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

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

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

inverse laplace

\frac{1}{( s + 1 ) ^{\frac{1}{2}}}

\frac{e ^{- t}}{π t ^{\frac{1}{2}}}

Solution steps

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

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s + 1 ) ^{\frac{1}{2}}} : a = - 1, F (s) = \frac{1}{s ^{\frac{1}{2}}}

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

L^{- 1} {\frac{1}{s ^{\frac{1}{2}}}} : \frac{1}{π t ^{\frac{1}{2}}}

= \frac{e ^{- t}}{π t ^{\frac{1}{2}}}

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