大阪大学(理系) 2025年　問題4


$次の問いに答えよ。$
\[(1)\ \ t > 0 \ \ のとき \quad -\dfrac{1}{t} < \int_t^{2t} \dfrac{\sin x}{x^2} dx < \dfrac{1}{t} \quad が成り立つことを示せ。\] \[(2)\ \ \lim_{t \rightarrow \infty} \int_t^{2t} \dfrac{\cos x}{x} dx =0 \quad を示せ。\] \[(3)\ \ f(x)=\sin \big(\dfrac{3x}{2}\big) \sin \big(\dfrac{x}{2}\big) \quad とおく。\lim_{t \rightarrow \infty} \int_1^{t} \dfrac{f(x)}{x} dx =\dfrac{1}{2}\int_1^2 \dfrac{\cos x}{x} dx \quad を示せ。\]


(1)

\begin{eqnarray*} & &\big|\int_t^{2t} \dfrac{\sin x}{x^2} dx\big|\\ \\ &\leqq&\int_t^{2t} \big|\dfrac{\sin x}{x^2}\big| dx\\ \\ &\leqq&\int_t^{2t} \dfrac{1}{x^2}dx\\ \\ &=&\big[-\dfrac{1}{x}\big]_t^{2t} \\ \\ &=&\dfrac{1}{t}-\dfrac{1}{2t}\\ \\ &<&\dfrac{1}{t}\\ \end{eqnarray*} \[よって \quad t > 0 \ \ だから \quad - \dfrac{1}{t} < \int_t^{2t} \dfrac{\sin x}{x^2} dx < \dfrac{1}{t} \]

(2)


\[\int_t^{2t} \dfrac{\cos x}{x} dx =\big[\dfrac{\sin x}{x}\big]_t^{2t}+\int_t^{2t} \dfrac{\sin x}{x^2} dx =\dfrac{\sin 2t}{2t} -\dfrac{\sin t}{t}+\int_t^{2t} \dfrac{\sin x}{x^2} dx \]
\[ 0 \leqq \big|\dfrac{\sin 2t}{2t}\big| \leqq \big|\dfrac{1}{2t}\big|,\qquad 0 \leqq \big|\dfrac{\sin t}{t}\big| \leqq \big|\dfrac{1}{t}\big|,\qquad (1)より \quad -\dfrac{1}{t} < \int_t^{2t} \dfrac{\sin x}{x^2} dx < \dfrac{1}{t} \quad だから\] \[はさみうちの原理により \quad t \longrightarrow \infty \quad のとき \quad \dfrac{\sin 2t}{2t} \longrightarrow 0 ,\qquad \dfrac{\sin t}{t} \longrightarrow 0,\qquad \int_t^{2t} \dfrac{\sin x}{x^2} dx \longrightarrow 0\] \[よって \quad \lim_{t \rightarrow \infty} \int_t^{2t} \dfrac{\cos x}{x} dx =0 \]

(3)


\[I=\int_1^{t} \dfrac{f(x)}{x} dx \quad とおくと\] \[I=\int_1^t \dfrac{\sin \big(\dfrac{3x}{2}\big) \sin \big(\dfrac{x}{2}\big)}{x}dx =-\dfrac{1}{2}\int_1^t \dfrac{\cos 2x -\cos x}{x}dx=-\dfrac{1}{2}\int_1^t \dfrac{\cos 2x}{x}dx + \dfrac{1}{2}\int_1^t \dfrac{\cos x}{x}dx\]
\[第1項の積分で、2x=u \quad とおくと \quad 2dx=du \qquad \begin{array}{c|c} x & 1\ \ \rightarrow t\\ \hline u & 2 \rightarrow 2t\\ \end{array} \]
\[\int_1^t \dfrac{\cos 2x}{x}dx=\int _2^{2t}\dfrac{\cos u}{\dfrac{u}{2}} \times \dfrac{du}{2}=\int _2^{2t}\dfrac{\cos u}{u} du=\int_2^{t}\dfrac{\cos u}{u}du +\int_t^{2t}\dfrac{\cos u}{u}du\] $よって$
\begin{eqnarray*} I &=&-\dfrac{1}{2}\big(\int_2^{t}\dfrac{\cos u}{u}du +\int_t^{2t}\dfrac{\cos u}{u}du \big)+\dfrac{1}{2}\int_1^t \dfrac{\cos x}{x}dx\\ \\ &=&-\dfrac{1}{2}\int_t^{2t}\dfrac{\cos u}{u}du + \dfrac{1}{2}\big(\int_1^t \dfrac{\cos x}{x}dx - \int_2^t\dfrac{\cos u}{u}du\big)\\ \\ &=&-\dfrac{1}{2}\int_t^{2t}\dfrac{\cos u}{u}du + \dfrac{1}{2}\big(\int_1^t \dfrac{\cos x}{x}dx + \int_t^2\dfrac{\cos x}{x}dx\big)\\ \\ &=&-\dfrac{1}{2}\int_t^{2t}\dfrac{\cos u}{u}du + \dfrac{1}{2}\int_1^2 \dfrac{\cos x}{x}dx \\ \end{eqnarray*}
$(2)より \quad t \longrightarrow \infty \quad のとき \quad 第1項 \longrightarrow 0 \quad だから$

\[\lim_{t \rightarrow \infty} \int_1^{t} \dfrac{f(x)}{x} dx =\dfrac{1}{2}\int_1^2 \dfrac{\cos x}{x} dx \]

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