九州大学(理系) 2026年　問題5


\[関数 \ \ f(x)=\int_x^{x+1} \log(4t^2+1)dt \ \ に関して、以下の問いに答えよ。\] $(1)\ \ f(x)\ が極値をとるときの \ x\ の値を求めよ。また、そのときの極値を求めよ。$
\[(2)\ \ 極限 \ \ \lim_{x \rightarrow} x\big\{f(x)-f(x-1)\big\}\ \ を求めよ。\]


(1)


$f'(x)=\log\big(4(x+1)^2+1\big)- \log(4x^2+1)=\log\dfrac{4x^2+8x+5}{4x^2+1}=\log\big(1+\dfrac{4(2x+1)}{4x^2+1}\big)$

$f'(x)=0 \ \ より \ \ 2x+1=0 \qquad \therefore x=-\dfrac{1}{2}$

$増減表$
\[ \begin{array}{c||c|c|c|c|c} x& \cdots & -\dfrac{1}{2} & \cdots \\ \hline f'(x)& - & 0 & + \\ \hline f(x) & \searrow & 極小 & \nearrow & \\ \end{array} \]
$f(x)\ は \ \ x=-\cfrac{1}{2}\ \ で極小となり、極小値は$

\begin{eqnarray*} & &f\big(-\cfrac{1}{2} \big)\\ \\ &=&\int_{\scriptsize{-\dfrac{1}{2}}}^{\scriptsize{\dfrac{1}{2}}}\log (4t^2+1)dt\\ \\ &=&2\int_0^{\scriptsize{\dfrac{1}{2}}}\log (4t^2+1)dt\\ \\ &=&2\big[t\log (4t^2+1)\big]_0^{\scriptsize{\dfrac{1}{2}}} - 2\int_0^{\scriptsize{\dfrac{1}{2}}} t \times \dfrac{8t}{4t^2+1}dt \\ \\ &=&\log 2 - \int_0^{\scriptsize{\dfrac{1}{2}}} \dfrac{16t^2}{4t^2+1}dt \\ \\ &=&\log 2 - \int_0^{\scriptsize{\dfrac{1}{2}}} \big(4-\dfrac{4}{4t^2+1}\big)dt \\ \\ &=&\log 2 - \big[4t\big]_0^{\scriptsize{\dfrac{1}{2}}} + \int_0^{\scriptsize{\dfrac{1}{2}}} \dfrac{1}{t^2+\dfrac{1}{4}}dt \\ \\ &=&\log 2 - 2 + \int_0^{\scriptsize{\dfrac{1}{2}}} \dfrac{1}{t^2+\dfrac{1}{4}}dt \\ \end{eqnarray*} \[ I=\int_0^{\scriptsize{\dfrac{1}{2}}} \dfrac{1}{t^2+\dfrac{1}{4}}dt\ \ は \ \ t=\dfrac{1}{2}\tan \theta \ \ とおくと \quad dt=\dfrac{1}{2} \dfrac{d\theta}{\cos^2\theta} \quad \begin{array}{c|c} t & 0\ \ \rightarrow \dfrac{1}{2} \\ \hline \theta & \ 0 \rightarrow \dfrac{\pi}{4} \\ \end{array} \]
\begin{eqnarray*} I &=&\int_0^{\scriptsize{\dfrac{\pi}{4}}} \dfrac{1}{\dfrac{1}{4} \tan^2 \theta + \dfrac{1}{4}} \times \dfrac{d\theta}{2\cos ^2\theta}\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{4}}} \dfrac{4}{ \tan^2 \theta + 1} \times \dfrac{d\theta}{2\cos ^2\theta}\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{4}}} 4\cos^2 \theta \times \dfrac{d\theta}{2\cos ^2\theta}\\ \\ &=&\int_0^{\scriptsize{\dfrac{\pi}{4}}} 2 d\theta\\ \\ &=&\dfrac{\pi}{2} \end{eqnarray*} $よって\ \ 極小値は \quad f\big(-\cfrac{1}{2} \big)=\log 2 -2 + \dfrac{\pi}{2}$


(2)


\begin{eqnarray*} f(x) &=&\int_x^{x+1} \log(4t^2+1)dt\\ \\ &=&\big[t\log(4t^2+1)\big]_x^{x+1} - \int_x^{x+1} t \times \dfrac{8t}{4t^2+1}dt\\ \\ &=&(x+1)\log(4(x+1)^2+1) -x\log(4x^2+1) - \int_x^{x+1} \dfrac{8t^2}{4t^2+1}dt\\ \\ &=&(x+1)\log(4(x+1)^2+1) -x\log(4x^2+1) - \int_x^{x+1} \big(2-\dfrac{2}{4t^2+1}\big)dt\\ \\ &=&(x+1)\log(4(x+1)^2+1) -x\log(4x^2+1) - 2 +2\int_x^{x+1}\dfrac{1}{4t^2+1}dt\\ \end{eqnarray*} $x \longrightarrow x-1 \ \ とおいて$

\[f(x-1)=x\log(4x^2+1) -(x-1)\log(4(x-1)^2+1) - 2 + 2\int_{x-1}^x\dfrac{1}{4t^2+1}dt\]
$したがって$

\begin{eqnarray*} & &x\big(f(x)-f(x-1)\big)\\ \\ &=&x\big\{(x+1)\log(4(x+1)^2+1) -x\log(4x^2+1) -2 +2\int_x^{x+1}\dfrac{1}{4t^2+1}dt\\ \\ & &\qquad -x\log(4x^2+1) +(x-1)\log(4(x-1)^2+1) +2 - 2\int_{x-1}^x\dfrac{1}{4t^2+1}dt\big\}\\ \\ &=&x^2\big\{\log(4x^2+8x+5) -2\log(4x^2+1) +\log(4x^2-8x+5)\big\} +x\big\{\log(4x^2+8x+5) -\log(4x^2-8x+5) \big\}\\ \\ & & \qquad +2\int_x^{x+1}\dfrac{1}{4t^2+1}dt - 2\int_{x-1}^x\dfrac{1}{4t^2+1}dt\big\}\\ \\ &=&x^2\log \dfrac{(4x^2+8x+5)(4x^2-8x+5)}{(4x^2+1)^2} +x\log \dfrac{4x^2+8x+5}{4x^2-8x+5} +2\int_x^{x+1}\dfrac{1}{4t^2+1}dt - 2\int_{x-1}^x\dfrac{1}{4t^2+1}dt\big\}\\ \\ &=&x^2\log \dfrac{(4x^2+5)^2-64x^2}{(4x^2+1)^2} +x\log \dfrac{4x^2+8x+5}{4x^2-8x+5} +2\int_x^{x+1}\dfrac{1}{4t^2+1}dt - 2\int_{x-1}^x\dfrac{1}{4t^2+1}dt\big\}\\ \\ &=&x^2\log \dfrac{16x^4-24x^2+25}{16x^4+8x^2+1} +x\log \dfrac{4x^2+8x+5}{4x^2-8x+5} +2\int_x^{x+1}\dfrac{1}{4t^2+1}dt - 2\int_{x-1}^x\dfrac{1}{4t^2+1}dt\big\}\\ \\ &=&x^2\log \big(1+\dfrac{-32x^2+24}{16x^4+8x^2+1}\big) +x\log \big(1+\dfrac{16x}{4x^2-8x+5}\big) +2\int_x^{x+1}\dfrac{1}{4t^2+1}dt - 2\int_{x-1}^x\dfrac{1}{4t^2+1}dt\big\}\\ \\ &=&\log \big(1+\dfrac{-8(4x^2-3)}{(4x^2+1)^2}\big)^{x^2} +\log \big(1+\dfrac{16x}{4x^2-8x+5}\big)^x +2\int_x^{x+1}\dfrac{1}{4t^2+1}dt - 2\int_{x-1}^x\dfrac{1}{4t^2+1}dt\big\}\\ \end{eqnarray*}
$第 \ 1\ 項について$

$p=\dfrac{(4x^2+1)^2}{8(4x^2-3)}\ \ とおくと \quad \dfrac{1}{p}=\dfrac{8(4x^2-3)}{(4x^2+1)^2} \ \ だから \quad x \longrightarrow \infty \ \ のとき \quad p \longrightarrow \infty,\quad \dfrac{1}{p} \longrightarrow 0$

$第 \ 1\ 項=\log \big(1-\dfrac{1}{p}\big)^{{-p} \times \scriptsize{\dfrac{x^2}{-p}}}=\log \big(1-\dfrac{1}{p}\big)^{-p \times \scriptsize{\dfrac{-8x^2(4x^2-3)}{(4x^2+1)^2}}}=\log \big\{\big(1-\dfrac{1}{p}\big)^{-p}\big\}^{\scriptsize{\dfrac{-8x^2(4x^2-3)}{(4x^2+1)^2}}}$

$\big(1-\dfrac{1}{p}\big)^{-p} \longrightarrow e,\qquad \dfrac{-8x^2(4x^2-3)}{(4x^2+1)^2} \longrightarrow -2 \ \ だから$

$第 \ 1\ 項　\longrightarrow \log e^{-2}=-2$

$第 \ 2\ 項について$

$q=\dfrac{4x^2-8x+5}{16x} \ \ とおくと \quad \dfrac{1}{q}=\dfrac{16x}{4x^2-8x+5} \ \ だから \quad 　x \longrightarrow \infty \ \ のとき \quad q \longrightarrow \infty,\quad \dfrac{1}{q} \longrightarrow 0$

$第 \ 2\ 項=\log \big(1+\dfrac{1}{q}\big)^{q \times \scriptsize{\dfrac{x}{q}}}=\log \big(1+\dfrac{1}{q}\big)^{q \times \scriptsize{\dfrac{16x^2}{4x^2-8x+5}}}=\log \big\{\big(1+\dfrac{1}{q}\big)^q\big\}^{\scriptsize{\dfrac{16x^2}{4x^2-8x+5}}}$

$\big(1+\dfrac{1}{q}\big)^{q} \longrightarrow e,\qquad \dfrac{16x^2}{4x^2-8x+5} \longrightarrow 4 \ \ だから$

$第 \ 2\ 項　\longrightarrow \log e^4=4$

$第 \ 3\ 項と第 \ 4\ 項については、積分の平均値の定理より$

\[\int_x^{x+1}\dfrac{1}{4t^2+1}dt =\dfrac{1}{4c^2+1} \ \ (x < c < x+1) \ \ を満たす \ c\ が少なくとも \ 1\ つ存在する。\]
$4x^2+1 < 4c^2+1 < 4(x+1)^2+1 \ \ だから \quad \dfrac{1}{4(x+1)^2+1} < \dfrac{1}{4c^2+1} < \dfrac{1}{4x^2+1}$

$x \longrightarrow \infty \ \ とすると \quad \dfrac{1}{4(x+1)^2+1} \longrightarrow 0 ,\quad \dfrac{1}{4x^2+1} \longrightarrow 0 \ \ だからはさみうちの原理により$

$\dfrac{1}{4c^2+1} \longrightarrow 0 \ \ したがって、第 \ 3\ 項 \longrightarrow 0$

$全く同様にして 第 \ 4\ 項　\longrightarrow 0$

$以上より$

\[\lim_{x \rightarrow \infty} x\big\{f(x)-f(x-1)\big\}=-2+4=2\]

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