サンプル問題　解答

まず$I_n$を次のように定める.
\begin{equation*} I_n=\int_{\pi}^{2\pi}\dfrac{|\sin{nx}|}{x-\cos^{2}{nx}}\,dx \end{equation*}
$nx=t$と置換すると,
\begin{align*} I_n&=\dfrac{1}{n}\int_{n\pi}^{2n\pi}\dfrac{|\sin{t}|}{\frac{t}{n}-\cos^{2}{t}}\,dt\\ \\ &=\dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{k\pi}^{(k+1)\pi}\dfrac{|\sin{t}|}{\frac{t}{n}-\cos^{2}{t}}\,dt\\ \\ &=\dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{0}^{\pi}\dfrac{|\sin{(t+k\pi)}|}{\frac{t+k\pi}{n}-\cos^{2}{(t+k\pi)}}\,dt\\ &=\dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{0}^{\pi}\dfrac{\sin{t}}{\frac{t+k\pi}{n}-\cos^{2}{t}}\,dt \end{align*}
次のように不等式評価する.
\begin{equation} \dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{0}^{\pi}\dfrac{\sin{t}}{\frac{(k+1)\pi}{n}-\cos^{2}{t}}\,dt< I_n<\dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{0}^{\pi}\dfrac{\sin{t}}{\frac{k\pi}{n}-\cos^{2}{t}}\,dt \end{equation}
右辺と左辺をそれぞれ$J,K$とする.つまり,
\begin{equation} J=\dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{0}^{\pi}\dfrac{\sin{t}}{\frac{k\pi}{n}-\cos^{2}{t}}\,dt,\qquad K=\dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{0}^{\pi}\dfrac{\sin{t}}{\frac{(k+1)\pi}{n}-\cos^{2}{t}}\,dt \end{equation}
である.$J,K$について,$-\cos{t}=u$と置換する.$\sin{t}dt=du$
に注意して,
\begin{equation*} J=\dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{-1}^{1}\dfrac{du}{\frac{k\pi}{n}-u^2}=\dfrac{2}{n}\sum_{k=n}^{2n-1}\int_{0}^{1}\dfrac{du}{\frac{k\pi}{n}-u^2} \end{equation*}
\begin{equation*} K=\dfrac{1}{n}\sum_{k=n}^{2n-1}\int_{-1}^{1}\dfrac{du}{\frac{(k+1)\pi}{n}-u^2}=\dfrac{2}{n}\sum_{k=n}^{2n-1}\int_{0}^{1}\dfrac{du}{\dfrac{(k+1)\pi}{n}-u^2} \end{equation*}
簡単のために,次を定める.
\begin{equation*} J^{\prime}=\int_{0}^{1}\dfrac{du}{\frac{k\pi}{n}-u^2},K^{\prime}=\int_{0}^{1}\dfrac{du}{\frac{(k+1)\pi}{n}-u^2} \end{equation*}
$J^{\prime}$では,$u=\sqrt{\dfrac{k\pi}{n}}y $ $K^{\prime}$では,$u=\sqrt{\dfrac{(k+1)\pi}{n}}y$と置換すると,
\begin{equation*} J^{\prime}=\int_{0}^{1}\dfrac{du}{\frac{k\pi}{n}-u^2} =\int_{0}^{\sqrt{\frac{n}{k\pi}}}\dfrac{\sqrt{\frac{k\pi}{n}}dy}{{\frac{k\pi}{n}}(1-y^2)} =\dfrac{1}{\sqrt{\frac{k\pi}{n}}}\int_{0}^{\sqrt{\frac{n}{k\pi}}} \dfrac{dy}{1-y^2} \end{equation*}
\begin{equation*} =\dfrac{1}{\sqrt{\frac{k\pi}{n}}}\int_{0}^{\sqrt{\frac{n}{k\pi}}} \dfrac{1}{2}\biggl(\dfrac{1}{1+y}-\dfrac{1}{1-y}\biggl)\,dy\\ \\ =\dfrac{1}{2\sqrt{\frac{k\pi}{n}}}\log{\Biggl(\dfrac{1+\sqrt{\frac{n}{k\pi}}}{1-\sqrt{\frac{n}{k\pi}}}\Biggl)} \end{equation*}
\begin{equation*} J^{\prime}=\dfrac{1}{2\sqrt{\frac{k\pi}{n}}}\log{\Biggl(\dfrac{\sqrt{\frac{k\pi}{n}}+1}{\sqrt{\frac{k\pi}{n}}-1}\Biggl)}\qquad K^{\prime}=\dfrac{1}{2\sqrt{\frac{(k+1)\pi}{n}}}\log{\Biggl(\dfrac{\sqrt{\frac{(k+1)\pi}{n}}+1}{\sqrt{\frac{(k+1)\pi}{n}}-1}\Biggl)} \end{equation*}
$J,K$に戻すと,
\begin{equation} J=\dfrac{1}{n}\sum_{k=n}^{2n-1}\dfrac{1}{\sqrt{\frac{k\pi}{n}}}\log{\Biggl(\dfrac{\sqrt{\frac{k\pi}{n}}+1}{\sqrt{\frac{k\pi}{n}}-1}\Biggl)} \end{equation}
\begin{equation*} K=\dfrac{1}{n}\sum_{k=n}^{2n-1}\dfrac{1}{\sqrt{\frac{(k+1)\pi}{n}}}\log{\Biggl(\dfrac{\sqrt{\frac{(k+1)\pi}{n}}+1}{\sqrt{\frac{(k+1)\pi}{n}}-1}\Biggl)} \end{equation*}
和の始めを$n+1$から,終わりを$2n$にする.
\begin{equation} K=\dfrac{1}{n}\sum_{k=n+1}^{2n}\dfrac{1}{\sqrt{\frac{k\pi}{n}}}\log{\Biggl(\dfrac{\sqrt{\frac{k\pi}{n}}+1}{\sqrt{\frac{k\pi}{n}}-1}\Biggl)} \end{equation}
より,
\begin{equation*} \dfrac{1}{n}\sum_{k=n+1}^{2n}\dfrac{1}{\sqrt{\frac{k\pi}{n}}}\log{\Biggl(\dfrac{\sqrt{\frac{k\pi}{n}}+1}{\sqrt{\frac{k\pi}{n}}-1}\Biggl)}< I_n<\dfrac{1}{n}\sum_{k=n}^{2n-1}\dfrac{1}{\sqrt{\frac{k\pi}{n}}}\log{\Biggl(\dfrac{\sqrt{\frac{k\pi}{n}}+1}{\sqrt{\frac{k\pi}{n}}-1}\Biggl)} \end{equation*}
区分求積法,はさみうちの原理より,
\begin{equation*} \lim_{n\to\infty}I_n=\int_{1}^{2}\dfrac{1}{\sqrt{\pi x}}\log{\biggl(\dfrac{\sqrt{\pi x}+1}{\sqrt{\pi x}-1}\biggl)}\,dx \end{equation*}
$\sqrt{\pi x}=t$と置換すると,
\begin{equation*} \dfrac{2}{\pi}\int_{\sqrt{\pi}}^{\sqrt{2\pi}}\log{\biggl(\dfrac{t+1}{t-1}\biggl)}\,dt=\dfrac{2}{\pi}\int_{\sqrt{\pi}}^{\sqrt{2\pi}}\log{(t+1)}-\log{(t-1)}\,dt \end{equation*}
\begin{equation*} =\dfrac{2}{\pi}\biggl[t\log{\biggl(\dfrac{t+1}{t-1}\biggl)}+\log{\biggl(\dfrac{t+1}{t-1}\biggl)}\biggl]_{\sqrt{\pi}}^{\sqrt{2\pi}} \end{equation*}
\begin{equation*} =\dfrac{2}{\pi}\log{\dfrac{(\sqrt{2\pi}+1)^{\sqrt{2\pi}+1}(\sqrt{\pi}-1)^{\sqrt{\pi}-1}}{(\sqrt{2\pi}-1)^{\sqrt{2\pi}-1}(\sqrt{\pi}+1)^{\sqrt{\pi}+1}} } \end{equation*}
以上より,次を得る.
\begin{equation*} \lim_{n\to\infty}\int_{\pi}^{2\pi}\dfrac{|\sin{nx}|}{x-\cos^{2}{nx}}\,dx=\dfrac{2}{\pi}\log{\dfrac{(\sqrt{2\pi}+1)^{\sqrt{2\pi}+1}(\sqrt{\pi}-1)^{\sqrt{\pi}-1}}{(\sqrt{2\pi}-1)^{\sqrt{2\pi}-1}(\sqrt{\pi}+1)^{\sqrt{\pi}+1}} } \end{equation*}