１ 三角関数の漸化式の利用
\begin{eqnarray*} I_{n-1}-I_n &=&\int_{0}^{\cfrac{\pi}{2}}x^2(\cos^{2n-2}x-\cos^{2n}x)dx \hspace{28em}\\ &=&\int_{0}^{\cfrac{\pi}{2}}x^2\cos^{2n-2}x(1-\cos^2x)dx \\ &=&\int_{0}^{\cfrac{\pi}{2}}(x^2\sin x) \cos^{2n-2}x \sin xdx \\ &=&\bigg[(x^2\sin x)\big(-\cfrac{1}{2n-1}\cos^{2n-1}x\big) \bigg]_0^{\cfrac{\pi}{2}}-\int_{0}^{\cfrac{\pi}{2}}(2x\sin x+x^2\cos x)\big(-\cfrac{1}{2n-1}\cos^{2n-1}x\big)dx \\ &=&\cfrac{1}{2n-1}\bigg\{\int_{0}^{\cfrac{\pi}{2}}(2x\sin x \cos^{2n-1}xdx+\int_{0}^{\cfrac{\pi}{2}}x^2\cos^{2n}xdx \bigg\}\\ &=&\cfrac{2}{2n-1}\int_{0}^{\cfrac{\pi}{2}}x\sin x \cos^{2n-1}xdx+\cfrac{1}{2n-1}I_n \\ \end{eqnarray*}
\begin{eqnarray*} I_{n-1}-\cfrac{2n}{2n-1}I_n &=&\cfrac{2}{2n-1}\int_{0}^{\cfrac{\pi}{2}}x\cos^{2n-1}x\sin xdx \hspace{24em}\\ &=&\cfrac{2}{2n-1}\bigg\{\bigg[x\big(-\cfrac{1}{2n}\cos^{2n}x\big)\bigg]_{0}^{\cfrac{\pi}{2}}+\int_{0}^{\cfrac{\pi}{2}}\cfrac{1}{2n}\cos^{2n}xdx\bigg\}\\ &=&\cfrac{2}{(2n-1)(2n)}\int_{0}^{\cfrac{\pi}{2}}\cos^{2n}xdx \hspace{20em} (1)\\ \end{eqnarray*} ここで \[\int_{0}^{\cfrac{\pi}{2}}\cos^{2n}xdx=\cfrac{2n-1}{2n}\ \cfrac{2n-3}{2n-2} \cdots \cfrac{3}{4}\ \cfrac{1}{2}\ \cfrac{\pi}{2}=\cfrac{(2n-1)!!}{(2n)!!} \ \cfrac{\pi}{2} \quad より \hspace{18em}\] \begin{eqnarray*} I_{n-1}-\cfrac{2n}{2n-1}I_n &=&\cfrac{2}{(2n-1)(2n)}\cfrac{(2n-1)!!}{(2n)!!} \ \cfrac{\pi}{2} \hspace{26em}\\ &=&\cfrac{2(2n-3)!!}{(2n)^2 (2n-2)!!} \ \cfrac{\pi}{2}\\ &=&\cfrac{(2n-3)!!}{(2n-2)!!} \ \cfrac{\pi}{4} \ \cfrac{1}{n^2}\\ \end{eqnarray*} $\hspace{6em} \therefore \cfrac{(2n-2)!!}{(2n-3)!!}\ I_{n-1}-\cfrac{(2n)!!}{(2n-1)!!}\ I_n=\cfrac{\pi}{4} \ \cfrac{1}{n^2} $
$この式にn=m,m-1,\cdots , 3,2 を代入すると$
$\hspace{6em}\cfrac{(2m-2)!!}{(2m-3)!!}\ I_{m-1}-\cfrac{(2m)!!}{(2m-1)!!}\ I_m=\cfrac{\pi}{4}\ \cfrac{1}{m^2}$
$\hspace{6em}\cfrac{(2m-4)!!}{(2m-5)!!}\ I_{m-2}-\cfrac{(2m-2)!!}{(2m-3)!!}\ I_{m-1}=\cfrac{\pi}{4}\ \cfrac{1}{(m-1)^2}$
$\hspace{16em} \vdots$
$\hspace{6em}\cfrac{2!!}{1!!}\ I_1-\cfrac{4!!}{3!!}\ I_2=\cfrac{\pi}{4}\ \cfrac{1}{2^2} $
$\quad 辺々加えて$
$\hspace{6em}2I_1-\cfrac{(2m)!!}{(2m-1)!!}\ I_m=\cfrac{\pi}{4}\bigl(\cfrac{1}{2^2}+\cfrac{1}{3^2}+\cdots + \cfrac{1}{m^2}\bigr) \hspace{17em}(2)$
$\quad (1)でn=1とおくと$
\[I_0-\cfrac{2}{1}\ I_1=\cfrac{2}{1 \times 2}\int_{0}^{\cfrac{\pi}{2}}\cos^2xdx \hspace{27em}\] $\hspace{6em}\therefore \ I_0-2I_1=\cfrac{1}{2}\ \times \cfrac{\pi}{2}=\cfrac{\pi}{4} \times \cfrac{1}{1^2} \hspace{24em}(3)$
$\quad (2)+(3) より$
$\hspace{6em}I_0-\cfrac{(2m)!!}{(2m-1)!!}\ I_m=\cfrac{\pi}{4}\bigl(\cfrac{1}{1^2}+\cfrac{1}{2^2}+\cdots + \cfrac{1}{m^2}\bigr) \hspace{18em}(4) $
\[I_0=\int_{0}^{\cfrac{\pi}{2}}x^2dx=\cfrac{\pi^3}{24} だから、後は \cfrac{(2m)!!}{(2m-1)!!}\ I_m が求まればよい。\hspace{12em}\] $ジョルダンの不等式 \hspace{1em} \cfrac{2}{\pi}\ x < \sin x \quad (0 < x <\cfrac{\pi}{2} \ ) より （有名な不等式なので証明は略）$
\begin{eqnarray*} \cfrac{(2m)!!}{(2m-1)!!} \ I_m &=&\cfrac{(2m)!!}{(2m-1)!!}\int_{0}^{\cfrac{\pi}{2}}x^2\cos^{2m}xdx \hspace{26em}\\ &\leqq &\cfrac{(2m)!!}{(2m-1)!!}\int_{0}^{\cfrac{\pi}{2}}\bigl(\cfrac{\pi}{2}\bigr)^2\sin^2x\cos^{2m}xdx \\ &=&\big(\cfrac{\pi}{2}\big)^2\cfrac{(2m)!!}{(2m-1)!!}\int_{0}^{\cfrac{\pi}{2}}(1-\cos^2x)\cos^{2m}xdx \\ &=&\cfrac{\pi^2}{4}\ \cfrac{(2m)!!}{(2m-1)!!}\{\int_{0}^{\cfrac{\pi}{2}}\cos^{2m}xdx-\int_{0}^{\cfrac{\pi}{2}}\cos^{2m+2}xdx \\ &=&\cfrac{\pi^2}{4}\ \cfrac{(2m)!!}{(2m-1)!!}\left\{\cfrac{(2m-1)!!}{(2m)!!}-\cfrac{(2m+1)!!}{(2m+2)!!}\right\} \cfrac{\pi}{2}\\ &=&\cfrac{\pi^3}{8}\big(1-\cfrac{2m+1}{2m+2}\big)\\ &=&\cfrac{\pi^3}{8}\ \cfrac{1}{2m+2}\\ \end{eqnarray*} したがって
$\hspace{6em}m \rightarrow \infty のとき \quad \cfrac{(2m)!!}{(2m-1)!!}\ I_m \rightarrow 0$だから (4)は
\[\cfrac{\pi^3}{24}=\cfrac{\pi}{4}\sum_{n=1}^\infty\cfrac{1}{n^2}\hspace{32em}\] \[\therefore \sum_{n=1}^\infty\cfrac{1}{n^2}=\cfrac{\pi^2}{6}\hspace{32em}\]
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