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# The Laplace Transform ## Definition Let $f(t)$ be a function defined for $t \geq 0$, the Laplace transform of $f(t)$, denoted by $F(s)$ or $\mathcal{L}\{f(t)\}$ is defined by $\qquad \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt$ where $s$ is a complex number. The integral must...

# The Laplace Transform ## Definition Let $f(t)$ be a function defined for $t \geq 0$, the Laplace transform of $f(t)$, denoted by $F(s)$ or $\mathcal{L}\{f(t)\}$ is defined by $\qquad \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} e^{-st} f(t) dt$ where $s$ is a complex number. The integral must converge for the Laplace transform to exist. ## Examples ### Example 1 $f(t) = 1$, $t \geq 0$ $\qquad \begin{aligned} \mathcal{L}\{1\} &= \int_{0}^{\infty} e^{-st} dt \\ &= \lim_{b \to \infty} \int_{0}^{b} e^{-st} dt \\ &= \lim_{b \to \infty} \left[ -\frac{1}{s} e^{-st} \right]_{0}^{b} \\ &= \lim_{b \to \infty} \left[ -\frac{1}{s} e^{-sb} + \frac{1}{s} \right] \\ &= \frac{1}{s}, \quad s > 0 \end{aligned}$ ### Example 2 $f(t) = e^{at}$, $t \geq 0$ $\qquad \begin{aligned} \mathcal{L}\{e^{at}\} &= \int_{0}^{\infty} e^{-st} e^{at} dt \\ &= \int_{0}^{\infty} e^{-(s-a)t} dt \\ &= \lim_{b \to \infty} \int_{0}^{b} e^{-(s-a)t} dt \\ &= \lim_{b \to \infty} \left[ -\frac{1}{s-a} e^{-(s-a)t} \right]_{0}^{b} \\ &= \lim_{b \to \infty} \left[ -\frac{1}{s-a} e^{-(s-a)b} + \frac{1}{s-a} \right] \\ &= \frac{1}{s-a}, \quad s > a \end{aligned}$ ### Example 3 $f(t) = \sin(at)$, $t \geq 0$ $\qquad \begin{aligned} \mathcal{L}\{\sin(at)\} &= \int_{0}^{\infty} e^{-st} \sin(at) dt \\ &= \lim_{b \to \infty} \int_{0}^{b} e^{-st} \sin(at) dt \end{aligned}$ Using integration by parts twice: $\qquad \int e^{-st} \sin(at) dt = -\frac{1}{s} e^{-st} \sin(at) + \frac{a}{s} \int e^{-st} \cos(at) dt$ $\qquad = -\frac{1}{s} e^{-st} \sin(at) + \frac{a}{s} \left[ -\frac{1}{s} e^{-st} \cos(at) - \frac{a}{s} \int e^{-st} \sin(at) dt \right]$ $\qquad \int e^{-st} \sin(at) dt = -\frac{1}{s} e^{-st} \sin(at) - \frac{a}{s^2} e^{-st} \cos(at) - \frac{a^2}{s^2} \int e^{-st} \sin(at) dt$ $\qquad \left( 1 + \frac{a^2}{s^2} \right) \int e^{-st} \sin(at) dt = -\frac{1}{s} e^{-st} \sin(at) - \frac{a}{s^2} e^{-st} \cos(at)$ $\qquad \left( \frac{s^2 + a^2}{s^2} \right) \int e^{-st} \sin(at) dt = -\frac{1}{s} e^{-st} \sin(at) - \frac{a}{s^2} e^{-st} \cos(at)$ $\qquad \int e^{-st} \sin(at) dt = \frac{-s e^{-st} \sin(at) - a e^{-st} \cos(at)}{s^2 + a^2}$ $\qquad \begin{aligned} \mathcal{L}\{\sin(at)\} &= \lim_{b \to \infty} \left[ \frac{-s e^{-st} \sin(at) - a e^{-st} \cos(at)}{s^2 + a^2} \right]_{0}^{b} \\ &= \lim_{b \to \infty} \left[ \frac{-s e^{-sb} \sin(ab) - a e^{-sb} \cos(ab)}{s^2 + a^2} - \frac{-s \cdot 1 \cdot \sin(0) - a \cdot 1 \cdot \cos(0)}{s^2 + a^2} \right] \\ &= \frac{a}{s^2 + a^2}, \quad s > 0 \end{aligned}$ ### Example 4 $f(t) = \cos(at)$, $t \geq 0$ $\qquad \mathcal{L}\{\cos(at)\} = \int_{0}^{\infty} e^{-st} \cos(at) dt$ Using integration by parts twice: $\qquad \begin{aligned} \int e^{-st} \cos(at) dt &= -\frac{1}{s} e^{-st} \cos(at) - \frac{a}{s} \int e^{-st} \sin(at) dt \\ &= -\frac{1}{s} e^{-st} \cos(at) - \frac{a}{s} \left[ \frac{-1}{s} e^{-st} \cos(at) + \frac{a}{s} \int e^{-st} \cos(at) dt \right] \\ &= -\frac{1}{s} e^{-st} \cos(at) + \frac{a}{s} e^{-st} \sin(at) - \frac{a^2}{s^2} \int e^{-st} \cos(at) dt \\ \left( 1 + \frac{a^2}{s^2} \right) \int e^{-st} \cos(at) dt &= -\frac{1}{s} e^{-st} \cos(at) + \frac{a}{s^2} e^{-st} \sin(at) \\ \left( \frac{s^2 + a^2}{s^2} \right) \int e^{-st} \cos(at) dt &= \frac{-s e^{-st} \cos(at) + a e^{-st} \sin(at)}{s^2} \\ \int e^{-st} \cos(at) dt &= \frac{-s e^{-st} \cos(at) + a e^{-st} \sin(at)}{s^2 + a^2} \end{aligned}$ $\qquad \begin{aligned} \mathcal{L}\{\cos(at)\} &= \lim_{b \to \infty} \left[ \frac{-s e^{-st} \cos(at) + a e^{-st} \sin(at)}{s^2 + a^2} \right]_{0}^{b} \\ &= \lim_{b \to \infty} \left[ \frac{-s e^{-sb} \cos(ab) + a e^{-sb} \sin(ab)}{s^2 + a^2} - \frac{-s \cdot 1 \cdot \cos(0) + a \cdot 1 \cdot \sin(0)}{s^2 + a^2} \right] \\ &= \frac{s}{s^2 + a^2}, \quad s > 0 \end{aligned}$