eを底とした指数凾数(at乗)のラプラス変換

ラプラス変換

$$\begin{array}{rcl}\mathfrak{L}\left[f\left(t\right)\right]& =& {\int }_0^{\mathrm{\infty }}f\left(t\right)e^{–st}\mathrm{d}t\end{array}$$

$e^{-at}$のラプラス変換

$$\begin{array}{rcl}f\left(t\right)& =& e^{at}\\ \mathfrak{L}\left[f\left(t\right)\right]& =& {\int }_0^{\mathrm{\infty }}f\left(t\right)e^{–st}\mathrm{d}t\\ & =& {\int }_0^{\mathrm{\infty }}{e}^{at}{e}^{–st}\mathrm{d}t\\ & =& {\int }_0^{\mathrm{\infty }}{e}^{-(s-a)t}\mathrm{d}t\\ & & \cdots u=-(s-a)t,t:0\to \mathrm{\infty },u:0\to -\mathrm{\infty }\\ & & \cdots \frac{\mathrm{d}u}{\mathrm{d}t}=-(s-a),\mathrm{d}t=\frac{-1}{s-a}\mathrm{d}u\\ & =& {\int }_0^{-\mathrm{\infty }}{e}^u\frac{-1}{s-a}\mathrm{d}u\\ & =& \frac{-1}{s-a}{\int }_0^{-\mathrm{\infty }}{e}^u\mathrm{d}u\\ & =& \frac{-1}{s-a}{\left[e^u\right]}_0^{-\mathrm{\infty }}\\ & =& \frac{-1}{s-a}[e^{-\mathrm{\infty }}-e^0]\\ & =& \frac{-1}{s-a}[0-1]\\ & =& \frac{-1}{s-a}(-1)\\ & =& \frac{1}{s-a}\end{array}$$