\r\n \r\n
\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n
(1)<\/strong><\/p>\r\n\\(t = 1+x\\) \u3068\u304a\u304f\u3068, \\(dt = dx\\) \u3067\u3042\u308a\r\n\\[\r\n\\begin{array}{c|ccc} x & 0 & \\rightarrow & a \\\\ \\hline t & 1 & \\rightarrow & 1+a \\end{array}\r\n\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\nI _ 0 (a) & = \\displaystyle\\int _ 1^{1+a} \\sqrt{t} \\, dt = \\left[ \\dfrac{2}{3} x^{\\frac{3}{2}} \\right] _ 1^{1+a} \\\\\r\n& = \\dfrac{2}{3} \\left\\{ ( 1+a )^{\\frac{3}{2}} -1 \\right\\}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\\begin{align}\r\na^{-\\frac{3}{2}} I _ 0 (a) & = \\dfrac{2}{3} \\left\\{ \\left( \\dfrac{1}{a} +1 \\right)^{\\frac{3}{2}} -\\dfrac{1}{a^{\\frac{3}{2}}} \\right\\} \\\\\r\n& \\rightarrow \\underline{\\dfrac{2}{3}} \\quad ( \\ a \\rightarrow \\infty \\text{\u306e\u3068\u304d} )\r\n\\end{align}\\]\r\n
(2)<\/strong><\/p>\r\n(1)<\/strong> \u3068\u540c\u69d8\u306b\u7f6e\u63db\u3059\u308b\u3068\r\n\\[\\begin{align}\r\nI _ n (a) & = \\displaystyle\\int _ 1^{1+a} t^{\\frac{1}{2}} (t-1)^{n} \\, dt \\\\\r\n& = \\displaystyle\\int _ 1^{1+a} t^{\\frac{3}{2}} (t-1)^{n-1} \\, dt -\\displaystyle\\int _ 1^{1+a} t^{\\frac{1}{2}} (t-1)^{n-1} \\, dt \\\\\r\n& = \\left[ t^{\\frac{3}{2}} \\cdot \\dfrac{(t-1)^n}{n} \\right] _ 1^{1+a} -\\dfrac{3}{2n} \\displaystyle\\int _ 1^{1+a} t^{\\frac{1}{2}} (t-1)^{n} \\, dt -I _ {n-1} (a) \\\\\r\n& = \\dfrac{1}{n} (1+a)^{\\frac{3}{2}} a^n -\\dfrac{3}{2n} I _ {n} (a) -I _ {n-1} (a)\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\\begin{align}\r\n(2n+3) I _ {n} (a) & = 2 (1+a)^{\\frac{3}{2}} a^n -2n I _ {n-1} (a) \\\\\r\n\\text{\u2234} \\quad I _ {n} (a) & = \\dfrac{2}{3+2n} (1+a)^{\\frac{3}{2}} a^n -\\dfrac{2n}{3+2n} I _ {n-1} (a)\r\n\\end{align}\\]\r\n(3)<\/strong><\/p>\r\n\u3059\u3079\u3066\u306e \\(0\\) \u4ee5\u4e0a\u306e\u6574\u6570 \\(n\\) \u306b\u3064\u3044\u3066\r\n\\[\r\n\\displaystyle\\lim _ {a \\rightarrow \\infty} a^{ -\\left( \\frac{3}{2} +n \\right)} I _ n (a) = \\dfrac{2}{3+2n} \\quad ... [\\text{A}]\r\n\\]\r\n\u3067\u3042\u308b\u3053\u3068\u3092\u6570\u5b66\u7684\u5e30\u7d0d\u6cd5\u3092\u7528\u3044\u3066\u793a\u3059.<\/p>\r\n
\r\n1*<\/strong>\u3000\\(n = 0\\) \u306e\u3068\u304d, (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a, \u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n2*<\/strong>\u3000\\(n = k-1 \\ ( k \\geqq 1 )\\) \u306e\u3068\u304d, [A] \u304c\u6210\u7acb\u3059\u308b, \u3059\u306a\u308f\u3061\r\n\\[\r\n\\displaystyle\\lim _ {a \\rightarrow \\infty} a^{- \\left( \\frac{1}{2} +k \\right)} I _ {k-1} (a) = \\dfrac{2}{1+2k}\r\n\\]\r\n\u3068\u4eee\u5b9a\u3059\u308b.
\r\n(2)<\/strong> \u306e\u7d50\u679c\u3088\u308a,\r\n\\[\\begin{align}\r\nI _ {k} (a) & = \\dfrac{2}{2k+3} (1+a)^{\\frac{3}{2}} a^{k} -\\dfrac{2k}{3+2k} I _ {k} (a) \\\\\r\na^{ -\\left( \\frac{3}{2} +n \\right)} I _ {k} (a) & = \\dfrac{2}{3+2k} \\left( \\dfrac{1}{a} +1 \\right)^{\\frac{3}{2}} -\\dfrac{1}{a} \\cdot \\dfrac{2k}{3+2k} a^{- \\left( \\frac{1}{2} +n \\right)} I _ {k-1} (a)\r\n\\end{align}\\]\r\n\u3053\u3053\u3067, \u8fba\u3005 \\(a \\rightarrow \\infty\\) \u3068\u3059\u308c\u3070\r\n\\[\\begin{gather}\r\n\\dfrac{2}{3+2k} \\left( \\dfrac{1}{a} +1 \\right)^{\\frac{3}{2}} \\rightarrow \\dfrac{2}{3+2k} , \\\\\r\n\\dfrac{1}{a} \\cdot \\dfrac{2k}{3+2k} a^{- \\left( \\frac{1}{2} +n \\right)} I _ {k-1} (a) \\rightarrow 0\r\n\\end{gather}\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n\\displaystyle\\lim _ {a \\rightarrow \\infty} a^{- \\left( \\frac{3}{2} +k \\right)} I _ {k} (a) = \\dfrac{2}{3+2k}\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(n=k\\) \u306e\u3068\u304d\u3082 [A] \u304c\u6210\u7acb\u3059\u308b.<\/p><\/li>\r\n<\/ol>\r\n1*<\/strong> 2*<\/strong> \u3088\u308a, \u3059\u3079\u3066\u306e \\(0\\) \u4ee5\u4e0a\u306e\u6574\u6570 \\(n\\) \u306b\u3064\u3044\u3066\r\n\\[\r\n\\displaystyle\\lim _ {a \\rightarrow \\infty} a^{- \\left( \\frac{3}{2} +n \\right)} I _ n (a) = \\underline{\\dfrac{2}{3+2n}}\r\n\\]\r\n\r\n \r\n « \u89e3\u7b54\u3092\u96a0\u3059 <\/a>\r\n <\/p>\r\n <\/div>","protected":false},"excerpt":{"rendered":"\\(a \\gt 0\\) \u306b\u5bfe\u3057 \\[\\begin{align} I _ 0 (a) & = \\displaystyle\\int _ {0}^a \\sqrt{1+x} \\, dx , \\\\ I _ n (a) & = \\d […]","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"footnotes":""},"categories":[95],"tags":[148,109],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/679"}],"collection":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/comments?post=679"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/679\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=679"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=679"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=679"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}