金沢大学(理系) 2026年　問題4


\[自然数 \ n\ に対し、I_n=\int_0^1 x^{\sqrt{n}} |\cos (n\pi x)|dx \ \ とおく。次の問いに答えよ。\] \[(1)\ \ k=1,\ 2,\ \cdots , \ 2n\ \ に対し、次の等式を示せ。\quad \int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} |\cos (n\pi x)|dx=\dfrac{1}{n\pi}\] \[(2)\ \ 次の不等式が成り立つことを示せ。 \quad I_n \leqq \dfrac{1}{n\pi} \sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}\] \[(3)\ \ 次の不等式が成り立つことを示せ。 \quad \dfrac{1}{n}\sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}} \leqq \dfrac{2}{\sqrt{n}+1}+\dfrac{1}{n}\] \[(4)\ \ 極限値 \ \ \lim_{n \rightarrow \infty} \sqrt{n}I_n \ \ を求めよ。\]


(1)


\[J_n=\int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} |\cos (n\pi x)|dx \ \ において\] \[n\pi x=t \ \ とおくと \quad dx=\dfrac{dt}{n\pi} \qquad \begin{array}{c|c} x & \dfrac{k-1}{2n} \ \ \rightarrow \dfrac{k}{2n} \quad \\ \hline t & \dfrac{k-1}{2}\pi \rightarrow \dfrac{k}{2}\pi \\ \end{array} \]
\[J_n=\int_{\scriptsize{\dfrac{k-1}{2}\pi}}^{\scriptsize{\dfrac{k}{2}\pi}} |\cos t | \times \dfrac{dt}{n\pi}=\dfrac{1}{n\pi} \int_{\scriptsize{\dfrac{k-1}{2}\pi}}^{\scriptsize{\dfrac{k}{2}\pi}} |\cos t |dt\]

　

$この定積分は \ \cos t \ の対称性により \ k\ の値によらず$

\[J_n=\dfrac{1}{n\pi} \int_0^{\scriptsize{\dfrac{\pi}{2}}} \cos t dt=\dfrac{1}{n\pi} \big[\sin t \big]_0^{\scriptsize{\dfrac{\pi}{2}}}=\dfrac{1}{n\pi}\]

(2)


$x^{\sqrt{n}}\ は \ x \geqq 0 \ で単調増加だから$

$k=1,\ 2,\ \cdots , \ 2n \ \ に対し \quad \dfrac{k-1}{2n} \leqq x \leqq \dfrac{k}{2n} \ \ のとき \quad \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}}\leqq x^{\sqrt{n}} \leqq \big(\dfrac{k}{2n}\big)^{\sqrt{n}}$

$両辺に\ \ |\cos (n\pi x)|\ \ をかけて$

$\big(\dfrac{k-1}{2n}\big)^{\sqrt{n}} |\cos (n\pi x)| \leqq x^{\sqrt{n}} |\cos (n\pi x)| \leqq \big(\dfrac{k}{2n}\big)^{\sqrt{n}} |\cos (n\pi x)| \hspace{5em} (A)$

$右側の不等式をつかって(1)より$

\[\int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} x^{\sqrt{n}} |\cos (n\pi x)|dx \leqq \int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}|\cos (n\pi x)|dx = \big(\dfrac{k}{2n}\big)^{\sqrt{n}} \int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} |\cos (n\pi x)|dx= \dfrac{1}{n\pi} \big(\dfrac{k}{2n}\big)^{\sqrt{n}} \]
$両辺の和をとって$

\[\sum_{k=1}^{2n} \int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} x^{\sqrt{n}} |\cos (n\pi x)|dx \leqq \sum_{k=1}^{2n} \dfrac{1}{n\pi} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}\] \[\int_0^1 x^{\sqrt{n}} |\cos (n\pi x)|dx \leqq \dfrac{1}{n\pi} \sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}\] \[\therefore I\ \ _n \leqq \dfrac{1}{n\pi} \sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}\]

(3)

　

\begin{eqnarray*} & & \dfrac{1}{n} \sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}\\ \\ &=& 2 \times \dfrac{1}{2n} \sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}\\ \\ &=& 2 \times \dfrac{1}{2n} \big\{ \big(\dfrac{1}{2n}\big)^{\sqrt{n}} + \big(\dfrac{2}{2n}\big)^{\sqrt{n}} + \cdots + \big(\dfrac{2n}{2n}\big)^{\sqrt{n}}\big\}\\ \\ &=& 2 \times \dfrac{1}{2n} \big\{\big(\dfrac{0}{2n}\big)^{\sqrt{n}} + \big(\dfrac{1}{2n}\big)^{\sqrt{n}} + \cdots + \big(\dfrac{2n-1}{2n}\big)^{\sqrt{n}}\big\}+ \dfrac{1}{n}\\ \\ &\leqq& 2 \times \int_0^1x^{\sqrt{n}}dx+\dfrac{1}{n}\\ \\ &=&2\big[\dfrac{x^{\sqrt{n}+1}}{\sqrt{n}+1}\big]_0^1+\dfrac{1}{n}\\ \\ &=&\dfrac{2}{\sqrt{n}+1}+\dfrac{1}{n} \end{eqnarray*}

(4)


$(2)の(A)の左側の不等式をつかって$

\begin{eqnarray*} & & \int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} x^{\sqrt{n}} |\cos (n\pi x)|dx \\ \\ &\geqq& \int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}}|\cos (n\pi x)|dx\\ \\ &=&\big(\dfrac{k-1}{2n}\big)^{\sqrt{n}} \int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} |\cos (n\pi x)|dx\\ \\ &=& \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}} \times \dfrac{1}{n\pi}\\ \\ &=& \dfrac{1}{n\pi} \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}} \end{eqnarray*}
$両辺の和をとって$

\[\sum_{k=1}^{2n} \int_{\scriptsize{\dfrac{k-1}{2n}}}^{\scriptsize{\dfrac{k}{2n}}} x^{\sqrt{n}} |\cos (n\pi x)|dx \geqq \sum_{k=1}^{2n} \dfrac{1}{n\pi} \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}}\] \[\int_0^1 x^{\sqrt{n}} |\cos (n\pi x)|dx \geqq \dfrac{1}{n\pi} \sum_{k=1}^{2n} \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}}\]
$ここで$

　

\begin{eqnarray*} & &\dfrac{1}{n} \sum_{k=1}^{2n} \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}}\\ \\ &=&2 \times \dfrac{1}{2n}\sum_{k=1}^{2n} \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}} \\ \\ &=&2 \times \dfrac{1}{2n}\sum_{k=0}^{2n-1} \big(\dfrac{k}{2n}\big)^{\sqrt{n}} \\ \\ &=&2 \times \dfrac{1}{2n} \big(\sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}- \big(\dfrac{2n}{2n}\big)^{\sqrt{n}}\big)\\ \\ &=&2 \times \dfrac{1}{2n} \sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}}- \dfrac{1}{n}\\ \\ &\geqq& 2\int_0^1 x^{\sqrt{n}}dx - \dfrac{1}{n}\\ \\ &=&\dfrac{2}{\sqrt{n}+1} - \dfrac{1}{n}\\ \end{eqnarray*}
\[(2)と(3)より \quad I_n \leqq \dfrac{1}{n\pi} \sum_{k=1}^{2n} \big(\dfrac{k}{2n}\big)^{\sqrt{n}} \leqq \dfrac{1}{\pi}\big(\dfrac{2}{\sqrt{n}+1} + \dfrac{1}{n}\big)\]
\[(4) より \quad I_n \geqq \dfrac{1}{n\pi} \sum_{k=1}^{2n} \big(\dfrac{k-1}{2n}\big)^{\sqrt{n}} \geqq \dfrac{1}{\pi}\big(\dfrac{2}{\sqrt{n}+1} - \dfrac{1}{n}\big)\]
$よって$

$\dfrac{1}{\pi} \big(\dfrac{2}{\sqrt{n}+1} - \dfrac{1}{n}\big) \leqq I_n \leqq \dfrac{1}{\pi} \big(\dfrac{2}{\sqrt{n}+1}+\dfrac{1}{n}\big)$

$\dfrac{1}{\pi} \big(\dfrac{2\sqrt{n}}{\sqrt{n}+1} - \dfrac{1}{\sqrt{n}}\big) \leqq \sqrt{n}I_n \leqq \dfrac{1}{\pi} \big(\dfrac{2\sqrt{n}}{\sqrt{n}+1}+\dfrac{1}{\sqrt{n}}\big)$

$n \longrightarrow \infty \ \ とすると \quad 左辺 \ \ \longrightarrow \dfrac{2}{\pi} ,\qquad 右辺 \ \ \longrightarrow \dfrac{2}{\pi}$

\[はさみ打ちの原理により \quad \lim_{n \rightarrow \infty} \sqrt{n}I_n =\dfrac{2}{\pi}\]

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