berljivx8

2021-12-10

Need to find the Laplace transform of this:
$f\left(t\right)={e}^{-2t}$

Timothy Wolff

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

servidopolisxv

There is no need of IBP.
${\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-2t}{e}^{-st}dt={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+2\right)t}dt=\frac{-1}{s+2}\left[0-1\right]=\frac{1}{s+2}$

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