berljivx8

2021-12-10

Need to find the Laplace transform of this:

Timothy Wolff

Beginner2021-12-11Added 26 answers

By the substitution $u=(2+s)twe\ge t{\int}_{0}^{\mathrm{\infty}}{e}^{-(2+s)t}dt=\frac{1}{2+s}{\int}_{0}^{\mathrm{\infty}}{e}^{-u}du=\frac{1}{2+s}$

servidopolisxv

Beginner2021-12-12Added 27 answers

There is no need of IBP.

$\int}_{0}^{\mathrm{\infty}}f\left(t\right){e}^{-st}dt={\int}_{0}^{\mathrm{\infty}}{e}^{-2t}{e}^{-st}dt={\int}_{0}^{\mathrm{\infty}}{e}^{-(s+2)t}dt=\frac{-1}{s+2}[0-1]=\frac{1}{s+2$