ラプラス変換
$$\begin{eqnarray}
\mathfrak{L}\left[ {f\left( t \right)} \right]
&=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t
\end{eqnarray}$$
\( \cos{\left(\omega t\right)} \)のラプラス変換
$$\begin{eqnarray}
f\left( t \right)&=&\cos{\left(\omega t\right)}
\\\mathfrak{L}\left[ {f\left( t \right)} \right]
&=& \int_0^\infty {\cos{\left(\omega t\right)}{e^{ –st}}}\mathrm{d}t
\\&=& \left[ \cos{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{-st}} \right]_0^{\infty}
-\int_0^\infty {-\omega \sin{\left(\omega t\right)} \cdot {\frac{–1}{s}e^{–st}}}\mathrm{d}t
\\&&\;\ldots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t
\right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分)
\\&&\;\ldots\;\int e^{at} \mathrm{d}t =\frac{1}{a}e^{at}+C
\\&&\;\ldots\;\frac{\mathrm{d}}{\mathrm{d}t}\cos{\left(\omega t\right)}=-\omega \sin{\left(\omega t\right)}
\\&=& \left[ \cos{\left(\omega \infty\right)} \cdot {\frac{–1}{s}e^{-s\infty}} - \cos{\left(\omega 0\right)}
\cdot {\frac{–1}{s}e^{-s0}}\right]
-\left(\frac{\omega}{s}\right)\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t
\\&=& \left[ 0 - \frac{–1}{s}\right] -\frac{\omega}{s}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t
\\&&\;\ldots\;e^{-\infty}=0,\;\cos{\left( 0\right)}=1,\;e^{0}=1
\\&=& \frac{1}{s} -\frac{\omega}{s}\int_0^\infty {\sin{\left(\omega t\right)} {e^{–st}}}\mathrm{d}t
\\&=& \frac{1}{s} -\frac{\omega}{s}\mathfrak{L}\left[ \sin{\left(\omega t\right)} \right]
\\&=& \frac{1}{s} -\frac{\omega}{s}\frac{\omega}{\left(s^2+\omega^2\right)}
\\&&\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/sin.html}{\mathfrak{L}\left[ \sin{\left(\omega t\right)} \right]=\frac{\omega}{\left(s^2+\omega^2\right)}}
\\&=& \frac{s^2+\omega^2}{s\left(s^2+\omega^2\right)} -\frac{\omega^2}{s\left(s^2+\omega^2\right)}
\\&=& \frac{s^2+\omega^2-\omega^2}{s\left(s^2+\omega^2\right)}
\\&=& \frac{s}{s^2+\omega^2}
\end{eqnarray}$$
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