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

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

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

inverse laplace

\frac{e ^{- s π}}{( 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)

\frac{d}{d x} (\frac{x ^{2}}{x ^{3} + 8})

\frac{d}{d x} (103 sin (x) - 10 cos (x))

\int \frac{x ^{2} + 2 x - 1}{( x + 1 ) ^{2}} d x

f (t) = (1 + t) (8 t^{2} - 5)

n = 1 \sum \infty (1 + \frac{6}{n})^{- n}

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