積分のラプラス変換

ラプラス変換

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

\( \int_0^t g(u) \mathrm{d}u \)のラプラス変換

$$\begin{eqnarray} f\left( t \right)&=&\int_0^t g(u) \mathrm{d}u \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=& \int_0^\infty {\left\{\int_0^t g(u) \mathrm{d}u\right\}{e^{ –st}}}\mathrm{d}t \\&=& \left[ \left\{\int_0^t g(u) \mathrm{d}u\right\} {\frac{–1}{s}e^{-st}} \right]_0^{\infty} -\int_0^\infty {g(t) \cdot {\frac{–1}{s}e^{–st}}}\mathrm{d}t \\&&\;\cdots\;\int_a^b{f'\left( t \right) g\left( t \right) }\mathrm{d}t=\left[f\left( t \right) g\left( t \right)\right]_a^b-\int_a^b{f\left( t \right) g'\left( t \right) }\mathrm{d}t\;\;(f':fの微分,\;g':gの部分) \\&&\;\cdots\;\int e^{at} \mathrm{d}t =\frac{1}{a}e^{at}+C \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2021/04/blog-post.html}{\frac{\mathrm{d}}{\mathrm{d}t}\int_0^t g\left(u\right)\mathrm{d}t =\frac{\mathrm{d}}{\mathrm{d}t}\left[G\left(t\right)-G\left(0\right)\right] =\frac{\mathrm{d}}{\mathrm{d}t}G\left(t\right)-\frac{\mathrm{d}}{\mathrm{d}t}G\left(0\right) =g\left(t\right)-0 =g\left(t\right)} \\&=& \left[ \left\{\int_0^\infty g(u) \mathrm{d}u\right\} {\frac{–1}{s}e^{-s\infty}} -\left\{\int_0^0 g(u)\mathrm{d}u\right\} {\frac{–1}{s}e^{-s0}} \right] +\frac{1}{s}\int_0^\infty {g(t) {e^{–st}}}\mathrm{d}t \\&&\;\cdots\;e^{-\infty}=0,\;\int_t^t f(x)\mathrm{d}x=\left[F(t)-F(t)\right]=0 \\&=& \left[0-0 \right] +\frac{1}{s}\int_0^\infty {g(t) {e^{–st}}}\mathrm{d}t \\&=& \frac{1}{s} \mathfrak{L}\left[ {g\left( t \right)} \right] \end{eqnarray}$$