What is inverse oflaplace x^{-1/2} ?

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

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inverse oflaplace x^{-1/2}

inverse laplace

x^{- \frac{1}{2}}

\frac{1}{π t ^{\frac{1}{2}}}

Solution steps

L^{- 1} {x^{- \frac{1}{2}}}

Expand

x^{- \frac{1}{2}} : \frac{1}{x ^{\frac{1}{2}}}

= L^{- 1} {\frac{1}{x ^{\frac{1}{2}}}}

= L^{- 1} {\frac{1}{π} \cdot \frac{( 2 \cdot 0 - 1 ) !! π}{2 ^{0} x ^{0 + \frac{1}{2}}}}

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)}

= \frac{1}{π} L^{- 1} {\frac{( 2 \cdot 0 - 1 ) !! π}{2 ^{0} x ^{0 + \frac{1}{2}}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{( 2 \cdot 0 - 1 ) !! π}{2 ^{0} x ^{0 + \frac{1}{2}}}} = t^{\frac{- 1}{2}}

= \frac{1}{π} t^{\frac{- 1}{2}}

= \frac{1}{π t ^{\frac{1}{2}}}

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

\int - 32 x + 38 d x

\int_{- \frac{1}{2}}^{\frac{1}{2}} 2 arccos (2 x) d x

\frac{d ^{2}}{d x ^{2}} (x^{3} - \frac{3}{x})

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

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