Laplace変換・逆変換　Laplace変換の例2 - 数学・物理・電気主任技術者試験、情報処理技術者試験等お勉強の記録

$\displaystyle{I= \int_0^\infty \cos(at) \exp(-st)dt, J=\int_0^\infty \sin(at) \exp(-st)dt}$

とおくと、

$(\cos(at)\exp(-st))'= -a\sin(at)\exp(-st) -s\cos(at)\exp(-st)$

$(\sin(at)\exp(-st))'=a\cos(at)\exp(-st)-s\sin(at)\exp(-st)$

より

$-a J -s I = [\cos(at)\exp(-st)]_0^\infty= -1$

$a I-s J = [\sin(at)\exp(-st)]_0^\infty = 0$

従って、

$-a^2 J -s^2 J = -a, -s^2 I -a^2 I = -s$

$\displaystyle{ I= \int_0^\infty \cos(at) \exp(-st)dt = \frac{s}{s^2+a^2} }$

$\displaystyle{ J= \int_0^\infty \sin(at) \exp(-st)dt = \frac{a}{s^2+a^2} }$

$\displaystyle{\mathcal{L}[f(at)](s)=\int_0^\infty f(at) \exp(-st)dt}$

$\displaystyle{u=at, dt=\frac{du}{a}}$ と変換すると

$\displaystyle{\mathcal{L}[f(at)](s)=\int_0^\infty f(u) \exp\left(-s\cdot\frac{u}{a}\right) \frac{du}{a} =\frac{1}{a} F\left(\frac{s}{a}\right)}$

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