Laplace transform of $ t^{1/2}$ and $ t^{-1/2}$ Prove the following Laplace transforms:
(a) $ \displaystyle{\mathcal{L} \{ t^{-1/2} \} = \sqrt{\frac{ \pi}{s}}} ,s>0 $
(b)  $ \displaystyle{\mathcal{L} \{ t^{1/2} \} =\frac{1}{2s} \sqrt{\frac{ \pi}{s}}} ,s>0 $
I did (a) as following:
(a) $ \displaystyle{\mathcal{L} \{ t^{-1/2} \}  = \int_{0}^{\infty} e^{-st} t^{-1/2}dt }$. Substituting $st=u$ and using the fact that $\displaystyle { \int_{0}^{\infty} e^{-u^2}du=\sqrt{\pi} }$ we are done.
Is there a similar way about (b)? Can we make a substitution to get in (a)?
edit: I know the formula $ \displaystyle  \mathcal{L} \{ t^n \} = \frac{\Gamma (n+1)}{s^{n+1}}, n>-1 ,s>0$ , but I would like to see a solution without this.
Thank's in advance!
 A: Like I mentioned earlier, there is the rule $\mathcal{L}\{tf(t)\}=-F'(s)$, here applicable with $f(t)=t^{-1/2}$. 
Or just directly apply $d/ds$ to part (a). Integration by-parts is equivalent ($u=t^{1/2},dv=e^{-ts}dt$).
A: For $t^{-1/2}$ we have
$$F(s)=\int\limits_0^\infty e^{-st} t^{-1/2}dt$$
Now make $st = u$ so that
$$F(s)=s^{-1/2} \int\limits_0^\infty e^{-u} u^{-1/2}du$$
Since the integral is $\Gamma(1/2)$ we get
$$F(s)=s^{-1/2} \sqrt \pi=\sqrt{\frac{\pi}{s}}$$
Why don't you want to prove the general case? Use the best tools you have when you can. We have
$$\mathcal{L}(t^n)=\int\limits_0^\infty e^{-st}t^n dt$$
We make $st = u$ and get
$$\mathcal{L}(t^n)=\frac{1}{s^{n+1}}\int\limits_0^\infty e^{-u}u^n du$$
Thus
$$\mathcal{L}(t^n)=\frac{\Gamma(n+1)}{s^{n+1}}$$
A: f (t)=t^-1/2
ANSWER:
clearly,f is not defined  at t=0,but it will be shown that laplace of (t^1/2) EXITS.
By definition we have 
 Let st =x
then
s dt=dx          or.       dt=1/s dx
t^-1/2=(x/s)^-1/2
substitute these values to formula of laplace
1/(s)^1/2 (sign of integral)0 to infinity e^-x.x^-1/2 dx 
1/(s)1/2(1/2) 
=(pie/s)     ANSEWR
A: simplification of the above 
st = u 
s dt = du 
dt = du/s
1/s integral(0- infinity ) e^-u (u/s)^-1/2 du
s^1/2 / s^1 = 1/s^1/2 therefore the answer is root pi by s since the multiplying term is s^1/2 again.
