Laplace Transform

Standard Form

$F(s)={\mathcal {L}}\left\{f(t)\right\}=\int _{0}^{\infty }e^{-st}f(t)\,dt$

Sample Functions

$F(s)={\mathcal {L}}\left\{1\right\}=\int _{0}^{\infty }e^{-st}\,dt=$ ${\frac {1}{s}}$

$F(s)={\mathcal {L}}\left\{t^{n}\right\}=\int _{0}^{\infty }e^{-st}t^{n}\,dt=$ ${\frac {n!}{s^{n+1}}}$

$F(s)={\mathcal {L}}\left\{e^{at}\right\}=\int _{0}^{\infty }e^{-st}e^{at}\,dt=$ ${\frac {1}{s-a}}$

$F(s)={\mathcal {L}}\left\{sin(\omega t)\right\}=\int _{0}^{\infty }e^{-st}sin(\omega t)\,dt=$ ${\frac {\omega }{s^{2}+\omega ^{2}}}$

$F(s)={\mathcal {L}}\left\{cos(\omega t)\right\}=\int _{0}^{\infty }e^{-st}cos(\omega t)\,dt=$ ${\frac {s}{s^{2}+\omega ^{2}}}$

$F(s)={\mathcal {L}}\left\{t^{n}g(t)\right\}=\int _{0}^{\infty }e^{-st}t^{n}g(t)\,dt=$ ${\frac {(-1)^{n}d^{n}G(s)}{ds^{n}}}{\mbox{ for}}~n\ {\mbox{= 1,2,...}}$

$F(s)={\mathcal {L}}\left\{tsin(\omega t)\right\}=\int _{0}^{\infty }e^{-st}tsin(\omega t)\,dt=$ ${\frac {2\omega s}{(s^{2}+\omega ^{2})^{2}}}$

$F(s)={\mathcal {L}}\left\{tcos(\omega t)\right\}=\int _{0}^{\infty }e^{-st}tcos(\omega t)\,dt=$ ${\frac {s^{2}-\omega ^{2}}{(s^{2}+\omega ^{2})^{2}}}$

$F(s)={\mathcal {L}}\left\{g(t)\right\}=\int _{0}^{\infty }e^{-st}g(t)\,dt=$ ${\frac {1}{a}}G\left({\frac {s}{a}}\right)$

$F(s)={\mathcal {L}}\left\{e^{at}g(t)\right\}=\int _{0}^{\infty }e^{-st}e^{at}g(t)\,dt=G(s-a)$

$F(s)={\mathcal {L}}\left\{e^{at}t^{n}\right\}=\int _{0}^{\infty }e^{-st}e^{at}t^{n}\,dt=$ ${\frac {n!}{(s-a)^{n+1}}}{\mbox{ for}}~n\ {\mbox{= 1,2,...}}$

$F(s)={\mathcal {L}}\left\{te^{-t}\right\}=\int _{0}^{\infty }e^{-st}te^{-t}\,dt=$ ${\frac {1}{(s+1)^{2}}}$

$F(s)={\mathcal {L}}\left\{1-e^{-t/T}\right\}=\int _{0}^{\infty }e^{-st}(1-e^{-t/T})\,dt=$ ${\frac {1}{s(1+Ts)}}$

$F(s)={\mathcal {L}}\left\{e^{at}sin(\omega t)\right\}=\int _{0}^{\infty }e^{-st}e^{at}sin(\omega t)\,dt=$ ${\frac {\omega }{(s-a)^{2}+\omega ^{2}}}$

$F(s)={\mathcal {L}}\left\{e^{at}cos(\omega t)\right\}=\int _{0}^{\infty }e^{-st}e^{at}cos(\omega t)\,dt=$ ${\frac {s-a}{(s-a)^{2}+\omega ^{2}}}$

$F(s)={\mathcal {L}}\left\{u(t)\right\}=\int _{0}^{\infty }e^{-st}u(t)\,dt=$ ${\frac {1}{s}}$

$F(s)={\mathcal {L}}\left\{u(t-a)\right\}=\int _{0}^{\infty }e^{-st}u(t-a)\,dt=$ ${\frac {e^{-as}}{s}}$

$F(s)={\mathcal {L}}\left\{u(t-a)g(t-a)\right\}=\int _{0}^{\infty }e^{-st}u(t-a)g(t-a)\,dt=e^{-as}G(s)$

$F(s)={\mathcal {L}}\left\{g'(t)\right\}=\int _{0}^{\infty }e^{-st}g'(t)\,dt=sG(s)-g(0)$

$F(s)={\mathcal {L}}\left\{g''(t)\right\}=\int _{0}^{\infty }e^{-st}g''(t)\,dt=s^{2}\cdot G(s)-s\cdot g(0)-g'(0)$

$F(s)={\mathcal {L}}\left\{g^{(n)}(t)\right\}=\int _{0}^{\infty }e^{-st}g^{(n)}(t)\,dt=s^{n}\cdot G(s)-s^{n-1}\cdot g(0)-s^{n-2}\cdot g'(0)-...-g^{(n-1)}(0)$

Failed to parse (unknown function "\t"): {\displaystyle F(s) = \mathcal{L} \left\{\int_0^{\t} g(t) \,dt \right\}=\int_0^{\t} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) }