Answer to Question #117355 in Math for Akshay

Question

Find the Laplace transform of f(t)=e^-t,where f(t+1)=f(t)

Answer 1

If $f(t+T)=f(t),$ then

$$F(s)=L(f(t))={\displaystyle\int_{0}^T e^{-st}f(t)dt\over 1-e^{-sT}}$$

$$f(t+1)=f(t),\ f(t)=e^{-t}$$

$$F(s)=L(f(t))={\displaystyle\int_{0}^1 e^{-st}e^{-t}dt\over 1-e^{-s(1)}}=$$

$$=-{1\over (1-e^{-1})(s+1)}\bigg[e^{-(s+1)t}\bigg]\begin{matrix} 1\\ 0 \end{matrix}=$$

$$={1-e^{-(s+1)}\over (1-e^{-1})(s+1)}$$