# Common Laplace Transforms $F(s) = L \left\{ f(t) \right\} = \int_{0}^{\infty} f(t)e^{-st} dt$

Common laplace transforms $\text{a is a constant}$ $L \left\{ a \right\} =$ $\frac{a}{s}$ $L \left\{ t\right\} =$ $\frac{1}{s^2}$ $L \left\{ t^n \right\} =$ $\frac{n!}{s^{n+1}}$ $L \left\{ e^{-at}\right\} =$ $\frac{1}{s+a}$ $L \left\{ te^{-at}\right\} =$ $\frac{1}{(s+a)^2}$ $L \left\{ 1 - e^{-at}\right\} =$ $\frac{a}{s(s+a)}$ $L \left\{ sin(\omega t)\right\} =$ $\frac{\omega}{s^2 + \omega^2}$ $L \left\{ cos(\omega t)\right\} =$ $\frac{s}{s^2 + \omega^2}$ $L \left\{ 1 - cos(\omega t)\right\} =$ $\frac{\omega^2}{s(s^2 + \omega^2)}$ $L \left\{ \omega t sin(\omega t) \right\} =$ $\frac{2 \omega^2 s}{(s^2 + \omega^2)^2}$ $L \left\{ sin(\omega t) - \omega t cos(\omega t) \right\} =$ $\frac{2 \omega^3}{(s^2 + \omega^2)^2}$ $L \left\{ sin(\omega t + \phi) \right\} =$ $\frac{sin(\phi) + \omega cos(\phi) }{s^2 + \omega^2}$ $L \left\{ e^{-at} sin(\omega t) \right\} =$ $\frac{\omega}{(s + a)^2 + \omega^2}$ $L \left\{ e^{-at} cos(\omega t) \right\} =$ $\frac{s+a}{(s + a)^2 + \omega^2}$ $L \left\{ e^{-at} (cos(\omega t) - \frac{a}{\omega}sin(\omega t )) \right\} =$ $\frac{s}{(s + a)^2 + \omega^2}$ $L \left\{ e^{-at} + \frac{a}{\omega}sin(\omega t ) - cos(\omega t) \right\} =$ $\frac{a^2 + \omega^2}{(s + a)(s^2 + \omega^2)}$ $L \left\{ sinh(\beta t \right\} =$ $\frac{\beta}{s^2- \beta^2}$ $L \left\{e^{-at} sinh(\beta t \right\} =$ $\frac{\beta}{(s+a)^2- \beta^2}$ $L \left\{cosh(\beta t) \right\} =$ $\frac{s}{s^2 - \beta^2}$ $L \left\{e^{-at} sinh(\beta t \right\} =$ $\frac{s+a}{(s+a)^2- \beta^2}$ $L \left\{e^{-at} f(t) \right\} =$ $F(s+a)$ $L \left\{tf(t) \right\} =$ $-F(s)$ $L \left\{ x \right\} =$ $\bar{x}$ $L \left\{ \dot{x} \right\} =$ $s \bar{x} - x(0)$ $L \left\{ \ddot{x} \right\} =$ $s^2 \bar{x} - sx(0) - \dot{x}(0)$ $\text{Unit Step Function:}$ $L \left\{ u(t-a) \right\} = s^{-1}e^{-as}$ $L \left\{ f(t-a)u(t-a) \right\} = F(s)e^{-sa}$ $\text{Dirac Delta Function:}$ $L \left\{ \delta (t) \right\} =$ $1$ $L \left\{ \delta (t-a) \right\} =$ $e^{-as}$ $\text{Convolution Integral:}$ $\int_{0}^{t} f(z) g(t-z) dz = F(s)G(s)$ $\int_{0}^{t} f(t - z)g(z) dz = F(s)G(s)$