What is inverse oflaplace 1/s*(e^{-2*s*pi}-1) ?

The answer to inverse oflaplace 1/s*(e^{-2*s*pi}-1) is \H(t-2pi)-\H(t)

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inverse oflaplace 1/s(e^{-2s*pi}-1)

inverse laplace

\frac{1}{s} \cdot (e^{- 2 \cdot s \cdot π} - 1)

H (t - 2 π) - H (t)

Solution steps

L^{- 1} {\frac{1}{s} (e^{- 2 π s} - 1)}

Expand

\frac{1}{s} (e^{- 2 s π} - 1) : \frac{e ^{- 2 π s}}{s} - \frac{1}{s}

= L^{- 1} {\frac{e ^{- 2 π s}}{s} - \frac{1}{s}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{e ^{- 2 π s}}{s}} - L^{- 1} {\frac{1}{s}}

L^{- 1} {\frac{e ^{- 2 π s}}{s}} : H (t - 2 π)

L^{- 1} {\frac{1}{s}} : H (t)

= H (t - 2 π) - H (t)

\frac{d P}{d t} = (P - P^{2})

\int_{4}^{2} (6 x^{2} - 2 x + 1) d x

\frac{\partial}{\partial x} (- 3 y (- 2 x - y - 9))

(2 t - 1)^{2}

3 δ (t) + 2 + 4 e^{- 3 t + 2}

What is inverse oflaplace 1/s*(e^{-2*s*pi}-1) ?
The answer to inverse oflaplace 1/s*(e^{-2*s*pi}-1) is \H(t-2pi)-\H(t)