三角関数のラプラス変換

${\displaystyle \mathcal{L}\left\{\sin \omega t\right\}=\frac{\omega }{s^2+{\omega }^2}}$

${\displaystyle \mathcal{L}\left\{\cos \omega t\right\}=\frac{s}{s^2+{\omega }^2}}$

ただし，$\mathrm{Re}s>0$ とする．

■証明：正弦関数

ラプラス変換の定義より

${\displaystyle \mathcal{L}\left\{\sin \omega t\right\}={\int }_0^{\infty }{e}^{-st}\sin \omega tdt}$

${\displaystyle ={{\displaystyle {\int }_0^{\infty }\left(-\frac{1}{s}{e}^{-st}\right)}}^{\text{'}}\sin \omega tdt}$

部分積分法を用いて

${\displaystyle ={\left[-\frac{1}{s}{e}^{-st}\sin \omega t\right]}_0^{\text{∞}}}$${\displaystyle -{\displaystyle {\int }_0^{\infty }\left(-\frac{1}{s}{e}^{-st}\right){\left(\sin \omega t\right)}^{\text{'}}dt}}$

${\displaystyle =\frac{1}{s}{\displaystyle {\int }_0^{\infty }{e}^{-st}\left(\omega \cos \omega t\right)}dt}$

${\displaystyle =\frac{1}{s}{\displaystyle {\int }_0^{\infty }{\left(-\frac{1}{s}{e}^{-st}\right)}^{\text{'}}\omega \cos \omega t}dt}$

もう一度，部分積分法を用いて

${\displaystyle {\displaystyle =\frac{1}{s}\{{\left[-\frac{1}{s}{e}^{-st}\omega \cos \omega t\right]}_0^{\text{∞}}}}$ ${\displaystyle -\left.{\displaystyle {\int }_0^{\infty }\left(-\frac{1}{s}{e}^{-st}\right){\left(\omega \cos \omega t\right)}^{\prime }dt}\right\}}$

${\displaystyle {\displaystyle =\frac{1}{s}\{{\left[-\frac{1}{s}{e}^{-st}\omega \cos \omega t\right]}_0^{\text{∞}}}}$ ${\displaystyle {\displaystyle -{\displaystyle {\int }_0^{\text{∞}}\frac{1}{s}{e}^{-st}{\omega }^2\sin \omega tdt}\}}}$

${\displaystyle =\frac{1}{s}\left(\frac{\omega }{s}-\frac{{\omega }^2}{s}{\displaystyle {\int }_0^{\infty }{e}^{-st}\sin \omega tdt}\right)}$

ここで

${\displaystyle \mathcal{L}\left\{\sin \omega t\right\}={\int }_0^{\infty }{e}^{-st}\sin \omega tdt}$

であることより

${\displaystyle \mathcal{L}\left\{\sin \omega t\right\}}$${\displaystyle =\frac{1}{s}\left(\frac{\omega }{s}-\frac{{\omega }^2}{s}\mathcal{L}\left\{\sin \omega t\right\}\right)}$

${\displaystyle \mathcal{L}\left\{\sin \omega t\right\}}$${\displaystyle =\frac{\omega }{s^2}-\frac{{\omega }^2}{s^2}\mathcal{L}\left\{\sin \omega t\right\}}$

この方程式を ${\displaystyle \mathcal{L}\left\{\sin \omega t\right\}}$ について解くと

${\displaystyle \mathcal{L}\left\{\sin \omega t\right\}\left(1+\frac{{\omega }^2}{s^2}\right)=\frac{\omega }{s^2}}$

${\displaystyle \mathcal{L}\left\{\sin \omega t\right\}=\frac{\omega }{s^2(1+\frac{{\omega }^2}{s^2})}}$

${\displaystyle =\frac{\omega }{s^2+{\omega }^2}}$

 

■証明：余弦関数

ラプラス変換の定義より

${\displaystyle \mathcal{L}\left\{\cos \omega t\right\}={\displaystyle {\int }_0^{\infty }{e}^{-st}\cos \omega tdt}}$

${\displaystyle ={{\displaystyle {\int }_0^{\infty }\left(-\frac{1}{s}{e}^{-st}\right)}}^{\text{'}}\cos \omega tdt}$

部分積分法を用いて

${\displaystyle ={\left[-\frac{1}{s}{e}^{-st}\cos \omega t\right]}_0^{\infty }-{\displaystyle {\int }_0^{\infty }\left(-\frac{1}{s}{e}^{-st}\right)}{\left(\cos \omega t\right)}^{\text{'}}dt}$

${\displaystyle ={\left[-\frac{1}{s}{e}^{-st}\cos \omega t\right]}_0^{\infty }-{\displaystyle {\int }_0^{\infty }\left(-\frac{1}{s}{e}^{-st}\right)}\left(-\omega \sin \omega t\right)dt}$

${\displaystyle =\frac{1}{s}-\frac{\omega }{s}{\displaystyle {\int }_0^{\infty }{e}^{-st}}\sin \omega tdt}$

${\displaystyle =\frac{1}{s}-\frac{\omega }{s}{{\displaystyle {\int }_0^{\infty }\left(-\frac{1}{s}{e}^{-st}\right)}}^{\text{'}}\sin \omega tdt}$

もう一度，部分積分法を用いて

${\displaystyle =\frac{1}{s}-\frac{\omega }{s}\{{\left[-\frac{1}{s}{e}^{-st}\sin \omega t\right]}_0^{\infty }}$ ${\displaystyle -{\displaystyle \underset{0}{\overset{\infty }{\int }}\left(-\frac{1}{s}{e}^{-st}\right)}{\left(\sin \omega t\right)}^{\prime }dt\}}$

${\displaystyle =\frac{1}{s}-\frac{\omega }{s}\{{\left[-\frac{1}{s}{e}^{-st}\sin \omega t\right]}_0^{\infty }}$${\displaystyle +{\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{1}{s}{e}^{-st}}\omega \cos \omega tdt\}}$

${\displaystyle =\frac{1}{s}-\frac{{\omega }^2}{s^2}{\displaystyle {\int }_0^{\infty }{e}^{-st}}\cos \omega tdt}$

ここで

${\displaystyle \mathcal{L}\left\{cos\omega t\right\}={\int }_0^{\infty }{e}^{-st}cos\omega tdt}$

であることより

${\displaystyle \mathcal{L}\left\{\cos \omega t\right\}=\frac{1}{s}-\frac{{\omega }^2}{s^2}\mathcal{L}\left\{\cos \omega t\right\}}$

この方程式を ${\displaystyle \mathcal{L}\left\{\cos \omega t\right\}}$ について解くと

${\displaystyle \mathcal{L}\left\{\cos \omega t\right\}\left(1+\frac{{\omega }^2}{s^2}\right)}$${\displaystyle =\frac{1}{s}}$

${\displaystyle \mathcal{L}\left\{\cos \omega t\right\}=\frac{1}{s\left(1+\frac{{\omega }^2}{s^2}\right)}}$

${\displaystyle =\frac{s}{s^2+{\omega }^2}}$

　

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　最終更新日： 2026年9月25日