blog, laplace

Propiedades de la transformada de Laplace. Laplace.

Propiedad de linealidad

Si c_1 y c_2 son constantes y f_1 (t) y f_2(t) son cuyas transformadas de Laplace son, respectivamente, F_1 (s) y F_2(s), entonces es

\displaystyle \mathcal{L} [c_1 f_1 (t) + c_2 f_2 (t)] = c_1 \mathcal{L} [f_1 (t)] + c_2 \mathcal{L} [f_2 (t)]

O también

\mathcal{L} [c_1 f_1(t) + c_2 f_2 (t)] = c_1 F_1 (s) + c_2 F_2 (s)


Demostración

\displaystyle \mathcal{L} [c_1 f_1(t) + c_2 f_2 (t)] = \int_{0}^{\infty}{e^{-st}[c_1 f_1 (t) + c_2 f_2 (t)] \ dt}

\displaystyle \mathcal{L} [c_1 f_1(t) + c_2 f_2 (t)] = \int_{0}^{\infty}{[e^{-st}c_1 f_1 (t) + e^{-st} c_2 f_2 (t)] \ dt}

\displaystyle \mathcal{L} [c_1 f_1(t) + c_2 f_2 (t)] = \int_{0}^{\infty}{e^{-st} c_1 f_1 (t) \ dt} + \int_{0}^{\infty}{e^{-st} c_2 f_2 (t) \ dt}

\displaystyle \mathcal{L} [c_1 f_1(t) + c_2 f_2 (t)] = c_1 \int_{0}^{\infty}{e^{-st} f_1 (t) \ dt} + c_2 \int_{0}^{\infty}{e^{-st} f_2 (t) \ dt}

\displaystyle \mathcal{L} [c_1 f_1 (t) + c_2 f_2 (t)] = c_1 \mathcal{L} [f_1 (t)] + c_2 \mathcal{L} [f_2 (t)]

\displaystyle \therefore \mathcal{L} [c_1 f_1(t) + c_2 f_2 (t)] = c_1 F_1 (s) + c_2 F_2 (s)


Primera propiedad de traslación

Si \mathcal{L} [f(t)] = F(s) entonces

\mathcal{L} [e^{at} \ f(t)] = F(s-a)


Demostración

\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L}[e^{at} f(t)] = \int_{0}^{\infty}{e^{-st} \cdot e^{at} f(t) \ dt}

\displaystyle \mathcal{L}[e^{at} f(t)] = \int_{0}^{\infty}{e^{-st + at} f(t) \ dt}

\displaystyle \mathcal{L}[e^{at} f(t)] = \int_{0}^{\infty}{e^{(-s + a)t} f(t) \ dt}

\displaystyle \mathcal{L}[e^{at} f(t)] = \int_{0}^{\infty}{e^{-(s - a)t} f(t) \ dt}

\displaystyle \therefore \mathcal{L}[e^{at} f(t)] =F(s-a)


Segunda propiedad de traslación

Si f(t) = F(s), para f(t-a) si t>a es

\mathcal{L} [f(t-a)] = e^{-as} F(s)


Demostración.

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{a}{e^{-st} f(t) \ dt} +   \int_{a}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{a}{e^{-st} (0) \ dt} +   \int_{a}^{\infty}{e^{-st} g(t-a) \ dt}

\displaystyle \mathcal{L} [f(t)] = \int_{a}^{\infty}{e^{-st} g(t-a) \ dt}

Tomando la sustitución t = u+a y dt=du(y también u=t-a), se tiene que

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{\infty}{e^{-s(u+a)} g(u) \ du}

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{\infty}{e^{-su - as} g(u) \ du}

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{\infty}{e^{-su} e^{-as} g(u) \ du}

\displaystyle \mathcal{L} [f(t)] = e^{-as} \int_{0}^{\infty}{e^{-su} g(u) \ du}

\displaystyle \mathcal{L} [f(t)] = e^{-as} G(s)


Propiedad del cambio de escala

Si \mathcal{L}[f(t)] = F(s), entonces

\displaystyle \mathcal{L} [f(at)] = \frac{1}{a} F(\frac{s}{a})


Demostración

\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L}[f(at)] = \int_{0}^{\infty}{e^{-st} f(at) \ dt}

Realizando la sustitución u = at y du = a \ dt (y también \displaystyle t = \frac{u}{a}, dt = 1/a \ du) en el segundo miembro, resulta que

\displaystyle \mathcal{L}[f(at)] = \int_{0}^{\infty}{e^{-s \ \frac{u}{a}} f(u) \ (\frac{1}{a} \ du)}

\displaystyle \mathcal{L}[f(at)] = \frac{1}{a} \int_{0}^{\infty}{e^{- \frac{su}{a}} f(u) \ du}

\displaystyle \therefore \mathcal{L}[f(at)] = \frac{1}{a} f \left(\frac{s}{a} \right)


Transformada de Laplace de las derivadas

1.- Si \mathcal{L}[f(t)] = F(s), entonces

\displaystyle \mathcal{L} \left[ \frac{d}{dt} f(t) \right] = s F(s) - f(0)

si f(t) es continua para 0 \le t \le N y de orden exponencial para t>N mientras que f(t) es seccionalmente continua para 0 \le t \le N.

2.- Si f(f) no satisface la continuidad en t=0, pero \displaystyle \lim_{}{f(t)} = f(0+) existe (aunque no sea igual a f(0), el cual, puede o no existir), entonces

\mathcal{L}[f'(t)] = s \cdot F(s) - f(0+)

3.- Si f(t) deja de ser continua en t=a, entonces

\mathcal{L}[f'(t)] = s \cdot F(s) - f(0) - e^{-as}[f(a+) - f(a-)]

donde f(a+) - f(a-) se llama el salto en la discontinuidad en t=a. Pueden hacerse modificaciones que sean adecuadas para más de un salto.

4.- Si \mathcal{L}[f(t)] = F(s), entonces

\mathcal{L}[f(t)] = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \cdots - s f^{(n-2)}(0) - f^{(n-1)}(0)

si f(t), \cdots , f^{(n-1)}(0) son continuas para 0\le t \le N y de orden exponencial para t>N, además, si f^{(n)}(t) es seccionalmente continua para 0 \le t \le N.


Demostración de la transformada de Laplace de la primera derivada de una función.

\displaystyle \mathcal{L}[f'(t)] = \int_{0}^{\infty}{e^{-st} f'(t) \ dt}

\displaystyle \mathcal{L}[f'(t)] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-st} f'(t) \ dt}}

\displaystyle \mathcal{L}[f'(t)] = \lim_{m \rightarrow \infty}{\left[\left. e^{-st} f(t) \right|_{0}^{m} - \int_{0}^{m}{(-s \cdot e^{-st}) f(t) \ dt} \right]}

\displaystyle \mathcal{L}[f'(t)] = \lim_{m \rightarrow \infty}{\left\{ \left[e^{-s \cdot m} f(m) - e^{-s \cdot 0} f(0) \right] + s \int_{0}^{m}{e^{-st} f(t) \ dt} \right\}}

\displaystyle \mathcal{L}[f'(t)] = \left[e^{-s \cdot \infty} f(\infty) - e^{0} f(0) \right] + s \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L}[f'(t)] = \left[0 - (1) \ f(0) \right] + s \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L}[f'(t)] = - f(0) + s \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L}[f'(t)] = - f(0) + s \ F(s)

\displaystyle \therefore \mathcal{L}[f'(t)] = s \ F(s) - f(0)


Demostración de la transformada de Laplace de la segunda derivada de una función.

\displaystyle \mathcal{L}[f''(t)] = \int_{0}^{\infty}{e^{-st} f''(t) \ dt}

Haciendo g(t) = f'(t) y g'(t) = f''(t), la expresión toma la siguiente forma

\displaystyle \mathcal{L}[g'(t)] = \int_{0}^{\infty}{e^{-st} g'(t) \ dt}

\displaystyle \mathcal{L}[g'(t)] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-st} g'(t) \ dt}}

\displaystyle \mathcal{L}[g'(t)] = \lim_{m \rightarrow \infty}{\left[\left. e^{-st} g(t) \right|_{0}^{m} - \int_{0}^{m}{(-s \cdot e^{-st}) g(t) \ dt} \right]}

\displaystyle \mathcal{L}[g'(t)] = \lim_{m \rightarrow \infty}{\left\{ \left[e^{-s \cdot m} g(m) - e^{-s \cdot 0} g(0) \right] + s \int_{0}^{m}{e^{-st} g(t) \ dt} \right\}}

\displaystyle \mathcal{L}[g'(t)] = \left[e^{-s \cdot \infty} g(\infty) - e^{0} g(0) \right] + s \int_{0}^{\infty}{e^{-st} g(t) \ dt}

\displaystyle \mathcal{L}[g'(t)] = \left[0 - (1) \ g(0) \right] + s \int_{0}^{\infty}{e^{-st} g(t) \ dt}

\displaystyle \mathcal{L}[g'(t)] = - g(0) + s \int_{0}^{\infty}{e^{-st} g(t) \ dt}

\displaystyle \mathcal{L}[g'(t)] = - g(0) + s \ G(s)

\displaystyle \mathcal{L}[f''(t)] = - f'(0) + s \ F'(s)

\displaystyle \mathcal{L}[f''(t)] = - f'(0) + s  \int_{0}^{\infty}{e^{-st} f'(t) \ dt}

\displaystyle \mathcal{L}[f''(t)] = \lim_{m \rightarrow \infty}{\left[- f'(0) + s  \int_{0}^{m}{e^{-st} f'(t) \ dt} \right]}

\displaystyle \mathcal{L}[f''(t)] = \lim_{m \rightarrow \infty}{\left\{- f'(0) + s \left[\left. e^{-st} f(t)\right|_{0}^{m} - \int_{0}^{m}{(-s \cdot e^{-st}) f(t) \ dt} \right] \right\}}

\displaystyle \mathcal{L}[f''(t)] = - f'(0) + s \left[\left. e^{-st} f(t)\right|_{0}^{\infty} - \int_{0}^{\infty}{(-s \cdot e^{-st}) f(t) \ dt} \right]

\displaystyle \mathcal{L}[f''(t)] = - f'(0) + s \left[\left. e^{-st} f(t)\right|_{0}^{\infty} + s \int_{0}^{\infty}{e^{-st} f(t) \ dt} \right]

\displaystyle \mathcal{L}[f''(t)] = - f'(0) + s \left\{\left[ e^{-s \cdot \infty} f(\infty) - e^{-s \cdot 0} f(0) \right]  + s \int_{0}^{\infty}{e^{-st} f(t) \ dt} \right\}

\displaystyle \mathcal{L}[f''(t)] = - f'(0) + s \left\{\left[ e^{-\infty} f(\infty) - e^{0} f(0) \right]  + s \int_{0}^{\infty}{e^{-st} f(t) \ dt} \right\}

\displaystyle \mathcal{L}[f''(t)] = - f'(0) + s \left\{\left[ (0)- (1) f(0) \right]  + s \int_{0}^{\infty}{e^{-st} f(t) \ dt} \right\}

\displaystyle \mathcal{L}[f''(t)] = - f'(0) + s \left[- f(0) + s \int_{0}^{\infty}{e^{-st} f(t) \ dt} \right]

\displaystyle \mathcal{L}[f''(t)] = - f'(0) - s f(0) + s^2 \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L}[f''(t)] = - f'(0) - s f(0) + s^2 F(s)

\displaystyle \therefore \mathcal{L}[f''(t)] =  s^2 F(s) - s f(0) - f'(0)


Transformada de Laplace de las integrales

Si \mathcal{L} [f(t)] = F(s), entonces

\displaystyle \mathcal{L} \left[\int_{0}^{t}{f(\tau) \ d\tau} \right] = \frac{F(s)}{s}


Demostración.

Sea \displaystyle g(t) = \int_{0}^{t}{f(\tau) \ d\tau}, g'(t) = f(t) y g(0) = 0. Entonces,

\displaystyle g'(t) = f(t)

\displaystyle \mathcal{L}[g'(t)] = \mathcal{L}[f(t)]

\displaystyle s G(s) - g(0) = F(s)

\displaystyle s G(s) - 0 = F(s)

\displaystyle s G(s) = F(s)

\displaystyle G(s) = \frac{F(s)}{s}

\displaystyle \mathcal{L} \left[g(t) \right] = \frac{F(s)}{s}

\displaystyle \therefore \mathcal{L} \left[ \int_{0}^{\infty}{f(\tau) \ d\tau} \right] = \frac{F(s)}{s}


Multiplicación por t^n

Si \mathcal{L}[f(t)] = F(s), entonces

\displaystyle \mathcal{L}[t^n f(t)] = {(-1)}^n \frac{d^n}{ds^n} F(s) = {(-1)}^n F^{(n)} (s)


Demostración para «t f(t)» (es decir, n=1).

\displaystyle F(s) = \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \frac{d}{ds}[F(s)] = \frac{d}{ds} \left[\int_{0}^{\infty}{e^{-st} f(t) \ dt} \right]

\displaystyle \frac{d}{ds}[F(s)] = \int_{0}^{\infty}{\frac{\partial}{\partial s} [e^{-st} f(t)] \ dt}

\displaystyle \frac{d}{ds}[F(s)] = \int_{0}^{\infty}{(-t \cdot e^{-st}) f(t) \ dt}

\displaystyle \frac{d}{ds}[F(s)] = - \int_{0}^{\infty}{t e^{-st} f(t) \ dt}

\displaystyle \frac{d}{ds}[F(s)] = - \int_{0}^{\infty}{e^{-st} [t \ f(t)] \ dt}

\displaystyle \frac{d}{ds}[F(s)] = -\mathcal{L} [t f(t)]

\displaystyle -\mathcal{L} [t f(t)] =  \frac{d}{ds}[F(s)]

\displaystyle \mathcal{L} [t f(t)] = - \frac{d}{ds}[F(s)]

\displaystyle \therefore \mathcal{L} [t f(t)] = - F'(s)


Demostración para «t^n f(t)» (es decir, n=k+1).

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L} [t f(t)] = - F(s)

\displaystyle \mathcal{L} [t f(t)] = (-1) F(s)

\displaystyle \mathcal{L} [t^k f(t)] =  (-1)^k F^{(k)}(s)

\displaystyle \int_{0}^{\infty}{e^{-st} \cdot t^k f(t) \ dt} =  (-1)^k F^{(k)}(s)

\displaystyle \frac{d}{ds} \left[\int_{0}^{\infty}{e^{-st} \cdot t^k f(t) \ dt} \right] =  \frac{d}{ds} \left[(-1)^k F^{(k)}(s) \right]

\displaystyle \int_{0}^{\infty}{\frac{\partial}{\partial s}[e^{-st} \cdot t^k f(t)] \ dt} =   \left[(-1)^k \cdot F^{(k+1)}(s) \right]

\displaystyle \int_{0}^{\infty}{(-t e^{-st}) \cdot t^k f(t) \ dt} = (-1)^k F^{(k+1)}(s)

\displaystyle - \int_{0}^{\infty}{e^{-st} \cdot t^{k+1} f(t) \ dt} = (-1)^k F^{(k+1)}(s)

\displaystyle (-1) \int_{0}^{\infty}{e^{-st} \cdot t^{k+1} f(t) \ dt} = (-1)^k F^{(k+1)}(s)

\displaystyle \int_{0}^{\infty}{e^{-st} \cdot t^{k+1} f(t) \ dt} = \frac{(-1)^k}{(-1)} F^{(k+1)}(s)

\displaystyle \int_{0}^{\infty}{e^{-st} \cdot t^{k+1} f(t) \ dt} = (-1)^{k-1} F^{(k+1)}(s)

\displaystyle \int_{0}^{\infty}{e^{-st} \cdot t^{k+1} f(t) \ dt} = (-1)^{k+1} F^{(k+1)}(s)

\displaystyle \mathcal{L}[t^{k+1} f(t)] = (-1)^{k+1} F^{(k+1)}(s)

\displaystyle \mathcal{L}[t^n f(t)] = (-1)^n F^n (s) = (-1)^n \frac{d^n}{ds^n} [F(s)]


División por t

Si \mathcal{L} \left[f(t) \right] = F(s), entonces

\displaystyle \mathcal{L} \left[\frac{f(t)}{t} \right] = \int^{\infty}_{s}{f(\sigma) \ d\sigma}

siempre que exista \displaystyle \lim_{t \rightarrow 0}{\frac{f(t)}{t}}.


\displaystyle g(t) = \frac{f(t)}{t}

\displaystyle t \cdot g(t) = f(t)

\displaystyle \mathcal{L}[t g(t)] =\mathcal{L} [f(t)]

\displaystyle - \frac{d}{ds} {\mathcal{L} [g(t)]} = \mathcal{L} [f(t)]

\displaystyle - \frac{d}{ds} [G(s)] = F(s)

\displaystyle d[G(s)] = - F(s) \ ds

\displaystyle \int_{\infty}^{s}{d[G(s)]} = - \int_{\infty}^{s}{F(\sigma) \ d\sigma}

\displaystyle G(s) = \int_{s}^{\infty}{F(\sigma) \ d\sigma}

\displaystyle \mathcal{L}[g(t)] = \int_{s}^{\infty}{F(\sigma) \ d\sigma}

\displaystyle \therefore \mathcal{L} \left[\frac{f(t)}{t} \right] = \int_{s}^{\infty}{F(\sigma) \ d\sigma}


Funciones periódicas

Sea f(t) con período T>0 tal que f(t+T) = f(t). Entonces

\displaystyle \mathcal{L}[f(t)] = \frac{\int^{T}_{0}{e^{-st} f(\tau) d\tau}}{1-e^{-sT}}

funcion periódica
Figura 1.3.1 Represencación gráfica de una función peródica (R. Spiegel, 1996).

Demostración.

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{\infty}{e^{-st} f(t) \ dt}

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{T}{e^{-st} f(t) \ dt} +  \int_{T}^{2T}{e^{-st} f(t) \ dt} + \int_{2T}^{3T}{e^{-st} f(t) \ dt} + \cdots

Tomando la sustitución: en la primera integral t=\tau y dt=d\tau, en la segunda integral t=\tau + T y dt=d\tau, y en la tercera integral t= \tau + 2T y dt=d\tau, resulta que

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} +  \int_{0}^{T}{e^{-s(\tau+T)} f(\tau +T) \ d\tau} + \int_{0}^{T}{e^{-s(\tau +2T)} f(\tau +2T) \ d\tau} +

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} +  \int_{0}^{T}{e^{-s\tau -sT} f(\tau +T) \ d\tau} + \int_{0}^{T}{e^{-s\tau -2sT} f(\tau +2T) \ d\tau} +

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} +  \int_{0}^{T}{e^{-s\tau} e^{-sT} f(\tau+T) \ d\tau} + \int_{0}^{T}{e^{-s\tau} e^{-2sT} f(\tau+2T) \ d\tau} +

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} + e^{-sT} \int_{0}^{T}{e^{-s\tau} f(\tau+T) \ d\tau} +  e^{-2sT} \int_{0}^{T}{e^{-s\tau} f(\tau+2T) \ d\tau} +

Como f(\tau+T)=f(\tau), f(\tau + 2T) = f(\tau), …, tiene la siguiente expresión

\displaystyle \mathcal{L} [f(t)] = \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} + e^{-sT} \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} +  e^{-2sT} \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} +

\displaystyle \mathcal{L} [f(t)] = (1 + e^{-sT} +  e^{-2sT} + \cdots) \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau}

\displaystyle \mathcal{L} [f(t)] = \left( \frac{1}{1-e^{-sT}} \right) \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} =  \frac{ \int_{0}^{T}{e^{-s\tau} f(\tau) \ d\tau} }{1-e^{-sT}}


Comportamiento de F(s) cuando s \rightarrow \infty

Si \mathcal{L}[f(t)]=F(s), entonces

\displaystyle \lim_{s \rightarrow \infty}{F(s)} = 0

Teorema del valor inicial

Si existen los límites indicados, entonces

\displaystyle \lim_{t \rightarrow 0}{f(t)} = \lim_{s \rightarrow \infty}{s F(s)}


Demostración.

\displaystyle \mathcal{L}[f'(t)] = s F(s) - f(0)

\displaystyle \int_{0}^{\infty}{e^{-st} f'(t) \ dt} = s F(s) - f(0)

\displaystyle \lim_{s \rightarrow \infty}{\left[\int_{0}^{\infty}{e^{-st} f'(t) \ dt} \right]} = \lim_{s \rightarrow \infty}{\left[s F(s) - f(0) \right]}

\displaystyle \lim_{s \rightarrow \infty}{\left[\int_{0}^{\infty}{e^{-st} f'(t) \ dt} \right]} = \lim_{s \rightarrow \infty}{s F(s)} - \lim_{s \rightarrow \infty}{f(0)}

\displaystyle 0 = \lim_{s \rightarrow \infty}{s F(s)} - f(0)

\displaystyle 0 = \lim_{s \rightarrow \infty}{s F(s)} - \lim_{t \rightarrow 0}{f(t)}

\displaystyle \lim_{t \rightarrow 0}{f(t)} =  \lim_{s \rightarrow \infty}{s F(s)}


Teorema del valor final

Si existen los valores indicados, entonces

\displaystyle \lim_{t \rightarrow \infty}{f(t)} = \lim_{s \rightarrow 0}{s F(s)}


Demostración.

\displaystyle \mathcal{L}[f'(t)] = s F(s) - f(0)

\displaystyle \int_{0}^{\infty}{e^{-st} f'(t) \ dt} = s F(s) - f(0)

\displaystyle \lim_{s \rightarrow 0}{\left[\int_{0}^{\infty}{e^{-st} f'(t) \ dt} \right]} = \lim_{s \rightarrow 0}{\left[s F(s) - f(0) \right]}

\displaystyle \lim_{s \rightarrow 0}{\left[\int_{0}^{\infty}{e^{-st} f'(t) \ dt} \right]} = \lim_{s \rightarrow 0}{s F(s)} - \lim_{s \rightarrow 0}{f(0)}

\displaystyle  \int_{0}^{\infty}{f'(t) \ dt} = \lim_{s \rightarrow 0}{s F(s)} - f(0)

\displaystyle \lim_{m \rightarrow \infty}{\int_{0}^{m}{f'(t) \ dt}} = \lim_{s \rightarrow 0}{s F(s)} - \lim_{s \rightarrow 0}{f(0)}

\displaystyle \lim_{m \rightarrow \infty}{[f(m) - f(0)]} = \lim_{s \rightarrow 0}{s F(s)} - f(0)

\displaystyle \lim_{m \rightarrow \infty}{f(m)} -  \lim_{m \rightarrow \infty}{f(0)} = \lim_{s \rightarrow 0}{s F(s)} - f(0)

\displaystyle \lim_{m \rightarrow \infty}{f(m)} -  f(0) = \lim_{s \rightarrow 0}{s F(s)} - f(0)

\displaystyle \lim_{m \rightarrow \infty}{f(m)} = \lim_{s \rightarrow 0}{s F(s)}

Cambiando la variable m por t

\displaystyle \therefore \lim_{t \rightarrow \infty}{f(t)} =  \lim_{s \rightarrow 0}{s F(s)}


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