x^3・exp(-α x^2)の広義積分[0,∞]

${\int }_0^{\mathrm{\infty }}{x}^3{e}^{-\alpha x^2}\mathrm{d}x$

$$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}{x}^3{e}^{-\alpha x^2}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}{\left(\frac{t}{\alpha }\right)}^2{e}^{-t}\frac{1}{2\sqrt{\alpha t}}\mathrm{d}t\\ & & \cdots t=\alpha x^2,x=0\to t=0,x=\mathrm{\infty }\to t=\mathrm{\infty },\\ & & \cdots x=\sqrt{\frac{t}{\alpha }}(\mathrm{積}\mathrm{分}\mathrm{範}\mathrm{囲}\mathrm{か}\mathrm{ら}x\ge 0\mathrm{と}\mathrm{す}\mathrm{る}(?))\\ & & \cdots x^3={\left(\sqrt{\frac{t}{\alpha }}\right)}^3={\left(\frac{t}{\alpha }\right)}^{\frac{3}{2}}\\ & & \cdots \frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha }}}\mathrm{d}t=\frac{1}{2\sqrt{\alpha t}}\mathrm{d}t\\ & =& {\int }_0^{\mathrm{\infty }}{\left(\frac{t}{\alpha }\right)}^{\frac{3}{2}}{e}^{-t}\frac{1}{2\sqrt{\alpha t}}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^{\frac{3}{2}}\sqrt{\alpha }}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{3}{2}}}{\sqrt{t}}{e}^{-t}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^2}{\int }_0^{\mathrm{\infty }}t{e}^{-t}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^2}{\int }_0^{\mathrm{\infty }}{t}^{2-1}{e}^{-t}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^2}\mathrm{\Gamma }\left(2\right)\cdots \mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}\mathrm{d}t(\mathfrak{Re}(z)>0)\\ & =& \frac{1}{2{\alpha }^2}(2-1)!\cdots \mathrm{\Gamma }(n)=(n-1)!(n\mathrm{は}\mathrm{自}\mathrm{然}\mathrm{数})\\ & =& \frac{1}{2{\alpha }^2}\cdot 1\\ & =& \frac{1}{2{\alpha }^2}\end{array}$$

x^4・exp(-α x^2) の広義積分[0,∞]

${\int }_0^{\mathrm{\infty }}{x}^4{e}^{-\alpha x^2}\mathrm{d}x$

$$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}{x}^4{e}^{-\alpha x^2}\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}{\left(\frac{t}{\alpha }\right)}^2{e}^{-t}\frac{1}{2\sqrt{\alpha t}}\mathrm{d}t\\ & & \cdots t=\alpha x^2,x=0\to t=0,x=\mathrm{\infty }\to t=\mathrm{\infty },\\ & & \cdots x^2=\frac{t}{\alpha },x^4=x^{2^2}={\left(\frac{t}{\alpha }\right)}^2\\ & & \cdots x=\sqrt{\frac{t}{\alpha }}(\mathrm{積}\mathrm{分}\mathrm{範}\mathrm{囲}\mathrm{か}\mathrm{ら}x\ge 0\mathrm{と}\mathrm{す}\mathrm{る}(?))\\ & & \cdots \frac{\mathrm{d}t}{\mathrm{d}x}=2\alpha x,\mathrm{d}x=\frac{1}{2\alpha x}\mathrm{d}t=\frac{1}{2\alpha \sqrt{\frac{t}{\alpha }}}\mathrm{d}t=\frac{1}{2\sqrt{\alpha t}}\mathrm{d}t\\ & =& {\int }_0^{\mathrm{\infty }}{\left(\frac{t}{\alpha }\right)}^2{e}^{-t}\frac{1}{2\sqrt{\alpha t}}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}{\int }_0^{\mathrm{\infty }}\frac{t^2}{\sqrt{t}}{e}^{-t}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}{\int }_0^{\mathrm{\infty }}{t}^2{t}^{-\frac{1}{2}}{e}^{-t}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}{\int }_0^{\mathrm{\infty }}{t}^{\frac{3}{2}}{e}^{-t}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}{\int }_0^{\mathrm{\infty }}{t}^{\frac{5}{2}-1}{e}^{-t}\mathrm{d}t\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}\mathrm{\Gamma }\left(\frac{5}{2}\right)\cdots \mathrm{\Gamma }(z)={\int }_0^{\mathrm{\infty }}{t}^{z-1}{e}^{-t}\mathrm{d}t(\mathfrak{Re}(z)>0)\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}\mathrm{\Gamma }(\frac{1}{2}+2)\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}\frac{(2\cdot 2-1)!!}{2^2}\sqrt{\pi }\cdots \mathrm{\Gamma }(\frac{1}{2}+n)=\frac{(2n-1)!!}{2^n}\sqrt{\pi }(!!\mathrm{は}\mathrm{二}\mathrm{重}\mathrm{階}\mathrm{乗})(n\mathrm{は}\mathrm{自}\mathrm{然}\mathrm{数})\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}\frac{3!!}{4}\sqrt{\pi }\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}\frac{3\cdot 1}{4}\sqrt{\pi }\cdots n\mathrm{の}\mathrm{二}\mathrm{重}\mathrm{階}\mathrm{乗}\mathrm{は}，1\mathrm{か}\mathrm{ら}n\mathrm{ま}\mathrm{で}n\mathrm{と}\mathrm{同}\mathrm{じ}\mathrm{偶}\mathrm{奇}\mathrm{性}\mathrm{を}\mathrm{持}\mathrm{つ}\mathrm{も}\mathrm{の}\mathrm{だ}\mathrm{け}\mathrm{を}\mathrm{全}\mathrm{て}\mathrm{掛}\mathrm{け}\mathrm{た}\mathrm{積}\\ & =& \frac{1}{2{\alpha }^2\sqrt{\alpha }}\frac{3\sqrt{\pi }}{4}\\ & =& \frac{3}{8{\alpha }^2}\sqrt{\frac{\pi }{\alpha }}\end{array}$$

逆行列の補題 / matrix inversion lemma / Sherman–Morrison–Woodbury formula

逆行列の補題 / matrix inversion lemma / Sherman–Morrison–Woodbury formula

$$\begin{array}{rcl}A& :& n\times n\text{matrix}\\ C& :& k\times k\text{matrix}\\ U& :& k\times n\text{matrix}\\ V& :& n\times k\text{matrix}\end{array}$$ の時， $$\begin{array}{rcl}{(A+UCV)}^{-1}& =& A^{-1}-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\end{array}$$

$(A+UCV)\left(\mathrm{右}\mathrm{辺}\right)=I_n$

$$\begin{array}{rcl}& & (A+UCV)\left(\mathrm{右}\mathrm{辺}\right)\\ & =& (A+UCV)\{A^{-1}-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\}\\ & =& (A+UCV)A^{-1}-(A+UCV)\left\{A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\right\}\\ & =& A{A}^{-1}+UCV{A}^{-1}-A\left\{A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\right\}-UCV\left\{A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\right\}\\ & =& I_n+UCV{A}^{-1}-A{A}^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}-UCV{A}^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\\ & =& I_n+U\{C-{(C^{-1}+V{A}^{-1}U)}^{-1}\}V{A}^{-1}-UCV{A}^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\\ & =& I_n+UC\{(\cancel{C^{-1}}+V{A}^{-1}U)-\cancel{C^{-1}}\}{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}-UCV{A}^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\\ & & \cdots X-Y^{-1}=XY{Y}^{-1}-X{X}^{-1}{Y}^{-1}=X(Y-X^{-1})Y^{-1}\\ & =& I_n+UC\left(V{A}^{-1}U\right){(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}-UCV{A}^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\\ & =& I_n+\cancel{UCV{A}^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}}-\cancel{UCV{A}^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}}\\ & =& I_n\end{array}$$

$\left(\mathrm{右}\mathrm{辺}\right)(A+UCV)=I_n$

$$\begin{array}{rcl}& & \left(\mathrm{右}\mathrm{辺}\right)(A+UCV)\\ & =& \{A^{-1}-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}\}(A+UCV)\\ & =& A^{-1}A+A^{-1}UCV-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}A-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}UCV\\ & =& I_n+A^{-1}UCV-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}UCV\\ & =& I_n+A^{-1}UCV-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}UCV\\ & =& I_n+A^{-1}U\{C-{(C^{-1}+V{A}^{-1}U)}^{-1}\}V-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}UCV\\ & =& I_n+A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}\{(\cancel{C^{-1}}+V{A}^{-1}U)-\cancel{C^{-1}}\}CV-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}UCV\\ & & \cdots X-Y^{-1}=Y^{-1}YX-Y^{-1}{X}^{-1}X=Y^{-1}(Y-X^{-1})X\\ & =& I_n+A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}\left(V{A}^{-1}U\right)CV-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}UCV\\ & =& I_n+\cancel{A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}UCV}-\cancel{A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1}UCV}\\ & =& I_n\end{array}$$

以上より

以上より右辺$(A^{-1}-A^{-1}U{(C^{-1}+V{A}^{-1}U)}^{-1}V{A}^{-1})$は$A+UCV$の逆行列である．

$C=I_k$の時

また$C=I_k$の時は，$C^{-1}=I_k$でもあるので， $$\begin{array}{rcl}{(A+UV)}^{-1}& =& A^{-1}-A^{-1}U{(I_k+V{A}^{-1}U)}^{-1}V{A}^{-1}\end{array}$$ となる．