Dirichlet積分をしてみた
今回の記事はめちゃくちゃ短いです．
タイトルにもある様にDirichlet積分をするだけです．
次の事が成り立つ．
\begin{equation}
\int_{t_1+t_2+\cdots+t_n=1\wedge t_1,t_2,...,t_n>0}t_1^{\lambda_1-1}t_2^{\lambda_2-1}\cdots t_n^{\lambda_n-1}(1-t_1-t_2-\cdots t_{n})^{\mu -1}dt_1dt_2\cdots dt_n=\frac{\Gamma(\lambda_1)\Gamma(\lambda_2)\cdots \Gamma(\lambda_n)\Gamma(\mu)}{\Gamma(\lambda_1+\lambda_2+\cdots\lambda_n+\mu)}
\end{equation}
求める積分を$I_n(\lambda_1,\lambda_2,...,\lambda_n;\mu)$と置く．
$n=1$の場合はベータ関数となるので明らかに成り立つ．
\begin{eqnarray}
I_1(\lambda_1;\mu)&=&\int_0^1t_1^{\lambda_1-1}(1-t_1)^{\mu-1}dt_1\\
&=&\frac{\Gamma(\lambda_1)\Gamma(\mu)}{\Gamma(\lambda_1+\mu)}
\end{eqnarray}
そこで，一般の1,2,...,nまで成り立つと仮定する．そして，n+1の時を考える．
\begin{eqnarray}
I_{n+1}(\lambda_1,\lambda_2,...,\lambda_n,\lambda_{n+1};\mu)&=&
\int_0^1dt_1t_1^{\lambda_1-1}\int_{t_2+t_3+\cdots t_n+t_{n+1}=1-t_1 \wedge t_2,t_3,...,t_n,t_{n+1}>0}dt_2dt_3\cdots dt_ndt_{n+1}t_2^{\lambda_2-1}t_{3}^{\lambda_3-1}\cdots t_n^{\lambda_n-1}t_{n+1}^{\lambda_{n+1}-1}(1-t_1-t_2-\cdots t_n-t_{n+1})^{\mu-1}\\
&=&
\int_0^1dt_1t_1^{\lambda_1-1}\int_0^{1-t_1}dt_2t_2^{\lambda_2-1}\int_{t_3+\cdots t_n+t_{n+1}=1-t_1-t_2 \wedge t_3,...,t_n,t_{n+1}>0}dt_3\cdots dt_ndt_{n+1}t_{3}^{\lambda_3-1}\cdots t_n^{\lambda_n-1}t_{n+1}^{\lambda_{n+1}-1}(1-t_1-t_2-\cdots t_n-t_{n+1})^{\mu-1}\\
&=&\cdots\\
&=&
\int_0^1dt_1t_1^{\lambda_1-1}\int_0^{1-t_1}dt_2t_2^{\lambda_2-1}\cdots\int_0^{1-t_1-t_2-\cdots t_{n-1}}t_n^{\lambda_n - 1}\int_0^{1-t_1-t_2-\cdots -t_{n-1}-t_n}dt_{n+1}t_{n+1}^{\lambda_{n+1}-1}(1-t_1-t_2-\cdots -t_n-t_{n+1})^{\mu-1}
\end{eqnarray}
より，$t_{n+1}=(1-t_1-t_2-\cdots -t_n)s$と置きなおすと，
\begin{eqnarray}
\int_0^{1-t_1-t_2-\cdots-t_{n-1}-t_n}dt_{n+1}t_{n+1}^{\lambda_{n+1}-1}(1-t_1-t_2-\cdots -t_n-t_{n+1})^{\mu-1}
&=&
(1-t_1-t_2-\cdots-t_{n-1}-t_n)^{\lambda_{n+1}+\mu-1}\int_0^1s^{\lambda_{n+1}-1}(1-s)^{\mu-1}\\
&=&
(1-t_1-t_2-\cdots-t_{n-1}-t_{n})^{\lambda_{n+1}+\mu-1}\frac{\Gamma(\lambda_{n+1})\Gamma(\mu)}{\Gamma(\lambda_{n+1}+\mu)}
\end{eqnarray}
を得るので，これを先の式に代入して次式を得る．
\begin{eqnarray}
I_{n+1}(\lambda_1,\lambda_2,...,\lambda_n,\lambda_{n+1};\mu)&=&
\int_0^1dt_1t_1^{\lambda_1-1}\int_0^{1-t_1}dt_2t_2^{\lambda_2-1}\cdots \int_0^{1-t_1-t_2-\cdots -t_{n-1}}dt_nt_n^{\lambda_n-1}(1-t_1-t_2-\cdots -t_n)^{\lambda_{n+1}+\mu-1}\\
&=&
I_{n}(\lambda_1,\lambda_2,...,\lambda_n;\lambda_{n+1}+\mu)\frac{\Gamma(\lambda_{n+1})\Gamma(\mu)}{\Gamma(\lambda_{n+1}\mu)}\\
&=&
\frac{\Gamma(\lambda_1)\Gamma(\lambda_2)\cdots \Gamma(\lambda_n)\Gamma(\lambda_{n+1}+\mu)}{\Gamma(\lambda_1+\lambda_2+\cdots \lambda_n+\lambda_{n+1}+\mu)}\frac{\Gamma(\lambda_{n+1})\Gamma(\mu)}{\Gamma(\lambda_{n+1}+\mu)}\\
&=&
\frac{\Gamma(\lambda_1)\Gamma(\lambda_2)\cdots \Gamma(\lambda_n)\Gamma(\lambda_{n+1})\Gamma(\mu)}{\Gamma(\lambda_1+\lambda_2+\cdots \lambda_n+\lambda_{n+1}+\mu)}
\end{eqnarray}
よって，$n+1$の場合も成り立つことが示せたので，本定理は示された．
