微分のラプラス変換

ラプラス変換

$$\begin{eqnarray} \mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \end{eqnarray}$$

\( \frac{\mathrm{d}f}{\mathrm{d}t} \)のラプラス変換

$$\begin{eqnarray} f\left( t \right)&=&\frac{\mathrm{d}g\left( t \right)}{\mathrm{d}t} \\\mathfrak{L}\left[ {g\left( t \right)} \right] &=& \int_0^\infty {\frac{\mathrm{d}g\left( t \right)}{\mathrm{d}t} {e^{ –st}}}\mathrm{d}t \\&=& \left[ g\left( t \right) {e^{-st}} \right]_0^{\infty} -\int_0^\infty {\frac{\mathrm{d}g\left( t\right)}{\mathrm{d}t} {\left(–s\right)e^{ –st}}}\mathrm{d}t \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分) \\&&\;\cdots\;\frac{\mathrm{d}}{\mathrm{d}t}e^{at}=ae^{at} \\&=& \left[ g\left( \infty \right) {e^{-s\infty}} - g\left( 0 \right) {e^{-s0}} \right] -\left(–s\right)\int_0^\infty {\frac{\mathrm{d}g\left( t\right)}{\mathrm{d}t} {e^{ –st}}}\mathrm{d}t \\&=& \left[ 0 - g\left( 0 \right) \cdot {1} \right] -\left(–s\right)\int_0^\infty {\frac{\mathrm{d}g\left( t\right)}{\mathrm{d}t} {e^{ –st}}}\mathrm{d}t \\&&\;\cdots\;e^{-\infty}=0,\;e^{0}=1 \\&=&-g\left( 0 \right) +s\int_0^\infty {\frac{\mathrm{d}g\left( t\right)}{\mathrm{d}t} {e^{ –st}}}\mathrm{d}t \\&=&-g\left( 0 \right)+s\mathfrak{L}\left[ {g\left( t \right)} \right] \\&=&s\mathfrak{L}\left[ {g\left( t \right)} \right]-g\left( 0 \right) \end{eqnarray}$$