\u6574\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066,\r\n\\[\r\nI _ n = \\displaystyle\\int _ {\\frac{\\pi}{4}}^{\\frac{\\pi}{2}} \\dfrac{\\cos ( (2n+1) x )}{\\sin x} \\, dx \\ .\r\n\\]\r\n\u3068\u3059\u308b.<\/p>\r\n
(1)<\/strong>\u3000\\(I _ 0\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(n\\) \u3092\u6b63\u306e\u6574\u6570\u3068\u3059\u308b\u3068\u304d, \\(I _ n -I _ {n-1}\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(I _ 5\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\[\\begin{align}\r\nI _ 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 \\\\\r\n& = \\left[ \\log \\left| \\sin x \\right| \\right] _ {\\frac{\\pi}{4}} ^{\\frac{\\pi}{2}} = 0 -\\log \\dfrac{1}{\\sqrt{2}} \\\\\r\n& = \\underline{\\dfrac{\\log 2}{2}} \\ .\r\n\\end{align}\\]\r\n (2)<\/strong><\/p>\r\n \u4e09\u89d2\u95a2\u6570\u306e\u548c\u7a4d\u306e\u516c\u5f0f\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\n\\cos & \\left( (2n+1) x \\right) -\\cos \\left( (2n-1) x \\right) \\\\\r\n& = -2 \\sin \\dfrac{(2n+1) x +(2n-1) x}{2} \\sin \\dfrac{(2n+1) x -(2n-1) x}{2} \\\\\r\n& = -2 \\sin 2nx \\sin x \\ .\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\nI _ n -I _ {n-1} & = \\displaystyle\\int _ {\\frac{\\pi}{4}} ^{\\frac{\\pi}{2}} \\left( -2 \\sin 2nx \\right) \\, dx \\\\\r\n& = \\dfrac{1}{n} \\left[ \\cos 2nx \\right] _ {\\frac{\\pi}{4}} ^{\\frac{\\pi}{2}} \\\\\r\n& = \\dfrac{\\cos n \\pi -\\cos \\frac{n \\pi}{2}}{n} \\ .\r\n\\end{align}\\]\r\n\u3053\u3053\u3067\r\n\\[\\begin{align}\r\n\\cos n \\pi & = \\left\\{ \\begin{array}{ll} 1 & ( \\ n \\text{\u304c\u5076\u6570\u306e\u3068\u304d} ) \\\\ -1 & ( \\ n \\text{\u304c\u5947\u6570\u306e\u3068\u304d} ) \\end{array} \\right. \\ .\r\n\\end{align}\\]\r\n\u307e\u305f, \\(k\\) \u3092\u975e\u8ca0\u6574\u6570\u3068\u3057\u3066\r\n\\[\\begin{align}\r\n\\cos \\dfrac{n \\pi}{2} & = \\left\\{ \\begin{array}{ll} 1 & ( \\ n = 4k \\ \\text{\u306e\u3068\u304d} ) \\\\ 0 & ( \\ n = 4k+1 \\ \\text{\u306e\u3068\u304d} ) \\\\ -1 & ( \\ n = 4k+2 \\ \\text{\u306e\u3068\u304d} ) \\\\ 0 & ( \\ n = 4k+3 \\ \\text{\u306e\u3068\u304d} ) \\\\ \\end{array} \\right. \\ .\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\nI _ n -I _ {n-1} = \\underline{\\left\\{ \\begin{array}{ll} 0 & ( \\ n = 4k \\ \\text{\u306e\u3068\u304d} ) \\\\ -\\dfrac{1}{n} & ( \\ n = 4k+1 \\ \\text{\u306e\u3068\u304d} ) \\\\ \\dfrac{2}{n} & ( \\ n = 4k+2 \\ \\text{\u306e\u3068\u304d} ) \\\\ -\\dfrac{1}{n} & ( \\ n = 4k+3 \\ \\text{\u306e\u3068\u304d} ) \\\\ \\end{array} \\right. \\quad} \\ .\r\n\\]\r\n (3)<\/strong><\/p>\r\n (2)<\/strong> \u306e\u7d50\u679c\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nI _ 5 & = I _ 0 +\\textstyle\\sum\\limits _ {n=1} ^5 ( I _ n -I _ {n-1} ) \\\\\r\n& = \\dfrac{\\log 2}{2} +0 -1 +1 -\\dfrac{1}{3} +0 -\\dfrac{1}{5} \\\\\r\n& = \\underline{\\dfrac{\\log 2}{2} -\\dfrac{8}{15}} \\ .\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u6574\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066, \\[ I _ n = \\displaystyle\\int _ {\\frac{\\pi}{4}}^{\\frac{\\pi}{2}} \\dfrac{\\cos ( (2n+1) x )}{\\sin x … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n