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

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

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inverse oflaplace-(e^{-pis})/(s^2+1)

inverse laplace

- \frac{e ^{- π s}}{s ^{2} + 1}

- H (t - π) sin (t - π)

Solution steps

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

= - H (t - π) sin (t - π)

(1950, 38300), (2000, 82500)

\int 7 sin (t) d t

f^{^{'}} (x) = x^{2} - 5

\int x sin (x) d x

\int \frac{1}{3 x ^{2} + 6 x + 78} d x

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