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# Answer to Question #223816 in Chemical Engineering for Lokika

Question #223816

L^{-1} e^{-s}/√s }

1
2021-08-10T05:54:45-0400

the time shifting property of Laplace transform:

"f(t-a)u(t-a)=L^{-1}\\{e^{-as}F(s)\\}, a>0"

where u(t) is the Heaviside step function and

"F(s)=L\\{f(t)\\}"

hence

"L^{-1}\\{e^{-s}\/\\sqrt{s}\\}=[a=1>0]=u(t-1)f(t-1)"

where

"F(s)=\\frac{1}{\\sqrt{s}}=\\Big[for \\space p=-\\frac 1 2 \\Big]=\\frac{\\Gamma(p+1)}{\\Gamma(p+1) s^{p+1}}=\\frac{\\Gamma(\\frac 1 2)}{\\Gamma(\\frac 1 2) \\sqrt{s}}"

The Gamma function is an extension of the normal factorial function and

"\\Gamma \\Big( \\frac 1 2 \\Big)=\\sqrt {\\pi}"

for pth power of Laplace transform:

"L^{-1} \\Bigg \\{ \\frac{\\Gamma(p+1)}{s^{p+1}} \\Bigg \\}=[Re(p)>-1]=t^p"

So

"f(t)=L^{-1} \\Bigg \\{ \\frac{\\Gamma(1\/2)}{\\Gamma(1\/2) \\sqrt{s}} \\Bigg \\}=\\frac{1}{\\sqrt {\\pi}}L^{-1} \\Bigg \\{ \\frac{\\Gamma(1\/2)}{ \\sqrt{s}} \\Bigg \\}=\\frac{1}{\\sqrt {\\pi t}}""L^{-1}\\{e^{-s}\/\\sqrt{s}\\}=u(t-1)f(t-1)=\\frac{u(t-1)}{\\sqrt {\\pi (t-1)}}"

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