Find the inverse Laplace transform
$F(t)=\mathcal{L}^{-1}(s^{-\frac{1}{2}}e^{-\frac{1}{s}})$
using each of the following techniques:
-
Expand the exponential in a Taylor series about s=∞, and take inverse Laplace transforms term by term (this is allowable since the series is uniformly convergent.).
-
Sum the resultant series in terms of elementary functions.
Best Answer
We know that $$ \displaystyle \mathcal{L} \{t^{n-1}\} = \frac{\Gamma(n)}{s^{n}}, n>0 ,s>0 $$ $$ \frac{1}{s^n} = \frac{\displaystyle \mathcal{L} \{t^{n-1}\}}{\Gamma(n)} $$ $$ \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^n}\} = \frac{t^{n-1}}{\Gamma(n)} $$ Therefore $$ \displaystyle \mathcal{L^{-1}} \{ s^{-1/2} e^{-1/s} \} = \displaystyle \mathcal{L^{-1}} \{ \frac{1}{s^{1/2}} - \frac{1}{s^{3/2}} + \frac{1}{(2!s^{5/2})} + ... + (-1)^{n}\frac{1}{n! s^{n+\frac{1}{2}}} \} $$ $$ = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^{n+\frac{1}{2}}}\} $$ $$ = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{t^{n-\frac{1}{2}}}{\Gamma(n+\frac{1}{2})} $$ But after this I don't know how to simplify