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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Cauchy formula for repeated integration ： ウィキペディア英語版 Cauchy formula for repeated integration The Cauchy formula for repeated integration, named after Augustin Louis Cauchy, allows one to compress ''n'' antidifferentiations of a function into a single integral (cf. Cauchy's formula). ==Scalar case== Let ''ƒ'' be a continuous function on the real line. Then the ''n''th repeated integral of ''ƒ'' based at ''a'', :$f^(x)\; =\; \backslash int\_a^x\; \backslash int\_a^\; \backslash cdots\; \backslash int\_a^)\; \backslash ,\; \backslash mathrm\backslash sigma\_\; \backslash cdots\; \backslash ,\; \backslash mathrm\backslash sigma\_2\; \backslash ,\; \backslash mathrm\backslash sigma\_1$, is given by single integration :$f^(x)\; =\; \backslash frac\; \backslash int\_a^x\backslash left(x-t\backslash right)^\; f(t)\backslash ,\backslash mathrmt$. A proof is given by induction. Since ''ƒ'' is continuous, the base case follows from the Fundamental theorem of calculus: :$\backslash fracx\}\; f^(x)\; =\; \backslash fracx\}\backslash int\_a^x\; f(t)\backslash ,\backslash mathrmt\; =\; f(x)$; where :$f^(a)\; =\; \backslash int\_a^a\; f(t)\backslash ,\backslash mathrmt\; =\; 0$. Now, suppose this is true for ''n'', and let us prove it for ''n+1''. Apply the induction hypothesis and switching the order of integration, :$$ \begin f^(x) &= \int_a^x \int_a^ \cdots \int_a^) \, \mathrm\sigma_ \cdots \, \mathrm\sigma_2 \, \mathrm\sigma_1 \\ &= \frac \int_a^x \int_a^\left(\sigma_1-t\right)^ f(t)\,\mathrmt\,\mathrm\sigma_1 \\ &= \frac \int_a^x \int_t^x\left(\sigma_1-t\right)^ f(t)\,\mathrm\sigma_1\,\mathrmt \\ &= \frac \int_a^x \left(x-t\right)^n f(t)\,\mathrmt \end The proof follows. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Cauchy formula for repeated integration」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.