一次変換だけで導く簡単な積分

[1]
\begin{eqnarray} \begin{pmatrix}e&f\\g&h\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\int_{A}^{B}f(x)dx&=&\begin{pmatrix}e&f\\g&h\end{pmatrix}det\begin{pmatrix}a&b\\c&d\end{pmatrix}\int_{\frac{dA-b}{-cA+a}}^{\frac{dB-b}{-cB+a}}f(\frac{ax+b}{cx+d})\frac{dx}{(cx+d)^{2}}\\ &=&det \begin{pmatrix}e&f\\g&h\end{pmatrix}det\begin{pmatrix}a&b\\c&d\end{pmatrix}\int_{\frac{h\frac{dA-b}{-cA+a}-f}{-g\frac{dA-b}{-cA+a}+e}}^{\frac{h\frac{dB-b}{-cB+a}-f}{-g\frac{dB-b}{-cB+a}+e}}f(\frac{e\frac{ax+b}{cx+d}+f}{g\frac{ax+b}{cx+d}+h})\frac{1}{(c\frac{ex+f}{gx+h}+d)^{2}}\frac{dx}{(gx+h)^{2}}\\ &=&det \begin{pmatrix}e&f\\g&h\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\int_{\frac{(hd+fc)A-(hb+fa)}{-(gd+ec)A+gb+ea}}^{\frac{(hd+fc)B-(hb+fa)}{-(gd+ec)B+gb+ea}}f(\frac{(ea+fc)x+eb+fd}{(ga+hc)x+gb+hd})\frac{dx}{\{(ec+gd)x+fc+hd\}^{2}} \end{eqnarray}
また以下の式が成り立つ。
\begin{equation} \begin{pmatrix}e&f\\g&h\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}ea+fc&eb+fd\\ga+hc&gb+hd\end{pmatrix} \end{equation}
\begin{eqnarray} \begin{pmatrix}e&f\\g&h\end{pmatrix}^{-1}\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}\begin{pmatrix}1&0\\0&1\end{pmatrix}&=&\frac{1}{(eh-fg)(ad-bc)}\begin{pmatrix}h&-f\\-g&e\end{pmatrix}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\\ &=&\frac{1}{(eh-fg)(ad-bc)}\begin{pmatrix}hd+fc&-(hb+fa)\\-(gd+ec)&gb+ea\end{pmatrix} \end{eqnarray}
\begin{eqnarray} \begin{pmatrix}0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}e&f\\g&h\end{pmatrix}=\begin{pmatrix}ec+gd&fc+hd\end{pmatrix} \end{eqnarray}
区間以外の部分に関しては準同型性は明らかなので積分区間のみ考察する。
[2]今度は二次元の正則行列の集まり$GL_{2}(\mathbb{C})$に下記の積を導入する。この積が群をなさない事を示す。
なお、この積を導入したものを$GL_{2}(\mathbb{C})^{\ast}$と書く事にする。
\begin{equation} \forall \begin{pmatrix}e&f\\g&h\end{pmatrix},\begin{pmatrix}a&b\\c&d\end{pmatrix}\in GL_{2}(\mathbb{C}):\begin{pmatrix}e&f\\g&h\end{pmatrix}\ast\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}e&f\\g&h\end{pmatrix}^{-1}\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1} \end{equation}
\begin{eqnarray} A\ast(B\ast C)&=&A\ast(B^{-1}C)\\ &=&A^{-1}(B^{-1}C)\\ &=&A^{-1}B^{-1}C \end{eqnarray}
\begin{eqnarray} (A\ast B)\ast C&=&(A^{-1}B)^{-1}\ast C\\ &=&B^{-1}AC \end{eqnarray}
[3]以上の計算から$GL_{2}(\mathbb{C})^{\ast}$は群をなさないので、そもそも準同型射$GL_{2}(\mathbb{C})\mapsto GL_{2}(\mathbb{C})^{\ast}$を構成できない。
この事から積分区間について準同型射を構成できないので証明終わり。

\begin{eqnarray} B(p,q)&=&\int_{0}^{1}x^{p-1}(1-x)^{q-1}dx\\ &=&(1-c)\int_{0}^{1}\frac{x^{p-1}\{(c-1)x+1-c\}^{q-1}}{(cx+1-c)^{p+q}}dx\\ &=&(1-c)^{-p}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(1+\frac{c}{1-c}x)^{p+q}}dx \end{eqnarray}
特に$c\mapsto -c$とすると
\begin{equation} B(p,q)=(1+c)^{-p}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(1-\frac{c}{c+1}x)^{p+q}}dx \end{equation}
さらに$c=\frac{\lambda}{1-\lambda}$とすると
\begin{equation} B(p,q)=(1-\lambda)^{p}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(1-\lambda x)^{p+q}}dx \end{equation}

\begin{eqnarray} B(p,q)&=&\frac{(1-\lambda)^{p}}{\lambda^{p+q}}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(\frac{1}{\lambda}-x)^{p+q}}dx\\ &=&\mu^{p+q}(1-\frac{1}{\mu})^{p}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(\mu-x)^{p+q}}dx\\ &=&\mu^{q}(\mu-1)^{p}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(\mu-x)^{p+q}}dx \end{eqnarray}

\begin{eqnarray} F_{n}(f)\coloneqq\int_{a}^{x}dx_{n}\int_{a}^{x_{n}}dx_{n-1}\cdots\int_{a}^{x_{2}}dx_{1}f(x_{1}) \end{eqnarray}
の様に置く。
\begin{eqnarray} F_{n}(f)(\mu)&=&\int_{-\infty}^{\mu}d\mu_{N}\int_{-\infty}^{\mu_{N}}d\mu_{N-1}\cdots\int_{-\infty}^{\mu_{2}}d\mu_{1}f(\mu_{1})\\ &=&\int_{a}^{\mu}d\mu_{1}f(\mu_{1})\int_{\mu_{1}}^{\mu}d\mu_{2}\cdots\int_{\mu_{n-1}}^{\mu}d\mu_{n}\\ &=&\frac{1}{\Gamma(N)}\int_{a}^{\mu}d\mu_{1}(\mu-\mu_{1})^{N-1}f(\mu_{1}) \end{eqnarray}

[1]
\begin{eqnarray} B(p,q)\int_{\mu}^{\infty}d\mu_{N}\int_{\mu_{N}}^{\infty}d\mu_{N-1}\cdots\int_{\mu_{2}}^{\infty}d\mu_{1}\frac{1}{\mu_{1}^{q}(\mu_{1}-1)^{p}}&=&B(p,q)\frac{1}{(N-1)!}\int_{\mu}^{\infty}d\mu_{1}\frac{(\mu_{1}-\mu)^{N-1}}{\mu_{1}^{q}(\mu_{1}-1)^{p}}\\ &=&(-1)^{n}\int_{\mu}^{\infty}d\mu_{N}\int_{\mu_{N}}^{\infty}d\mu_{N-1}\cdots\int_{\mu_{2}}^{\infty}d\mu_{1}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(\mu_{1}-x)^{p+q}}dx\\ &=&\frac{1}{(p+q-1)(p+q-2)\cdots(p+q-N)}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(\mu_{1}-x)^{p+q-N}}dx\\ &=&\frac{(p+q-N-1)!}{(p+q-1)!}\int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(\mu-x)^{p+q-N}}dx \end{eqnarray}
[2]
\begin{equation} \int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(\mu-x)^{p+q-N}}dx=N\begin{pmatrix}p+q-1\\N\end{pmatrix}B(p,q)\int_{\mu}^{\infty}d\mu_{1}\frac{(\mu_{1}-\mu)^{N-1}}{\mu_{1}^{q}(\mu_{1}-1)^{p}} \end{equation}
[3]
\begin{eqnarray} \int_{\mu}^{\infty}d\mu_{1}\frac{(\mu_{1}-\mu)^{N-1}}{\mu_{1}^{q}(\mu_{1}-1)^{p}}&=&\mu^{N-q}\int_{1}^{\infty}d\nu_{1}\frac{(\nu_{1}-1)^{N-1}}{\nu_{1}^{q}(\mu\nu_{1}-1)^{p}}\\ &=&\mu^{N-q}\int_{0}^{1}d\nu_{2}\frac{\nu_{2}^{q}(\frac{1}{\nu_{2}}-1)^{N-1}}{(\frac{\mu}{\nu_{2}}-1)^{p}}\frac{1}{\nu_{2}^{2}}\\ &=&\mu^{N-q}\int_{0}^{1}d\nu_{2}\nu_{2}^{p+q-N-1}(1-\nu_{2})^{N-1}(\mu-\nu_{2})^{-p}\\ &=&\mu^{N-p-q}\int_{0}^{1}d\nu_{2}\nu_{2}^{p+q-N-1}(1-\nu_{2})^{N-1}(1-\frac{1}{\mu}\nu_{2})^{-p}\\ &=&\mu^{N-p-q}B(p+q-N,N){}_{2}F_{1}(p+q-N,p;p+q;\frac{1}{\mu}) \end{eqnarray}
[4]
\begin{equation} \int_{0}^{1}\frac{x^{p-1}(1-x)^{q-1}}{(\mu-x)^{p+q-N}}dx=N\begin{pmatrix}p+q-1\\N\end{pmatrix}B(p,q)B(p+q-N,N){}_{2}F_{1}(p+q-N,p;p+q;\frac{1}{\mu}) \end{equation}