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}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

Where

H (t)

is Heaviside step function

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

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

= H (t - π) sin (t - π)

y = \frac{4 x}{( x ^{2} + 1 )}

\int \frac{2}{9 x ^{2} - 1} d x

\frac{d}{d x} (3 + sin (2 x))

x^{2} - 4 x, 2 x

\int 1 + 2 x 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)