次の問いに答えよ.
(1) 不定積分 \[ \displaystyle\int \sqrt{1 -e^{-2x}} \, dx \] を置換 \(\sqrt{1 -e^{-2x}} = t\) を用いて求めよ.
(2) 極限 \[ \displaystyle\lim _ {\alpha \rightarrow \infty} \displaystyle\int _ 0^{\alpha} \left( 1 -\sqrt{1 -e^{-2x}} \right) \, dx \] を求めよ.
【 解 答 】
(1)
\(\sqrt{1 -e^{-2x}} = t\) より \[\begin{align} e^{-2x} & = 1-t^2 \\ \text{∴} \quad x & = -\dfrac{1}{2} \log ( 1-t^2 ) \end{align}\] なので \[ \dfrac{dx}{dt} = -\dfrac{-2t}{2 (1-t^2)} = \dfrac{t}{1-t^2} \] よって \[\begin{align} \displaystyle\int \sqrt{1 -e^{-2x}} \, dx & =\displaystyle\int t \cdot \dfrac{t}{1-t^2} \, dt \\ & = \displaystyle\int \left( \dfrac{1}{1-t^2} -1 \right) \, dt \\ & = \displaystyle\int \left\{ \dfrac{1}{2(1+t)} +\dfrac{1}{2(1-t)} -1 \right\} \, dt \\ & = \dfrac{1}{2} \log |1+t| -\dfrac{1}{2} \log |1-t| -t +C \\ & = \dfrac{1}{2} \log \left| \dfrac{1 +\sqrt{1 -e^{-2x}}}{1 -\sqrt{1 -e^{-2x}}} \right| -\sqrt{1 -e^{-2x}} +C \\ & = \dfrac{1}{2} \log \left| \dfrac{\left( 1 +\sqrt{1 -e^{-2x}} \right)^2}{e^{-2x}} \right| -\sqrt{1 -e^{-2x}} +C \\ & = \underline{\log \left( 1 +\sqrt{1 -e^{-2x}} \right) +x -\sqrt{1 -e^{-2x}} +C} \end{align}\] ただし, \(C\) は積分定数.
(2)
(1) の結果より \[\begin{align} \displaystyle\int _ 0^{\alpha} & \left( 1 -\sqrt{1 -e^{-2x}} \right) \, dx \\ & = \left[ \sqrt{1 -e^{-2x}} -\log \left( 1 +\sqrt{1 -e^{-2x}} \right) \right] _ 0^{\alpha} \\ & = \sqrt{1 -e^{-2\alpha}} -\log \left( 1 +\sqrt{1 -e^{-2\alpha}} \right) \\ & \rightarrow \sqrt{1} -\log \left( 1 +\sqrt{1} \right) \quad ( \alpha \rightarrow \infty \text{のとき} ) \\ & = \underline{1-\log 2} \end{align}\]