\(0 \leqq x \leqq \pi\) に対して, 関数 \(f(x)\) を \[ f(x) = \displaystyle\int _ {0}^{\frac{\pi}{2}} \dfrac{\cos |t-x|}{1 +\sin |t-x|} \, dt \] と定める. \(f(x)\) の \(0 \leqq x \leqq \pi\) における最大値と最小値を求めよ.
【 解 答 】
\[\begin{align} \cos |t-x| & = \cos (t-x) \\ \sin |t-x| & = \left\{ \begin{array}{ll} \sin (t-x) & \left( \ t \geqq x \text{のとき} \right) \\ -\sin (t-x) & \left( \ t \lt x \text{のとき} \right) \end{array} \right. \end{align}\] また \[\begin{align} \displaystyle\int \dfrac{\cos (t-x)}{1 \pm \sin (t-x)} \, dt & = \displaystyle\int \dfrac{\pm \left\{ 1 \pm \sin (t-x) \right\}'}{1 \pm \sin (t-x)} \, dt \\ & = \pm \log \left\{ 1 \pm \sin (t-x) \right\} +C \quad ( \ C \text{は積分定数} ) \end{align}\] 以上を用いて, 場合分けして考える.
1* \(0 \leqq x \leqq \dfrac{\pi}{2}\) のとき \[\begin{align} f(x) & =\displaystyle\int _ {0}^{x} \dfrac{\cos (t-x)}{1 -\sin (t-x)} \, dt +\displaystyle\int _ {x}^{\frac{\pi}{2}} \dfrac{\cos (t-x)}{1 +\sin (t-x)} \, dt \\ & = \left[ -\log \left\{ 1 - \sin (t-x) \right\} \right] _ {0}^{x} +\left[ \log \left\{ 1 + \sin (t-x) \right\} \right] _ {x}^{\frac{\pi}{2}} \\ & = \log ( 1+\sin x ) +\log \left\{ 1 +\sin \left( \dfrac{\pi}{2} -x \right) \right\} \\ & = \log \underline{( 1+\sin x )( 1+\cos x )} _ {[1]} \end{align}\] 関数 \(\log x\) は単調増加なので, 下線部 [1] を \(g(x)\) とおいて, これの増減を調べればよい. \[\begin{align} g'(x) & = \cos x ( 1+\cos x ) -\sin x ( 1+\sin x ) \\ & = ( \cos x -\sin x )( \cos x +\sin x +1 ) \\ & = \cos \left( x +\dfrac{\pi}{4} \right) \left\{ \sin \left( x +\dfrac{\pi}{4} \right) +1 \right\} \end{align}\] \(g'(x)=0\) を解くと \[ x = \dfrac{\pi}{4} \] \(0 \leqq x \leqq \dfrac{\pi}{2}\) における \(g(x)\) の増減は下表のとおり. \[ \begin{array}{c|ccccc} x & 0 & \cdots & \frac{\pi}{4} & \cdots & \frac{\pi}{2}\\ \hline g'(x) & & + & 0 & - & \\ \hline g(x) & 2 & \nearrow & \left( 1 +\frac{\sqrt{2}}{2} \right)^2 & \searrow & 2 \end{array} \] ここで \[\begin{align} f \left( \dfrac{\pi}{4} \right) & = \log \left( 1 +\frac{\sqrt{2}}{2} \right)^2 \\ & = 2 \log \dfrac{2 +\sqrt{2}}{2} \end{align}\]
2* \(\dfrac{\pi}{2} \lt x \leqq \pi\) のとき \[\begin{align} f(x) & = \displaystyle\int _ {0}^{\frac{\pi}{2}} \dfrac{\cos (t-x)}{1 -\sin (t-x)} \, dt \\ & = \left[ -\log \left\{ 1 - \sin (t-x) \right\} \right] _ {0}^{\frac{\pi}{2}} \\ & = \log ( 1+\sin x ) -\log \left\{ 1 -\sin \left( \dfrac{\pi}{2} -x \right) \right\} \\ & = \log \dfrac{1 +\sin x}{1 -\cos x} \end{align}\] \(\dfrac{\pi}{2} \lt x \leqq \pi\) において, 関数 \(\log x\) は単調増加, 関数 \(1 +\sin x\) は単調減少, 関数 \(1 -\cos x\) は単調増加なので, \(f(x)\) は単調減少関数である.
ここで \[ f( \pi ) = \log \dfrac{1}{2} = -\log 2 \]
1* 2* より, 求める最大値と最小値は \[ \underline{\left\{ \begin{array}{ll} \text{最大値} : \quad & f \left( \dfrac{\pi}{4} \right) = 2 \log \dfrac{2 +\sqrt{2}}{2} \\ \text{最小値} : \quad & f( \pi ) = -\log 2 \end{array} \right.} \]