整数 \(n\) に対して, \[ I _ n = \displaystyle\int _ {\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{\cos ( (2n+1) x )}{\sin x} \, dx \ . \] とする.
(1) \(I _ 0\) を求めよ.
(2) \(n\) を正の整数とするとき, \(I _ n -I _ {n-1}\) を求めよ.
(3) \(I _ 5\) を求めよ.
【 解 答 】
(1)
\[\begin{align} I _ 0 & = \displaystyle\int _ {\frac{\pi}{4}} ^{\frac{\pi}{2}} \dfrac{\cos x}{\sin x} \, dx = \displaystyle\int _ {\frac{\pi}{4}} ^{\frac{\pi}{2}} \dfrac{( \sin x )'}{\sin x} \, dx \\ & = \left[ \log \left| \sin x \right| \right] _ {\frac{\pi}{4}} ^{\frac{\pi}{2}} = 0 -\log \dfrac{1}{\sqrt{2}} \\ & = \underline{\dfrac{\log 2}{2}} \ . \end{align}\]
(2)
三角関数の和積の公式を用いれば \[\begin{align} \cos & \left( (2n+1) x \right) -\cos \left( (2n-1) x \right) \\ & = -2 \sin \dfrac{(2n+1) x +(2n-1) x}{2} \sin \dfrac{(2n+1) x -(2n-1) x}{2} \\ & = -2 \sin 2nx \sin x \ . \end{align}\] なので \[\begin{align} I _ n -I _ {n-1} & = \displaystyle\int _ {\frac{\pi}{4}} ^{\frac{\pi}{2}} \left( -2 \sin 2nx \right) \, dx \\ & = \dfrac{1}{n} \left[ \cos 2nx \right] _ {\frac{\pi}{4}} ^{\frac{\pi}{2}} \\ & = \dfrac{\cos n \pi -\cos \frac{n \pi}{2}}{n} \ . \end{align}\] ここで \[\begin{align} \cos n \pi & = \left\{ \begin{array}{ll} 1 & ( \ n \text{が偶数のとき} ) \\ -1 & ( \ n \text{が奇数のとき} ) \end{array} \right. \ . \end{align}\] また, \(k\) を非負整数として \[\begin{align} \cos \dfrac{n \pi}{2} & = \left\{ \begin{array}{ll} 1 & ( \ n = 4k \ \text{のとき} ) \\ 0 & ( \ n = 4k+1 \ \text{のとき} ) \\ -1 & ( \ n = 4k+2 \ \text{のとき} ) \\ 0 & ( \ n = 4k+3 \ \text{のとき} ) \\ \end{array} \right. \ . \end{align}\] よって \[ I _ n -I _ {n-1} = \underline{\left\{ \begin{array}{ll} 0 & ( \ n = 4k \ \text{のとき} ) \\ -\dfrac{1}{n} & ( \ n = 4k+1 \ \text{のとき} ) \\ \dfrac{2}{n} & ( \ n = 4k+2 \ \text{のとき} ) \\ -\dfrac{1}{n} & ( \ n = 4k+3 \ \text{のとき} ) \\ \end{array} \right. \quad} \ . \]
(3)
(2) の結果を用いれば \[\begin{align} I _ 5 & = I _ 0 +\textstyle\sum\limits _ {n=1} ^5 ( I _ n -I _ {n-1} ) \\ & = \dfrac{\log 2}{2} +0 -1 +1 -\dfrac{1}{3} +0 -\dfrac{1}{5} \\ & = \underline{\dfrac{\log 2}{2} -\dfrac{8}{15}} \ . \end{align}\]