Question

# Find the inverse Laplace transform of frac{e^{-s}}{s(s+1)}

Laplace transform
Find the inverse Laplace transform of
$$\frac{e^{-s}}{s(s+1)}$$

2021-02-20

Step 1
To determine : The inverse Laplace transform of
$$\left(\frac{e^{-s}}{s(s+1)}\right)$$
Step 2
Solution :
Using Heaviside step function
,if $$[L^{-1}\left[F(s)\right]=f(t) \text{ then } L^{-1}\left[e^{-as}F(s)\right]=f(t-a)u(t-a) \text{ for any } a\geq0 \text{ where } u(t-a) \text{ is the step function.}$$
$$\text{The given function is }\frac{e^{-s}}{s(s+1)}$$
$$\text{Here, }F(s)=\frac{1}{s(s+1)}\ \text{and }a=1$$
$$\text{Now, }u(t-1)L^{-1}\frac{1}{s(s+1)}(t-1)$$
$$\text{Using }L^{-1}\left[F(s)\right]=f(t)\ ,\text{then }L^{-1}\bigg(\frac{F(s)}{s}\bigg)=\int_{0}^{t}f(t)dt$$
$$L^{-1}\bigg(\frac{1}{s(s+1)}\bigg)=\int_{0}^{t}e^{-t}dt$$
$$L^{-1}\bigg(\frac{1}{s(s+1)}\bigg)=\left[\frac{e^{-t}}{-1}\right]_{0}^{t}=-\left[e^{-t}-e^{-0}\right]=-\left[e^{-t}-1\right]=1-e^{-t}$$
$$\text{Hence, the Inverse Laplace transform is } u(t-1)(1-e^{-(t-1)})$$