Laplace transform of elementary functions

Posted December 8th, 2007 by Structure

Some useful Laplace transform
We use for the unit step function notation h in honor of Oliver Heaviside(1850-1925).

$\mathcal{L}(h(t))(s)=\frac{1}{s}\:\: ;<br /> \mathcal{L}(th(t))(s)=\frac{1!}{s^2}\:\: ;<br /> \mathcal{L}(t^2h(t))(s)=\frac{2!}{s^3}\:\: ;<br /> \mathcal{L}(t^3h(t))(s)=\frac{3!}{s^4}\:\: ;<br /> \mathcal{L}(t^nh(t))(s)=\frac{n!}{s^{n+1}}$

$\mathcal{L}(t^{\alpha}h(t))(s)=\frac{\Gamma (\alpha+1)}{s^{\alpha+1}}\:\: for \:\: \alpha>-1$

$\mathcal{L}(\cos t \:h(t))(s)=\frac{s}{s^2+1}$

Proof

$\mathcal{L}(\sin t\: h(t))(s)=\frac{1}{s^2+1}$

Proof

$\mathcal{L}(\cosh t \:h(t))(s)=\frac{s}{s^2-1}$

Proof

$\mathcal{L}(\sinh t \:h(t))(s)=\frac{1}{s^2-1}$

Proof

$\mathcal{L}(\cos \omega t \:h(t))(s)=\frac{s}{s^2+\omega^2}$

$\mathcal{L}(\sin \omega t \:h(t))(s)=\frac{\omega}{s^2+\omega^2}$

$\mathcal{L}(\cosh \omega t\:h(t))(s)=\frac{s}{s^2-\omega^2}$

$\mathcal{L}(\sinh \omega t \:h(t))(s)=\frac{\omega}{s^2-\omega^2}$

$\mathcal{L}(e^{\lambda t}\cos \omega t \:h(t))(s)=\frac{s-\lambda}{(s-\lambda)^2+\omega^2}$

$\mathcal{L}(e^{\lambda t}\sin \omega t\: h(t))(s)=\frac{\omega}{(s-\lambda)^2+\omega^2}$

$\mathcal{L}(e^{\lambda t}\cosh \omega t\:h(t))(s)=\frac{s-\lambda}{(s-\lambda)^2-\omega^2}$

$\mathcal{L}(e^{\lambda t}\sinh \omega t \:h(t))(s)=\frac{\omega}{(s-\lambda)^2-\omega^2}$

$\mathcal{L}(\frac{1}{2\sqrt{\pi}}\:t^{-\frac{3}{2}}e^{-\frac{1}{4t}}h(t))(s)=e^{-\sqrt s}$

where $\sqrt s$ is positive on positive reals.

$\mathcal{L}(|\sin t| h(t))(s)=\int_{0}^{+\infty}\sin te^{-st}dt=\frac{1}{1-e^{-s\pi}}\int_{0}^{\pi}\sin t e^{-st}dt=\frac{1+e^{-s\pi}}{1-e^{-s\pi}}\frac{1}{s^2+1}$

Proof

Laplace transform of elementary functions

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