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青青子衿 发表于 2018-7-5 19:35

几个带绝对值的函数的定积分公式

[i=s] 本帖最后由 青青子衿 于 2018-7-27 11:08 编辑 [/i]

\[
\int_0^x \big|\,t\,\big|^n{\rm d}t=\dfrac{{\rm sgn}(x)}{n+1}x^{n+1}=\dfrac{1}{n+1}x^{n}\big|x\big|
\qquad\begin{cases}
x\in\mathbb{R}\quad n\in \mathbb{Z^+}\\
x\ne0\quad n\in \mathbb{Z}\backslash(\mathbb{N}\bigcup\{-1\})\end{cases}\]
\[ \int_1^x \dfrac{1}{\big|\,t\,\big|}{\rm d}t={\rm sgn}(x)\ln x \qquad(x\ne0)\]
\[ \int_0^x a^{|\,t\,|}{\rm d}t=\dfrac{{\rm sgn}(x)}{\ln a}\left(a^{|\,x\,|}-1\right) \qquad(a>0,a\ne1)\]
\[ \int_0^x a^{-|\,t\,|}{\rm d}t=\dfrac{{\rm sgn}(x)}{\ln a}\left(1-a^{-|\,x\,|}\right) \qquad(a>0,a\ne1)\]
\[ \int_1^x \big|\,\ln t\,\big|{\rm d}t=x\big|\,\ln x\,\big|-(x-1){\rm sgn}(\ln x)+1\qquad(x\ne 0)\]
\[ \int_0^x \big|\sin t\big|{\rm d}t=1-\cos\left[x-\pi\,{\rm floor}\left(\dfrac{x}{\pi}\right)\right]+2\,{\rm floor}\left(\dfrac{x}{\pi}\right)\]
\[ \begin{align*}
\int_0^x \big|\cos t\big|{\rm d}t &= \sin\left[x-\pi\,{\rm round}\left(\dfrac{x}{\pi}\right)\right]+2\,{\rm round}\left(\dfrac{x}{\pi}\right) \\
&= \sin\left[x-\pi\,{\rm floor}\left(\dfrac{x}{\pi}+\dfrac{1}{2}\right)\right]+2\,{\rm floor}\left(\dfrac{x}{\pi}+\dfrac{1}{2}\right)\end{align*}\]
Absolute-value functions
[url]https://en.wikipedia.org/wiki/Lists_of_integrals[/url]

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