次の問いに答えよ.
(1) 定積分 \[ \displaystyle\int _ {1}^{e} x^{\frac{1}{n}} \log x \, dx \quad ( n =1, 2, 3, \cdots ) \] を求めよ.
(2) 極限 \[ \displaystyle\lim _ {n \rightarrow \infty} n \left( \displaystyle\int _ {1}^{e} x^{\frac{1}{n}} \log x \, dx -1 \right) \] を求めよ.
【 解 答 】
(1)
求める積分を \(I\) とおく. \[\begin{align} I & = \left[ \dfrac{n x^{\frac{n+1}{n}}}{n+1} \log x \right] _ 1^e -\dfrac{n}{n+1} \displaystyle\int _ {1}^{e} x^{\frac{1}{n}} \, dx \\ & = \dfrac{n e^{\frac{n+1}{n}}}{n+1} -\dfrac{n}{n+1} \left[ \dfrac{n x^{\frac{n+1}{n}}}{n+1} \right] _ 1^e \\ & = \dfrac{n}{(n+1)^2} \left\{ (n+1) e^{\frac{n+1}{n}} -n \left( e^{{\frac{n+1}{n}}} -1 \right) \right\} \\ & = \underline{\dfrac{n \left( e^{\frac{n+1}{n}} +n \right)}{(n+1)^2}} \end{align}\]
(2) \[\begin{align} n (I-1) & = n \left\{ \dfrac{n \left( e^{\frac{n+1}{n}} +n \right)}{(n+1)^2} -1 \right\} \\ & = \dfrac{n^2 \left( e^{\frac{n+1}{n}} +n \right) -n (n+1)^2}{(n+1)^2} \\ & = \dfrac{n^2 e^{1+\frac{1}{n}} -2n^2 -n}{(n+1)^2} \\ & = \dfrac{e^{1+\frac{1}{n}} -2 -\frac{1}{n}}{\left( 1+\frac{1}{n} \right)^2} \\ & \rightarrow e-2 \quad ( \ n \rightarrow \infty \text{のとき} ) \end{align}\] よって \[ \displaystyle\lim _ {n \rightarrow \infty} n \left( \displaystyle\int _ {1}^{e} x^{\frac{1}{n}} \log x \, dx -1 \right) = \underline{e-2} \]