Beta函数

$ \operatorname B:(m,n)\mapsto\int_0^1x^m(1-x)^n\mathrm dx

①の覚え方

$ \int_0^1x^{\boxed{\textcolor{red}{m}}}(1-x)^{\boxed{\textcolor{orange}{n}}}\mathrm dx^{\boxed{\textcolor{lightgreen}{1}}}=\frac{\boxed{\textcolor{red}{m}}!\boxed{\textcolor{orange}{n}}!}{\left((\boxed{\textcolor{red}{m}}+\boxed{\textcolor{orange}{n}}+\boxed{\textcolor{lightgreen}{1}}\right)!}

②の覚え方

$ \int_\alpha^\beta(x-\alpha)^m(\beta-x)^n\mathrm dx=\frac{m!n!}{(m+n+1)!}(\beta-\alpha)^{m+n+1}

分母と同様

$ \int_0^1x^m(1-x)^n\mathrm dx=\frac{m!n!}{(m+n+1)!}

$ \iff\int_0^1(x-0)^m(1-x)^n\mathrm dx=\frac{m!n!}{(m+n+1)!}(1-0)^{m+n+1}

から類推する

性質

$ \because\operatorname B(m,n)=\int_0^1x^m(1-x)^n\mathrm dx

$ = -\int_{1-x=1}^{1-x=0}(1-(1-x))^m(1-x)^n\mathrm d(1-x)

$ = -\int_1^0(1-x)^mx^n\mathrm dx

$ =\operatorname B(n,m)

$ \because\operatorname B(m,n)=\int_0^1x^m(1-x)^n\mathrm dx

$ =\int_{x=0}^{x=1}(1-x)^n\frac1{m+1}\mathrm d(x^{m+1})

$ =\frac1{m+1}((1-1)^n\cdot 1^{m+1}-(1-0)^n\cdot0^{m+1})-\frac1{m+1}\int_{x=0}^{x=1}x^{m+1}\mathrm d((1-x)^n)

$ =0+\frac n{m+1}\int_0^1x^{m+1}(1-x)^{n-1}\mathrm dx

$ =\frac{n}{m+1}\operatorname B(m+1,n-1)

と

$ \operatorname B(m,0)=\int_0^1x^m\mathrm dx

$ = \frac1{m+1}

より、

$ \operatorname B(m,n)=\frac{n}{m+1}\operatorname B(m+1,n-1)

$ =\frac{n^{\underline{2}}}{(m+2)^{\underline 2}}\operatorname B(m+2,n-2)

$ =\frac{n^{\underline{n}}}{(m+n)^{\underline n}}\operatorname B(m+n,0)

$ =\frac{m!n!}{(m+n+1)!}

$ \because\int_\alpha^\beta(x-\alpha)^m(\beta-x)^n\mathrm dx=\int_{x-\alpha=0}^{x-\alpha=\beta-\alpha}(x-\alpha)^m(\beta-\alpha-(x-\alpha))^n\mathrm d(x-\alpha)

$ = \int_0^{\beta-\alpha}x^m(\beta-\alpha-x)^n\mathrm dx

$ =(\beta-\alpha)^{m+n+1}\int_{\frac{x}{\beta-\alpha}=0}^{\frac{x}{\beta-\alpha}=1}\left(\frac{x}{\beta-\alpha}\right)^m\left(1-\frac{x}{\beta-\alpha}\right)^n\mathrm d\frac{x}{\beta-\alpha}

$ = (\beta-\alpha)^{m+n+1}\int_0^1x^m(1-x)^n\mathrm dx

$ = (\beta-\alpha)^{m+n+1}\operatorname B(m,n)