\\(e\\) \u306f\u81ea\u7136\u5bfe\u6570\u306e\u5e95\u3068\u3059\u308b.\r\n\\(t \\gt e\\) \u306b\u304a\u3044\u3066\u95a2\u6570 \\(f(t) , g(t)\\) \u3092\u6b21\u306e\u3088\u3046\u306b\u5b9a\u3081\u308b.\r\n\\[\r\nf(t) = \\displaystyle\\int _ 1^e \\dfrac{t^2 \\log x}{t-x} \\, dx , \\ g(t) =\\displaystyle\\int _ 1^e \\dfrac{x^2 \\log x}{t-x} \\, dx\r\n\\]\r\n
(1)<\/strong>\u3000\\(f(t) -g(t)\\) \u3092 \\(t\\) \u306e \\(1\\) \u6b21\u5f0f\u3067\u8868\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(1 \\leqq x \\leqq e\\) \u304b\u3064 \\(t \\gt e\\) \u306e\u3068\u304d, \\(\\dfrac{1}{t-x} \\leqq \\dfrac{1}{t-e}\\) \u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u7528\u3044\u3066, \\(\\displaystyle\\lim _ {t \\rightarrow \\infty} g(t) = 0\\) \u3092\u793a\u305b.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(\\displaystyle\\lim _ {t \\rightarrow \\infty} \\left( f(t) -\\dfrac{bt^2}{t-a} \\right) =0\\) \u3068\u306a\u308b\u5b9a\u6570 \\(a , b\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\[\r\n\\dfrac{t^2 \\log x}{t-x} -\\dfrac{x^2 \\log x}{t-x} = (t+x) \\log x\r\n\\]\r\n\u307e\u305f, \\(F(x)= \\displaystyle\\int (t+x) \\log x \\, dx\\) \u3068\u304a\u304f\u3068\r\n\\[\\begin{align}\r\nF(x) & = \\dfrac{1}{2} (t+x)^2 \\log x -\\dfrac{1}{2} \\displaystyle\\int \\left( \\dfrac{t^2}{x} +2t +x \\right) \\, dx \\\\\r\n& = \\dfrac{1}{2} \\left\\{ (t+x)^2 \\log x -t^2 \\log x -2tx -\\dfrac{x^2}{2} \\right\\} \\\\\r\n& = \\dfrac{tx}{2} \\left( \\log x -1 \\right) +\\dfrac{x^2}{4} \\left( 2 \\log x -1 \\right) \\quad ( \\ C \\text{\u306f\u7a4d\u5206\u5b9a\u6570} )\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\\begin{align}\r\nf(t) -g(t) & = F(e) -F(1) \\\\\r\n& = \\dfrac{e^2}{4} -\\left( -\\dfrac{t}{2} -\\dfrac{1}{4} \\right) \\\\\r\n& = \\underline{\\dfrac{t}{2}+\\dfrac{e^2+1}{4}}\r\n\\end{align}\\]\r\n (2)<\/strong><\/p>\r\n \\(x^2 \\log x\\) \u306f, \\(1 \\leqq x \\leqq e\\) \u306b\u304a\u3044\u3066\u5358\u8abf\u5897\u52a0\u306a\u306e\u3067, \u3053\u306e\u533a\u9593\u306b\u304a\u3044\u3066\r\n\\[\r\n0 \\leqq \\dfrac{x^2 \\log x}{t-x} \\leqq \\dfrac{e^2}{t-e}\r\n\\]\r\n\u8fba\u3005\u3092 \\(x\\) \u306b\u3064\u3044\u3066 \\(1 \\rightarrow e\\) \u306e\u7bc4\u56f2\u3067\u7a4d\u5206\u3059\u308b\u3068\r\n\\[\\begin{align}\r\n0 & \\leqq \\displaystyle\\int _ 1^e \\dfrac{x^2 \\log x}{t-x} \\, dx \\leqq \\displaystyle\\int _ 1^e \\dfrac{e^2}{t-e} \\, dx \\\\\r\n\\text{\u2234} \\quad 0 & \\leqq g(t) \\leqq \\dfrac{e^2 (e-1)}{t-e}\r\n\\end{align}\\]\r\n\\(\\displaystyle\\lim _ {t \\rightarrow \\infty} \\dfrac{e^2 (e-1)}{t-e} =0\\) \u306a\u306e\u3067, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u3088\u308a\r\n\\[\r\n\\displaystyle\\lim _ {t \\rightarrow \\infty} g(x) = 0\r\n\\]\r\n (3)<\/strong><\/p>\r\n (1) (2)<\/strong> \u306e\u7d50\u679c\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{gather}\r\n\\displaystyle\\lim _ {t \\rightarrow \\infty} \\left( f(t) -\\dfrac{bt^2}{t-a} \\right) = \\dfrac{e^2+1}{4} -\\dfrac{(b-1)t^2 +at}{t-a} = 0 \\\\\r\n\\text{\u2234} \\quad \\displaystyle\\lim _ {t \\rightarrow \\infty} \\left\\{ \\dfrac{(b-1)t^2}{t-a} +\\dfrac{a}{1-\\frac{a}{t}} \\right\\} = \\dfrac{e^2+1}{4}\r\n\\end{gather}\\]\r\n\u3053\u308c\u304c\u6210\u7acb\u3059\u308b\u305f\u3081\u306e\u6761\u4ef6\u306f\r\n\\[\r\nb-1 = 0 , \\ a = \\dfrac{e^2+1}{4}\r\n\\]\r\n\u3088\u3063\u3066\r\n\\[\r\na = \\underline{\\dfrac{e^2+1}{4}} , \\ b =\\underline{1}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(e\\) \u306f\u81ea\u7136\u5bfe\u6570\u306e\u5e95\u3068\u3059\u308b. \\(t \\gt e\\) \u306b\u304a\u3044\u3066\u95a2\u6570 \\(f(t) , g(t)\\) \u3092\u6b21\u306e\u3088\u3046\u306b\u5b9a\u3081\u308b. \\[ f(t) = \\displaystyle\\int _ 1^e \\dfrac{t^2 \\ … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n