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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 Integral of inverse functions ： ウィキペディア英語版 Integral of inverse functions In mathematics, integrals of inverse functions can be computed by means of a formula that expresses the antiderivatives of the inverse $f^$ of a continuous and invertible function $f$, in terms of $f^$ and an antiderivative of $f$. This formula was published in 1905 by Charles-Ange Laisant. ==Statement of the theorem== Let $I\_1$ and $I\_2$ be two intervals of $\backslash mathbb$. Assume that $f:\; I\_1\backslash to\; I\_2$ is a continuous and invertible function, and let $f^$ denote its inverse $I\_2\backslash to\; I\_1$. Then $f$ and $f^$ have antiderivatives, and if $F$ is an antiderivative of $f$, the possible antiderivatives of $f^$ are: :$\backslash int\; f^(y)\backslash ,dy=\; y\; f^(y)-F\backslash circ\; f^(y)+C,$ where $C$ is an arbitrary real number. In his 1905 article, Laisant gave three proofs. First, under the additional hypothesis that $f^$ is differentiable, one may differentiate the above formula, which completes the proof immediately. His second proof was geometric. If $f(a)=c$ and $f(b)=d$, the theorem can be written: :$\backslash int\_c^df^(y)\backslash ,dy+\backslash int\_a^bf(x)\backslash ,dx=bd-ac.$ The figure on the right is a proof without words of this formula. Laisant does not discuss the hypotheses necessary to make this proof rigorous, but it can be made explicit with the help of the Darboux integral〔 (or Fubini's theorem if a demonstration based on the Lebesgue integral is desired). Laisant's third proof uses the additional hypothesis that $f$ is differentiable. Beginning with $f^(f(x))\; =\; x$, one multiplies by $f\text{'}(x)$ and integrates both sides. The right-hand side is calculated using integration by parts to be $xf(x)\; -\; \backslash textstyle\backslash int\; f(x)\backslash ,dx$, and the formula follows. Nevertheless, it can be shown that this theorem holds even if $f$ or $f^$ is not differentiable:〔〔 it suffices, for example, to use the Stieltjes integral in the previous argument. On the other hand, even though general monotonic functions are differentiable almost everywhere, the proof of the general formula does not follow, unless $f^$ is absolutely continuous.〔 It is also possible to check that for every $y$ in $I\_2$, the derivative of the function $y\; \backslash to\; y\; f^(y)\; -F(f^(y))$ is equal to $f^(y)$.〔This very simple proof of the general theorem, the only one that does not make use of integrals, was communicated by the French mathematician and Wikipedian Anne Bauval in the corresponding pages in French. It seems to have escaped the persons who published proofs of this result.〕 In other words: :$\backslash forall\; x\backslash in\; I\_1\backslash quad\backslash lim\_\backslash frac=x.$ To this end, it suffices to apply the mean value theorem to $F$ between $x$ and $x+h$, taking into account that $f$ is monotonic. 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「Integral of inverse functions」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.