一些带绝对值的函数的拉普拉斯变换

青青子衿

\[ \int_0^{+\infty} \big|\sin(ax)\big|e^{-sx}{\rmd}x =\dfrac{a \coth\left(\dfrac{\pi s}{2a}\right)}{s^2+a^2}\]
\[ \int_0^{+\infty} \big|\cos(ax)\big|e^{-sx}{\rmd}x =\dfrac{s+a\,{\rm{csch}}\left(\dfrac{\pi s}{2a}\right)}{s^2+a^2}\]
\[ \int_0^{+\infty} x\big|\sin(ax)\big|e^{-sx}{\rmd}x =\dfrac{2\,as \coth\left(\dfrac{\pi s}{2a}\right)}{\left(s^2+a^2\right)^2}+\dfrac{\pi\,{\rm{csch}}^2\left(\dfrac{\pi s}{2a}\right)}{2\left(s^2+a^2\right)}\]
\[ \int_0^{+\infty} x\big|\cos(ax)\big|e^{-sx}{\rmd}x =\dfrac{2\,as\,{\rm{csch}}\left(\dfrac{\pi s}{2a}\right)}{\left(s^2+a^2\right)^2}+\dfrac{\pi\,\coth\left(\dfrac{\pi s}{2a}\right){\rm{csch}}\left(\dfrac{\pi s}{2a}\right)}{2\left(s^2+a^2\right)}+\dfrac{s^2-a^2}{s^2+a^2}\]
\[ \int_0^{+\infty} \big|\sin(ax^2)\big|e^{-sx}{\rmd}x =?\]
\[ \int_0^{+\infty} \big|\cos(ax^2)\big|e^{-sx}{\rmd}x =?\]
\[ \int_0^{+\infty} x\big|\sin(ax^2)\big|e^{-sx}{\rmd}x =?\]
\[ \int_0^{+\infty} x\big|\cos(ax^2)\big|e^{-sx}{\rmd}x =?\]
\[ \int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =?\]

青青子衿

回复 1# 青青子衿

\[\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =?\]...
青青子衿 发表于 2018-7-7 21:52

\[ -\dfrac{{\rmd}}{{\rmd}s}\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =\int_0^{+\infty} \big|\sin(ax)\big|e^{-sx}{\rmd}x =\dfrac{a \coth\left(\dfrac{\pi s}{2a}\right)}{s^2+a^2}\]
\[ \dfrac{{\rmd}}{{\rmd}s}\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =-\dfrac{a \coth\left(\dfrac{\pi s}{2a}\right)}{s^2+a^2}\]
\[ \int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =\int_{s}^{+\infty}\dfrac{a \coth\left(\dfrac{\pi t}{2a}\right)}{t^2+a^2}{\rmd}t\]
______________________________________________________________________________________
\[\int_0^a\cos\left(yx\right)\operatorname{sgn}\left[\sin\left(ax\right)\right]{\rmd}y=\frac{\left|\sin ax\right|}{x}\]
\[\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =\int_0^a\int_0^{+\infty}\cos\left(yx\right)\operatorname{sgn}\left[\sin\left(ax\right)\right]e^{-sx}{\rmd}x{\rmd}y=?\]

1. FullSimplify[LaplaceTransform[Cos[y x] Sign[Sin[a x]], x, s,Assumptions -> a > 0,Assumptions -> y > 0]]

Copy the Code

\[\int_0^a\cos\left(yx\right)\operatorname{sgn}\left[\sin\left(yx\right)\right]{\rmd}y=\frac{\left|\sin ax\right|}{x}\]
\[
\begin{align*}
\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x
&=\int_0^a\int_0^{+\infty}\cos\left(yx\right)\operatorname{sgn}\left[\sin\left(yx\right)\right]e^{-sx}{\rmd}x{\rmd}y\\
&=\int_0^a\dfrac{s \coth\left(\dfrac{\pi s}{2y}\right)}{y^2+s^2}{\rmd}y
\end{align*}
\]

1. FullSimplify[LaplaceTransform[Cos[y x] Sign[Sin[y x]], x, s, Assumptions -> y > 0]]

Copy the Code

\[\color{red}{\int_0^{+\infty} \dfrac{\big|\sin(ax)\big|}{x}e^{-sx}{\rmd}x =\int_{s}^{+\infty}\dfrac{a \coth\left(\dfrac{\pi t}{2a}\right)}{t^2+a^2}{\rmd}t=\int_0^a\dfrac{s \coth\left(\dfrac{\pi s}{2y}\right)}{y^2+s^2}{\rmd}y
}\]

artofproblemsolving.com/community/c7h1715374