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

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

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

inverse laplace

\frac{1 + e ^{- s π}}{s ^{2} + 1}

sin (t) + H (t - π) sin (t - π)

Solution steps

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

Expand

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

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

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

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

= sin (t) + H (t - π) sin (t - π)

y = 3 e^{3 e^{x} + x}

tan (x), 2 sin (x), [- \frac{π}{3}, \frac{π}{3}]

\frac{\partial}{\partial x} (e^{- 2 x})

\frac{d}{d x} ((4 x)^{3 x})

y = 0, y = x^{2}, y = x + 2

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