(1)

\(\alpha\in\mathbb{C}\setminus\left\{ 0\right\} \)とする。
\begin{align*} B\left(z;\alpha,\beta\right) & =\int_{0}^{z}t^{\alpha-1}\left(1-t\right)^{\beta-1}dt\\ & =\sum_{k=0}^{\infty}C\left(\beta-1,k\right)\int_{0}^{z}t^{\alpha-1}\left(-t\right)^{k}dt\\ & =\sum_{k=0}^{\infty}C\left(\beta-1,k\right)\left(-1\right)^{k}\int_{0}^{z}t^{\alpha+k-1}dt\\ & =\sum_{k=0}^{\infty}C\left(\beta-1,k\right)\left(-1\right)^{k}\left[\frac{t^{\alpha+k}}{a+k}\right]_{t=0}^{t=z}\\ & =\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{C\left(\beta-1,k\right)}{\alpha+k}z^{\alpha+k}\\ & =z^{\alpha}\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{C\left(\beta-1,k\right)}{\alpha+k}z^{k}\\ & =z^{\alpha}\sum_{k=0}^{\infty}\left(-1\right)^{k}\frac{P\left(\beta-1,k\right)}{k!\left(\alpha+k\right)}z^{k}\\ & =z^{\alpha}\sum_{k=0}^{\infty}\frac{Q\left(1-\beta,k\right)}{k!\left(\alpha+k\right)}z^{k}\\ & =z^{\alpha}\sum_{k=0}^{\infty}\frac{P\left(k-\beta,k\right)}{k!\left(\alpha+k\right)}z^{k}\\ & =z^{\alpha}\sum_{k=0}^{\infty}\frac{C\left(k-\beta,k\right)}{\alpha+k}z^{k} \end{align*}

(2)

\(z\ne1,n\in\mathbb{N}\)とする。
\begin{align*} B\left(z;n,0\right) & =\int_{0}^{z}\frac{z^{n-1}}{1-z}dz\\ & =\int_{0}^{z}z^{n-1}\sum_{k-0}^{\infty}z^{k}dz\\ & =\sum_{k-0}^{\infty}\int_{0}^{z}z^{n+k-1}dz\\ & =\sum_{k=0}^{\infty}\left[\frac{z^{n+k}}{n+k}\right]_{0}^{z}\\ & =\sum_{k=0}^{\infty}\frac{z^{n+k}}{n+k}\\ & =\sum_{k=n}^{\infty}\frac{z^{k}}{k}\\ & =\sum_{k=1}^{\infty}\frac{z^{k}}{k}-\sum_{k=1}^{n-1}\frac{z^{k}}{k}\\ & =\Li_{1}\left(z\right)-\sum_{k=1}^{n-1}\frac{z^{k}}{k} \end{align*}

(2)-2

\begin{align*} B\left(z;n,0\right) & =z^{n}\sum_{k=0}^{\infty}\frac{C\left(k-0,k\right)}{n+k}z^{k}\cmt{\because B\left(z;\alpha,\beta\right)=z^{\alpha}\sum_{k=0}^{\infty}\frac{C\left(k-\beta,k\right)}{\alpha+k}z^{k}}\\ & =\sum_{k=0}^{\infty}\frac{1}{n+k}z^{k+n}\\ & =\left(\sum_{k=1}^{\infty}\frac{1}{k}z^{k}-\sum_{k=1}^{n-1}\frac{1}{k}z^{k}\right)\\ & =\Li_{1}\left(z\right)-\sum_{k=1}^{n-1}\frac{1}{k}z^{k} \end{align*}