Answer to Question #223816 in Chemical Engineering for Lokika

Question

L^{-1} e^{-s}/√s }

Answer 1

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

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

Answer: $L^{-1}\{e^{-s}/\sqrt{s}\}=\frac{u(t-1)}{\sqrt {\pi (t-1)}}$