ヘロンの公式

余弦定理

$$\begin{eqnarray} |BH|^2&=&c^2-|AH|^2 \\&=&c^2-c^2 \cos^2(A) \\&=&a^2-|CH|^2 \\&=&a^2-\{b-c \cos(A)\}^2 \\&=&a^2-b^2+2bc\cos(A)-c^2\cos^2(A) \end{eqnarray}$$ $$\begin{eqnarray} c^2\cancel{-c^2 \cos^2(A)}&=&a^2-b^2+2bc\cos(A)\cancel{-c^2\cos^2(A)} \\c^2&=&a^2-b^2+2bc\cos(A) \\b^2+c^2-a^2&=&2bc\cos(A) \\\cos(A)&=&\frac{b^2+c^2-a^2}{2bc} \end{eqnarray}$$

ヘロンの公式

$$\begin{eqnarray} S^2&=&\left(\frac{1}{2}|AC||BH|\right)^2 \\&=&\left\{\frac{1}{2}bc\sin{(A)}\right\}^2 \\&=&\frac{1}{4}b^2c^2\sin^2{(A)} \\&=&\frac{1}{4}b^2c^2\left\{1-\cos^2{(A)}\right\}\;\cdots\;\sin^2{(x )}+\cos^2{(x)}=1 \\&=&\frac{1}{4}b^2c^2-\frac{1}{4}b^2c^2\cos^2{(A)} \\&=&\frac{1}{4}b^2c^2-\frac{1}{4}b^2c^2\left(\frac{b^2+c^2-a^2}{2bc}\right)^2\;\cdots\;\cos{(A)}=\frac{b^2+c^2-a^2}{2bc} \\&=&\frac{1}{4}b^2c^2-\frac{1}{4}\cancel{b^2c^2}\frac{\left(b^2+c^2-a^2\right)^2}{4\cancel{b^2c^2}} \\&=&\frac{1}{4}b^2c^2-\frac{1}{16}\left(b^2+c^2-a^2\right)^2 \\&=&\frac{1}{16}\left\{4b^2c^2-\left(b^2+c^2-a^2\right)^2\right\} \\&=&\frac{1}{16}\left\{\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\right\} \\&=&\frac{1}{16}\left[\left\{2bc+\left(b^2+c^2-a^2\right)\right\}\left\{2bc-\left(b^2+c^2-a^2\right)\right\}\right] \\&=&\frac{1}{16}\left\{\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\right\} \\&=&\frac{1}{16}\left[\left\{\left(b^2+2bc+c^2\right)-a^2\right\}\left\{a^2-\left(b^2-2bc+c^2\right)\right\}\right] \\&=&\frac{1}{16}\left[\left\{\left(b+c\right)^2-a^2\right\}\left\{a^2-\left(b-c\right)^2\right\}\right]\;\cdots\;(a \pm b)^2=a^2\pm2ab+b^2 \\&=&\frac{1}{16}\left[\left\{\left(b+c+a\right)\left(b+c-a\right)\right\}\left\{\left(a+b-c\right)\left(a-\left(b-c\right)\right)\right\}\right] \\&=&\frac{1}{16}\left[\left\{\left(b+c+a\right)\left(b+c-a\right)\right\}\left\{\left(a+b-c\right)\left(a-b+c\right)\right\}\right] \\&=&\frac{1}{16}\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right) \\&=&\frac{a+b+c}{2}\frac{-a+b+c}{2}\frac{a-b+c}{2}\frac{a+b-c}{2} \\&=&\frac{a+b+c}{2}\frac{-a+b+c\color{red}{+a-a}}{2}\frac{a-b+c\color{red}{+b-b}}{2}\frac{a+b-c\color{red}{+c-c}}{2} \\&=&\frac{a+b+c}{2}\frac{a+b+c-2a}{2}\frac{a+b+c-2b}{2}\frac{a+b+c-2c}{2} \\&=&\frac{a+b+c}{2}\left(\frac{a+b+c}{2}-a\right)\left(\frac{a+b+c}{2}-b\right)\left(\frac{a+b+c}{2}-c\right) \\&=&s\left(s-a\right)\left(s-b\right)\left(s-c\right)\;\cdots\;s=\frac{a+b+c}{2} \end{eqnarray}$$ $$\begin{eqnarray} S&=&\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\;\cdots\;s=\frac{a+b+c}{2} \end{eqnarray}$$

バーゼル問題

オリジナル: https://youtu.be/6olRd_dDS80

準備1．\(2nA_n=\left(2n-1\right)A_{n-1}\)

$$\begin{eqnarray} A_n&=&\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\int_0^{\frac{\pi}{2}}\cos{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&\left[\sin{\left(x\right)\cdot\cos^{2n-1}{\left(x\right)}}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}\sin{\left(x\right)}\left\{-\left(2n-1\right)\cos^{2n-2}{\left(x\right)}\sin{\left(x\right)}\right\}\mathrm{d}x \\&&\;\dots\;(fg)^\prime=f^\prime g+fg^\prime,f^\prime g=(fg)^\prime-fg^\prime,\int f^\prime g=\int(fg)^\prime-\int fg^\prime,\int f^\prime g=\left[fg\right]-\int fg^\prime \\&&\;\dots\;f^\prime(x)=\cos{\left(x\right)} \\&&\;\dots\;f(x)=\sin{\left(x\right)} \\&&\;\dots\;g(x)=\cos^{2n-1}{\left(x\right)}=u^{2n-1}(x)\;\cdots\;u=\cos{\left(x\right)} \\&&\;\dots\;g^\prime(x)=\frac{\mathrm{d}g}{\mathrm{d}x}=\frac{\mathrm{d}g}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} =\left(2n-1\right)u^{2n-2}(x)\cdot\left(-\sin{\left(x\right)}\right) =-\left(2n-1\right)\cos^{2n-2}{\left(x\right)}\sin{\left(x\right)} \\&=&\left[0-0\right]+\left(2n-1\right)\int_0^{\frac{\pi}{2}}\color{red}{\sin^2{\left(x\right)}}\color{black}{}\cos^{2n-2}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\color{red}{\left(1-\cos^2{\left(x\right)}\right)}\color{black}{}\cos^{2n-2}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\left(\cos^{2n-2}{\left(x\right)}-\cos^{2n}{\left(x\right)}\right)\mathrm{d}x \\&=&\left(2n-1\right)\int_0^{\frac{\pi}{2}}\cos^{2(n-1)}{\left(x\right)}\mathrm{d}x-\left(2n-1\right)\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\left(2n-1\right)A_{n-1}-\left(2n-1\right)A_n \\A_n+\left(2n-1\right)A_n&=&\left(2n-1\right)A_{n-1} \\2nA_n&=&\left(2n-1\right)A_{n-1} \end{eqnarray}$$

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準備2．\(A_n=(2n-1)nB_{n-1}-2n^2B_{n}\)

$$\begin{eqnarray} A_n&=&\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\left[x\cdot\cos^{2n}{\left(x\right)}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}x\left\{2n\cos^{2n-1}{\left(x\right)\left(-\sin{\left(x\right)}\right)}\right\}\mathrm{d}x \\&&\;\dots\;(fg)^\prime=f^\prime g+fg^\prime,f^\prime g=(fg)^\prime-fg^\prime,\int f^\prime g=\int(fg)^\prime-\int fg^\prime,\int f^\prime g=\left[fg\right]-\int fg^\prime \\&&\;\dots\;f^\prime(x)=1 \\&&\;\dots\;f(x)=x \\&&\;\dots\;g(x)=\cos^{2n}{\left(x\right)}=u^{2n}\;\cdots\;u=\cos{\left(x\right)} \\&&\;\dots\;g^\prime(x)=\frac{\mathrm{d}g}{\mathrm{d}x}=\frac{\mathrm{d}g}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} =2nu^{2n-1}\cdot\left(-\sin{\left(x\right)}\right) =2n\cos^{2n-1}{\left(x\right)}\left(-\sin{\left(x\right)}\right) \\&=&\left[0-0\right]+2n\int_0^{\frac{\pi}{2}}x\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&2n\int_0^{\frac{\pi}{2}}x\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\mathrm{d}x \\&=&2n\left[\left[\frac{1}{2}x^2\cdot\sin{\left(x\right)}\cos^{2n-1}{\left(x\right)}\right]_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}\frac{1}{2}x^2\left\{ \cos{\left(x\right)}\cdot\cos^{2n-1}{\left(x\right)} +\sin{\left(x\right)}\cdot(2n-1)\cos^{2n-2}{\left(x\right)}\left(-\sin{\left(x\right)}\right) \right\}\mathrm{d}x \right] \\&=&-2n\frac{1}{2}\int_0^{\frac{\pi}{2}}x^2\left\{ \cos^{2n}{\left(x\right)}-(2n-1)\color{red}{\sin^2{\left(x\right)}}\color{black}{}\cos^{2n-2}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\color{red}{\left\{1-\cos^2{\left(x\right)}\right\}}\color{black}{}\cos^{2n-2}{\left(x\right)} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\left\{\cos^{2n-2}{\left(x\right)}-\cos^2{\left(x\right)}\cos^{2n-2}{\left(x\right)}\right\} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left[ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\left\{\cos^{2n-2}{\left(x\right)}-\cos^{2n}{\left(x\right)}\right\} \right]\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left\{ x^2\cos^{2n}{\left(x\right)}-(2n-1)x^2\cos^{2n-2}{\left(x\right)}+(2n-1)x^2\cos^{2n}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}}\left\{ -(2n-1)x^2\cos^{2n-2}{\left(x\right)}+2nx^2\cos^{2n}{\left(x\right)} \right\}\mathrm{d}x \\&=&-n\int_0^{\frac{\pi}{2}} -(2n-1)x^2\cos^{2(n-1)}{\left(x\right)} \mathrm{d}x -n\int_0^{\frac{\pi}{2}} 2nx^2\cos^{2n}{\left(x\right)} \mathrm{d}x \\&=&(2n-1)n\int_0^{\frac{\pi}{2}} x^2\cos^{2(n-1)}{\left(x\right)} \mathrm{d}x -2n^2\int_0^{\frac{\pi}{2}} x^2\cos^{2n}{\left(x\right)} \mathrm{d}x \\&=&(2n-1)nB_{n-1}-2n^2B_{n}\;\cdots\;B_n=\int_0^{\frac{\pi}{2}}x^2\cos^{2n}{\left(x\right)}\mathrm{d}x \end{eqnarray}$$

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準備3．\(\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0\)

$$\begin{eqnarray} B_n&=&\int_0^{\frac{\pi}{2}}x^2\cos^{2n}{\left(x\right)}\mathrm{d}x \\&\leq&\int_0^{\frac{\pi}{2}}\left\{\frac{\pi}{2} \sin{\left(x\right)}\right\}^2\cos^{2n}{\left(x\right)}\mathrm{d}x \\&&\;\cdots\;\frac{2}{\pi}x\leq\sin{\left(x\right)}\;\cdots\;x\in\left[0,\frac{\pi}{2}\right] \\&&\;\cdots\;x\leq\frac{\pi}{2}\sin{\left(x\right)} \\&&\;\cdots\;(0,0)と(\frac{\pi}{2},1)を通る直線\frac{2}{\pi}xは，区間\left[0,\frac{\pi}{2}\right]では常に\sin{\left(x\right)}以下である(図を参照)． \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\sin^{2}{\left(x\right)}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\color{red}{\sin^{2}{\left(x\right)}}\color{black}{}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\color{red}{\left\{1-\cos^{2}{\left(x\right)}\right\}}\color{black}{}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\left\{1-\cos^{2}{\left(x\right)}\right\}\cos^{2n}{\left(x\right)}\mathrm{d}x \\&=&\frac{\pi^2}{4}\int_0^{\frac{\pi}{2}}\left\{\cos^{2n}{\left(x\right)}-\cos^{2n+2}{\left(x\right)}\right\}\mathrm{d}x \\&=&\frac{\pi^2}{4}\left[\int_0^{\frac{\pi}{2}}\cos^{2n}{\left(x\right)}\mathrm{d}x-\int_0^{\frac{\pi}{2}}\cos^{2(n+1)}{\left(x\right)}\mathrm{d}x\right] \\&=&\frac{\pi^2}{4}\left(A_n-A_{n+1}\right) \\&=&\frac{\pi^2}{4}\left(A_n-\frac{2(n+1)-1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{2(n+1)-\left\{2(n+1)-1\right\}}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{2(n+1)-2(n+1)+1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\left(\frac{1}{2(n+1)}A_{n}\right) \\&=&\frac{\pi^2}{4}\frac{A_{n}}{2(n+1)} \\B_n&\leq&\frac{\pi^2}{4}\frac{A_n}{2(n+1)} \\B_n\frac{1}{A_n}&\leq&\frac{\pi^2}{4}\frac{A_n}{2(n+1)}\frac{1}{A_n}\;\cdots\;A_n\geq0 \\\frac{B_n}{A_n}&\leq&\frac{\pi^2}{4}\frac{1}{2(n+1)} \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&\leq&\lim_{n\rightarrow\infty}\frac{\pi^2}{4}\frac{1}{2(n+1)}=0 \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&\geq&0 \;\cdots\;A_n,B_nは共に被積分凾数\left(\cos^{2n}{\left(x\right)}, x^2\cos^{2n}{\left(x\right)}\right)が偶数乗のため，常に値は0以上． \\&&\;\cdots\;よって，その積分結果も常に0以上となる． \\\lim_{n\rightarrow\infty}\frac{B_n}{A_n}&=&0\;\cdots\;前述の2つから，はさみうちの原理より極限は0． \end{eqnarray}$$

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\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)

$$\begin{eqnarray} \\\sum_{n=1}^{\infty}\frac{1}{n^2}&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{1}{n^2} \\&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\left\{\frac{(2n-1)B_{n-1}}{nA_n}-\frac{2B_{n}}{A_n}\right\} \\&&\;\cdots\;A_n=(2n-1)nB_{n-1}-2n^2B_{n}\;(準備2．) \\&&\;\cdots\;\frac{A_n}{n^2}=\frac{(2n-1)nB_{n-1}}{n^2}-\frac{2n^2B_{n}}{n^2} \\&&\;\cdots\;\frac{A_n}{n^2}=\frac{(2n-1)B_{n-1}}{n}-2B_{n} \\&&\;\cdots\;\frac{1}{n^2}=\frac{(2n-1)B_{n-1}}{nA_n}-\frac{2B_{n}}{A_n} \\&=&\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{2B_{n-1}}{A_{n-1}}-\frac{2B_{n}}{A_n} \\&&\;\cdots\;2nA_n=\left(2n-1\right)A_{n-1}\;(準備1．) \\&&\;\cdots\;\frac{2}{A_{n-1}}nA_n=\left(2n-1\right) \\&&\;\cdots\;\frac{2}{A_{n-1}}=\frac{\left(2n-1\right)}{nA_n} \\&=&2\lim_{N\rightarrow\infty} \sum_{n=1}^{N}\frac{B_{n-1}}{A_{n-1}}-\frac{B_{n}}{A_n} \\&=&2\lim_{N\rightarrow\infty} \left\{ \left(\frac{B_{1-1}}{A_{1-1}}-\frac{B_{1}}{A_1}\right) +\left(\frac{B_{2-1}}{A_{2-1}}-\frac{B_{2}}{A_2}\right) +\left(\frac{B_{3-1}}{A_{3-1}}-\frac{B_{3}}{A_3}\right) +\cdots+\left(\frac{B_{N-1}}{A_{N-1}}-\frac{B_{N}}{A_N}\right) \right\} \\&=&2\lim_{N\rightarrow\infty} \left\{ \frac{B_{0}}{A_{0}}\cancel{-\frac{B_{1}}{A_1} +\frac{B_{1}}{A_{1}}}\cancel{-\frac{B_{2}}{A_2} +\frac{B_{2}}{A_{2}}}\cancel{-\frac{B_{3}}{A_3} +\cdots}+\cancel{\cdots+\frac{B_{N-1}}{A_{N-1}}}-\frac{B_{N}}{A_N} \right\} \;\cdots\;Telescoping Series \\&=&2\left(\frac{B_{0}}{A_{0}}-\lim_{N\rightarrow\infty}\frac{B_{N}}{A_N}\right) \\&=&2\left(\frac{B_{0}}{A_{0}}-0\right) \;\cdots\;\lim_{n\rightarrow\infty}\frac{B_n}{A_n}=0\;(準備3．) \\&=&2\frac{B_{0}}{A_{0}} \\&=&\left.2\frac{\int_0^{\frac{\pi}{2}}x^2\cos^{2\cdot n}{\left(x\right)}\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\cos^{2\cdot n}{\left(x\right)}\mathrm{d}x}\right|_{n=0} \\&=&2\frac{\int_0^{\frac{\pi}{2}}x^2\cos^{2\cdot0}{\left(x\right)}\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\cos^{2\cdot0}{\left(x\right)}\mathrm{d}x} \\&=&2\frac{\int_0^{\frac{\pi}{2}}x^2\mathrm{d}x}{\int_0^{\frac{\pi}{2}}\mathrm{d}x} \\&=&2\frac{\left[\frac{1}{3}x^3\right]_0^{\frac{\pi}{2}}}{\left[x\right]_0^{\frac{\pi}{2}}} \\&=&2\frac{\left[\frac{1}{3}\left(\frac{\pi}{2}\right)^3-\frac{1}{3}0^3\right]}{\left[\frac{\pi}{2}-0\right]} \\&=&2\frac{\frac{\pi^3}{24}}{\frac{\pi}{2}} \\&=&2\frac{\pi^2}{12} \\&=&\frac{\pi^2}{6} \end{eqnarray}$$

t分布の期待値と分散

t分布の期待値

\begin{eqnarray} \mathrm{E}\left[X\right]&=&\int_{-\infty}^{\infty}x\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \;\ldots\;\href{https://shikitenkai.blogspot.com/2022/10/t.html}{t分布の確率密度凾数:\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}}} \\&=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \end{eqnarray} \begin{eqnarray} 彼積分凾数(x)&=&x\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\彼積分凾数(-x)&=&(-x)\left(1+\frac{(-x)^2}{n}\right)^{-\frac{n+1}{2}} \\&=&-x\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&-彼積分凾数(x)\;\ldots\;奇凾数 \end{eqnarray} \begin{eqnarray} \mathrm{E}\left[X\right] &=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}}\cdot0 \\&&\;\ldots\;奇凾数の上端と下端の絶対値が等しい定積分は0． \\&&\;\ldots\;\int_{-a}^{a}f(x)\mathrm{d}x=0,f(x)が奇凾数の場合 \\&=&0 \end{eqnarray}

t分布の分散

\begin{eqnarray} \mathrm{E}\left[X^2\right] &=&\int_{-\infty}^{\infty}x^2\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&=&\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x^2 \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \end{eqnarray} \begin{eqnarray} 彼積分凾数(x)&=&x^2\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\彼積分凾数(-x)&=&(-x)^2\left(1+\frac{(-x)^2}{n}\right)^{-\frac{n+1}{2}} \\&=&x^2\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&彼積分凾数(x)\;\ldots\;偶凾数 \end{eqnarray} \begin{eqnarray} \mathrm{E}\left[X^2\right]&=&\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x^2 \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&=&2\int_{0}^{\infty}x^2\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&&\;\ldots\;偶凾数の上端と下端の絶対値が等しい定積分は2\int_{0}^{a}f(x)\mathrm{d}x． \\&&\;\ldots\;\int_{-a}^{a}f(x)\mathrm{d}x=2\int_{0}^{a}f(x)\mathrm{d}x,f(x)が偶凾数の場合 \\&=&2\int_{1}^{0}\left(n\left(t^{-1}-1\right)\right)\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{n\left(t^{-1}-1\right)}{n}\right)^{-\frac{n+1}{2}} n^{\frac{1}{2}}\frac{1}{2}\left(t^{-1}-1\right)^{-\frac{1}{2}}\left(-t^{-2}\right) \mathrm{d}t \\&&\;\ldots\;x^2=n\left(t^{-1}-1\right) \\&&\;\ldots\;x=n^{\frac{1}{2}}\left(t^{-1}-1\right)^{\frac{1}{2}}=n^{\frac{1}{2}}s^{\frac{1}{2}}\;\ldots\;s=t^{-1}-1 \\&&\;\ldots\;x:0\rightarrow\infty, t:1\rightarrow0 \\&&\;\ldots\;\frac{\mathrm{d}{x}}{\mathrm{d}{t}}=\frac{\mathrm{d}{x}}{\mathrm{d}{s}}\frac{\mathrm{d}{s}}{\mathrm{d}{t}} =n^{\frac{1}{2}}\frac{1}{2}s^{\frac{1}{2}-1}\frac{\mathrm{d}{s}}{\mathrm{d}{t}}=n^{\frac{1}{2}}\frac{1}{2}\left(t^{-1}-1\right)^{-\frac{1}{2}}\left(-t^{-2}\right) \\&=&\cancel{2}\frac{1}{\cancel{\sqrt{n}}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}}\cancel{\frac{1}{2}} n\cancel{n^{\frac{1}{2}}}(-1) \int_{1}^{0} \left(1+\frac{\cancel{n}\left(t^{-1}-1\right)}{\cancel{n}}\right)^{-\frac{n+1}{2}} \left(t^{-1}-1\right)^{1-\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} \left(\cancel{1}+t^{-1}\cancel{-1}\right)^{-\frac{n+1}{2}} \left(t^{-1}-1\right)^{\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n+1}{2}} \left(t^{-1}-1\right)^{\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{t}{t}\left(t^{-1}-1\right)\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{1}{t}\left(1-t\right)\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{1}{t}\right)^\frac{1}{2}\left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} t^{-\frac{1}{2}}\left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{4}{2}} \left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-1-1} \left(1-t\right)^{\frac{3}{2}-1} \mathrm{d}t \\&=&\frac{n}{ \frac{ \Gamma{\left(\frac{1}{2}\right)} \Gamma{\left(\frac{n}{2}\right)} }{ \Gamma{ \left(\frac{1}{2}+\frac{n}{2}\right) } }} \frac{ \Gamma{\left(\frac{n}{2}-1\right)} \Gamma{\left(\frac{3}{2}\right)} }{ \Gamma{ \left(\frac{n}{2}-1+\frac{3}{2}\right) } } \;\ldots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{ \beta{\left(a,b\right)}=\frac{ \Gamma{\left(a\right)} \Gamma{\left(b\right)} }{ \Gamma{ \left(a+b\right) } }=\int_0^1 x^{a-1}(1-x)^{b-1}\mathrm{d}x} \\&=&n \frac{ \cancel{\Gamma{ \left(\frac{1}{2}+\frac{n}{2}\right) }} }{ \Gamma{\left(\frac{1}{2}\right)} \color{green}{\Gamma{\left(\frac{n}{2}\right)}} } \frac{ \Gamma{\left(\frac{n}{2}-1\right)} \color{blue}{\Gamma{\left(\frac{3}{2}\right)}} }{ \cancel{\Gamma{ \left(\frac{n}{2}+\frac{1}{2}\right) }} } \\&=&n \frac{1}{ \cancel{\Gamma{\left(\frac{1}{2}\right)}} \color{green}{\left(\frac{n}{2}-1\right)\cancel{\Gamma{\left(\frac{n}{2}-1\right)}}} } \frac{ \cancel{\Gamma{\left(\frac{n}{2}-1\right)}} \color{blue}{\frac{1}{2}\cancel{\Gamma{\left(\frac{1}{2}\right)}}} }{1} \;\ldots\;\href{https://shikitenkai.blogspot.com/2020/08/s1ss.html}{\Gamma\left(z+1\right)=z\Gamma\left(z\right)} \\&=&n \frac{1}{ \left(\frac{n}{2}-1\right) } \frac{ \frac{1}{2} }{1} \\&=&\frac{n}{n-2} \end{eqnarray} \begin{eqnarray} \mathrm{V}\left[X\right]&=&\mathrm{E}\left[X^2\right]-\mathrm{E}\left[X\right]^2 \\&=&\frac{n}{n-2}-0^2 \\&=&\frac{n}{n-2} \end{eqnarray}