}^{n_1-1}\\ &=\sum_{1\leq n_1}\frac{1}{n_1}\int_{0}^{x}\frac{dt_r}{1-t_r}\cdots\int_{0}^{t_4}\frac{dt_3}{1-t_3}\int_{0}^{t_3}\frac{{t_2}^{n_1}}{1-t_2}dt_2\\ &=\sum_{1\leq n_1,n_2}\frac{1}{n_1}\int_{0}^{x}\frac{dt_r}{1-t_r}\cdots\int_{0}^{t_4}\frac{dt_3}{1-t_3}\int_{0}^{t_3}{t_2}^{n_1+n_2-1}dt_2\\ &=\sum_{1\leq n_1,n_2}\frac{1}{n_1(n_1+n_2)}\int_{0}^{x}\frac{dt_r}{1-t_r}\cdots\int_{0}^{t_4}\frac{{t_3}^{n_1+n_2}}{1-t_3}dt_3\\ &\vdots\\ &=\sum_{1\leq n_1,n_2\cdots n_r}\frac{x^{n_1+n_2+\cdots+n_r}}{n_1(n_1+n_2)\cdots(n_1+n_2+\cdots+n_r)}\\ &=\sum_{0\lt n_1 \lt n_2\lt \cdots \lt n_r}\frac{x^{n_r}}{n_1 n_2 \cdots n_r} \quad\blacksquare \end{align}

\begin{align} \frac{1}{(r-1)!}\int_{0}^{1}\frac{\Big(-\log{(1-x)}\Big)^{r-1}}{x}dx&=\int_{0}^{1}\sum_{0\lt n_1 \lt n_2\lt \cdots \lt n_{r-1}}\frac{x^{n_{r-1}-1}}{n_1 n_2 \cdots n_{r-1}}dx \quad(\because補題2)\\ &=\sum_{0\lt n_1 \lt n_2\lt \cdots \lt n_{r-1}}\frac{1}{n_1 n_2 \cdots n_{r-1}^2}\\ &=\zeta(\lbrace 1 \rbrace^{r-2},2) \end{align}
一方で
\begin{align} \frac{1}{(r-1)!}\int_{0}^{1}\frac{\Big(-\log{(1-x)}\Big)^{r-1}}{x}dx&=\frac{1}{(r-1)!}\int_{0}^{\infty}\frac{t^{r-1}}{1-e^{-t}}e^{-t}dt \quad(x=1-e^{-t})\\ &=\frac{1}{(r-1)!}\int_{0}^{\infty}\frac{t^{r-1}}{e^{t}-1}dt\\ &=\frac{1}{(r-1)!}\Gamma(r)\zeta(r) \quad(補題1)\\ &=\zeta(r) \end{align}
したがって
$$\zeta(\lbrace 1 \rbrace^{r-2},2)=\zeta(r) \blacksquare$$