eを底とした指数凾数(at乗)とcos凾数との積に対するラプラス変換

ラプラス変換

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

\(e^{at}\cos{\left(b t\right)}\)のラプラス変換

$$\begin{eqnarray} f\left( t \right) &=& e^{at}\cos{\left(b t\right)} \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \\&=&\int_0^\infty {e^{at}\cos{\left(b t\right)}\;{e^{–st}}}\mathrm{d}t \\&=&\int_0^\infty { e^{-st+at}\cos{\left(b t\right)} }\mathrm{d}t\;\ldots\;e^Ae^B=e^{A+B} \\&=&\int_0^\infty { e^{-\left(s-a\right)t}\cos{\left(b t\right)} }\mathrm{d}t \\&=&\int_0^\infty { e^{-ut}\cos{\left(b t\right)} }\mathrm{d}t\;\ldots\;u=s-a \\&=&\mathfrak{L}\left[ {\cos{\left(b t\right)}} \right]\;\ldots\;uをsとみなして利用する \\&=&\frac{u}{u^2+b^2}\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/05/cos.html}{\mathfrak{L}\left[ {\cos{\left(\omega t\right)}} \right]＝\frac{s}{s^2+\omega^2}} \\&=&\frac{s-a}{\left(s-a\right)^2+b^2} \end{eqnarray}$$

eを底とした指数凾数(at乗)とsin凾数との積に対するラプラス変換

ラプラス変換

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

\(e^{at}\sin{\left(b t\right)}\)のラプラス変換

$$\begin{eqnarray} f\left( t \right) &=& e^{at}\sin{\left(b t\right)} \\\mathfrak{L}\left[ {f\left( t \right)} \right] &=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t \\&=&\int_0^\infty {e^{at}\sin{\left(b t\right)}\;{e^{–st}}}\mathrm{d}t \\&=&\int_0^\infty { e^{-st+at}\sin{\left(b t\right)} }\mathrm{d}t\;\ldots\;e^Ae^B=e^{A+B} \\&=&\int_0^\infty { e^{-\left(s-a\right)t}\sin{\left(b t\right)} }\mathrm{d}t \\&=&\int_0^\infty { e^{-ut}\sin{\left(b t\right)} }\mathrm{d}t\;\ldots\;u=s-a \\&=&\mathfrak{L}\left[ {\sin{\left(b t\right)}} \right]\;\ldots\;uをsとみなして利用する \\&=&\frac{b}{u^2+b^2}\;\ldots\;\href{https://shikitenkai.blogspot.com/2021/04/sin.html}{\mathfrak{L}\left[ {\sin{\left(\omega t\right)}} \right]＝\frac{\omega}{s^2+\omega^2}} \\&=&\frac{b}{\left(s-a\right)^2+b^2} \end{eqnarray}$$

log(1+x)の冪級数

\(\log{\left(1+x\right)}\)の冪級数

$$\begin{eqnarray} \log{\left(1+x\right)} &=&\int_0^x \frac{1}{1+t} \mathrm{d}t \;\ldots\;\frac{\mathrm{d}}{\mathrm{d}x}\log{x}=\frac{1}{x}より．ただしx\gt-1 \\&=&\int_0^x \frac{1+t-t}{1+t} \mathrm{d}t \;\ldots\;分子にt-tを加える \\&=&\int_0^x \frac{1+t}{1+t}+\frac{-t}{1+t} \mathrm{d}t \\&=&\int_0^x 1+\frac{-t}{1+t} \mathrm{d}t \\&=&\int_0^x 1\mathrm{d}t-\int_0^x \frac{t}{1+t}\mathrm{d}t \\&=&\left[t\right]_0^x -\int_0^x \frac{t}{1+t}\mathrm{d}t \\&=&\left[x-0\right] -\int_0^x \frac{t}{1+t}\mathrm{d}t \\&=&x -\int_0^x \frac{t}{1+t}\mathrm{d}t \\&=&x -\int_0^x \frac{t+t^2-t^2}{1+t}\mathrm{d}t \;\ldots\;分子にt^2-t^2を加える \\&=&x -\int_0^x \frac{t\left(1+t\right)-t^2}{1+t}\mathrm{d}t \\&=&x -\int_0^x \frac{t\left(1+t\right)}{1+t}+\frac{-t^2}{1+t}\mathrm{d}t \\&=&x -\int_0^x t+\frac{-t^2}{1+t}\mathrm{d}t \\&=&x-\int_0^x t\mathrm{d}t -\int_0^x\frac{-t^2}{1+t}\mathrm{d}t \\&=&x-\int_0^x t\mathrm{d}t +\int_0^x\frac{t^2}{1+t}\mathrm{d}t \\&=&x-\left[\frac{1}{2}t^2\right]_0^x +\int_0^x\frac{t^2}{1+t}\mathrm{d}t \\&=&x-\left[\frac{1}{2}x^2-\frac{1}{2}0^2\right] +\int_0^x\frac{t^2}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2 +\int_0^x\frac{t^2}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2 +\int_0^x\frac{t^2+t^3-t^3}{1+t}\mathrm{d}t \;\ldots\;分子にt^3-t^3を加える \\&=&x-\frac{1}{2}x^2 +\int_0^x\frac{t^2\left(1+t\right)-t^3}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2 +\int_0^x\frac{t^2\left(1+t\right)}{1+t}\frac{-t^3}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2 +\int_0^x t^2 +\frac{-t^3}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\int_0^x t^2\mathrm{d}t -\int_0^x \frac{t^3}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\left[\frac{1}{3}t^3\right]_0^x -\int_0^x \frac{t^3}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\left[\frac{1}{3}x^3-\frac{1}{3}0^3\right] -\int_0^x \frac{t^3}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3 -\int_0^x \frac{t^3}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3 -\int_0^x \frac{t^3+t^4-t^4}{1+t}\mathrm{d}t \;\ldots\;分子にt^4-t^4を加える \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3 -\int_0^x \frac{t^3\left(1+t\right)-t^4}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3 -\int_0^x t^3+\frac{-t^4}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3 -\int_0^x t^3\mathrm{d}t-\int_0^x\frac{-t^4}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3-\int_0^x t^3\mathrm{d}t +\int_0^x\frac{t^4}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3-\left[\frac{1}{4}t^4\right]_0^x +\int_0^x\frac{t^4}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3-\left[\frac{1}{4}x^4-\frac{1}{4}0^4\right] +\int_0^x\frac{t^4}{1+t}\mathrm{d}t \\&=&x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4 +\int_0^x\frac{t^4}{1+t}\mathrm{d}t \\&=&\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k+1}}{k}x^k \end{eqnarray}$$