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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 Integration using parametric derivatives ： ウィキペディア英語版 Integration using parametric derivatives In mathematics, integration by parametric derivatives is a method of integrating certain functions. For example, suppose we want to find the integral : $\backslash int\_0^\backslash infty\; x^2\; e^\; \backslash ,\; dx.$ Since this is a product of two functions that are simple to integrate separately, repeated integration by parts is certainly one way to evaluate it. However, we may also evaluate this by starting with a simpler integral and an added parameter, which in this case is ''t'' = 3: : $$ \begin & \int_0^\infty e^ \, dx = \left(\frac \right )_0^\infty = \left( \lim_ \frac \right) - \left( \frac \right) \\ & = 0 - \left( \frac \right) = \frac. \end This converges only for ''t'' > 0, which is true of the desired integral. Now that we know : $\backslash int\_0^\backslash infty\; e^\; \backslash ,\; dx\; =\; \backslash frac,$ we can differentiate both sides twice with respect to ''t'' (not ''x'') in order to add the factor of ''x''2 in the original integral. : $$ \begin & \frac \int_0^\infty e^ \, dx = \frac \frac \\() & \int_0^\infty \frac e^ \, dx = \frac \frac \\() & \int_0^\infty \frac \left (-x e^\right) \, dx = \frac \left(-\frac\right) \\() & \int_0^\infty x^2 e^ \, dx = \frac. \end This is the same form as the desired integral, where ''t'' = 3. Substituting that into the above equation gives the value: : $\backslash int\_0^\backslash infty\; x^2\; e^\; \backslash ,\; dx\; =\; \backslash frac\; =\; \backslash frac.$ 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「Integration using parametric derivatives」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.