Finding Inverse Laplace Transform using Taylor Series 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.
 A: 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
A: you should use the gamma function to find the inverse of each of the 1/sqrt(s) instead of the basic integer factorial. then take out a common factor of sqrt(t) and find a general series sum from n=0 to infinity. you also seem to be missing a 1/n! in the series. 
then use the given gamma formula from the standard formulae to sub for the Gamma(n+1/2) function. you can then use wolfram to get a standard function that the series will converge to. 
Then suggest what I can do for part b :)
A: $\frac{\cos 2(\sqrt{t})}{\sqrt{\pi t}}$
