一些带绝对值的函数的拉普拉斯变换
\[ \int_0^{+\infty} \big|\sin(ax)\big|e^{-sx}{\rmd}x =\dfrac{a \coth\left(\dfrac{\pi s}{2a}\right)}{s^2+a^2}\]\[ \int_0^{+\infty} \big|\cos(ax)\big|e^{-sx}{\rmd}x =\dfrac{s+a\,{\rm{csch}}\left(\dfrac{\pi s}{2a}\right)}{s^2+a^2}\]
\[ \int_0^{+\infty} x\big|\sin(ax)\big|e^{-sx}{\rmd}x =\dfrac{2\,as \coth\left(\dfrac{\pi s}{2a}\right)}{\left(s^2+a^2\right)^2}+\dfrac{\pi\,{\rm{csch}}^2\left(\dfrac{\pi s}{2a}\right)}{2\left(s^2+a^2\right)}\]
\[ \int_0^{+\infty} x\big|\cos(ax)\big|e^{-sx}{\rmd}x =\dfrac{2\,as\,{\rm{csch}}\left(\dfrac{\pi s}{2a}\right)}{\left(s^2+a^2\right)^2}+\dfrac{\pi\,\coth\left(\dfrac{\pi s}{2a}\right){\rm{csch}}\left(\dfrac{\pi s}{2a}\right)}{2\left(s^2+a^2\right)}+\dfrac{s^2-a^2}{s^2+a^2}\]
\[ \int_0^{+\infty} \big|\sin(ax^2)\big|e^{-sx}{\rmd}x =?\]
\[ \int_0^{+\infty} \big|\cos(ax^2)\big|e^{-sx}{\rmd}x =?\]
\[ \int_0^{+\infty} x\big|\sin(ax^2)\big|e^{-sx}{\rmd}x =?\]
\[ \int_0^{+\infty} x\big|\cos(ax^2)\big|e^{-sx}{\rmd}x =?\]
\[ \int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =?\] [i=s] 本帖最后由 青青子衿 于 2019-3-11 17:50 编辑 [/i]
[b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=27410&ptid=5472]1#[/url] [i]青青子衿[/i] [/b]
[quote] \[\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =?\]...
[size=2][color=#999999]青青子衿 发表于 2018-7-7 21:52[/color] [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=27410&ptid=5472][img]http://kuing.orzweb.net/images/common/back.gif[/img][/url][/size][/quote]
\[ -\dfrac{{\rmd}}{{\rmd}s}\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =\int_0^{+\infty} \big|\sin(ax)\big|e^{-sx}{\rmd}x =\dfrac{a \coth\left(\dfrac{\pi s}{2a}\right)}{s^2+a^2}\]
\[ \dfrac{{\rmd}}{{\rmd}s}\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =-\dfrac{a \coth\left(\dfrac{\pi s}{2a}\right)}{s^2+a^2}\]
\[ \int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =\int_{s}^{+\infty}\dfrac{a \coth\left(\dfrac{\pi t}{2a}\right)}{t^2+a^2}{\rmd}t\]
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\[\int_0^a\cos\left(yx\right)\operatorname{sgn}\left[\sin\left(ax\right)\right]{\rmd}y=\frac{\left|\sin ax\right|}{x}\]
\[\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =\int_0^a\int_0^{+\infty}\cos\left(yx\right)\operatorname{sgn}\left[\sin\left(ax\right)\right]e^{-sx}{\rmd}x{\rmd}y=?\][code]FullSimplify[LaplaceTransform[Cos[y x] Sign[Sin[a x]], x, s,Assumptions -> a > 0,Assumptions -> y > 0]][/code]\[\int_0^a\cos\left(yx\right)\operatorname{sgn}\left[\sin\left(yx\right)\right]{\rmd}y=\frac{\left|\sin ax\right|}{x}\]
\[
\begin{align*}
\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x
&=\int_0^a\int_0^{+\infty}\cos\left(yx\right)\operatorname{sgn}\left[\sin\left(yx\right)\right]e^{-sx}{\rmd}x{\rmd}y\\
&=\int_0^a\dfrac{s \coth\left(\dfrac{\pi s}{2y}\right)}{y^2+s^2}{\rmd}y
\end{align*}
\][code]FullSimplify[LaplaceTransform[Cos[y x] Sign[Sin[y x]], x, s, Assumptions -> y > 0]][/code]\[\color{red}{\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =\int_{s}^{+\infty}\dfrac{a \coth\left(\dfrac{\pi t}{2a}\right)}{t^2+a^2}{\rmd}t=\int_0^a\dfrac{s \coth\left(\dfrac{\pi s}{2y}\right)}{y^2+s^2}{\rmd}y
}\]
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