関数 \[ f(x) = 2 \log \left( 1+e^x \right) -x -\log 2 \] を考える.
(1) \(f(x)\) の第 \(2\) 次導関数を \(f''(x)\) とする. 等式 \[ \log f''(x) = -f(x) \] が成り立つことを示せ.
(2) 定積分 \(\displaystyle\int _ 0^{\log 2} \left( x-\log 2 \right) e^{-f(x)} \, dx\) を求めよ.
【 解 答 】
(1)
\[\begin{align} f'(x) & = 2 \cdot \dfrac{e^x}{1+e^x} -1 \\ & = 2 -\dfrac{2}{1+e^x} -1 \\ & = 1 -\dfrac{2}{1+e^x} , \\ f''(x) & = \dfrac{2 e^x}{\left( 1+e^x \right)^2} \end{align}\] よって \[\begin{align} \log f''(x) & = \log 2 +\log e^x -2 \log \left( 1+e^x \right) \\ & = -2 \log \left( 1+e^x \right) +x +\log 2 \\ & = -f(x) \end{align}\]
(2)
(1) の結果より \[ f''(x) = e^{-f(x)} \] これを用いれば \[\begin{align} \displaystyle\int _ 0^{\log 2} & \left( x-\log 2 \right) e^{-f(x)} \, dx = \displaystyle\int _ 0^{\log 2} \left( x-\log 2 \right) f''(x) \, dx \\ & = \left[ \left( x-\log 2 \right) f'(x) \right] _ 0^{\log 2} - \int _ 0^{\log 2} 1 \cdot f'(x) \, dx \\ & = 0 \cdot f'( \log 2 ) -\left( -\log 2 \right) f'(0) -\left[ f(x) \right] _ 0^{\log 2} \\ & = \log 2 \cdot 0 - \left[ f(x) \right] _ 0^{\log 2} \quad ( \ \text{∵} \ f'(0) = 1 -\dfrac{2}{1+e^0} = 0 \ ) \\ & = f(0) -f( \log 2 ) \\ & = \left( 2 \log 2 -\log 2 \right) - \left\{ 2 \log (1+2) -\log 2 -\log 2 \right\} \\ & = \underline{3 \log 2 - 2 \log 3} \end{align}\]