I am trying to calculate the Laplace transform of :
\begin{equation} f(t)=\frac{\sin t}{\sqrt{t}} \end{equation}
It necessitates to make the following conversion $\sin t=\frac{e^{it}-e^{-it}}{2i}$, then we obtain, by the Laplace transform formula $\int_0^\infty f(t)e^{-st}dt$:
\begin{equation} f(s)=\int_0^\infty t^{-\frac{1}{2}}\sin t\ e^{-st}\ dt=\int_0^\infty t^{-\frac{1}{2}}\frac{e^{it}-e^{-it}}{2i}\ e^{-st}\ dt \end{equation}
We then split it up in two integrals:
\begin{equation} f(s)=\frac{1}{2i}\int_0^\infty t^{-\frac{1}{2}}e^{(-s+i)t} dt-\frac{1}{2i}\int_0^\infty t^{-\frac{1}{2}}e^{(-s-i)t} dt \end{equation}
So here the primary choice from the Laplace transform table would be $t^ne^{at}=\frac{n!}{(s-a)^{n+1}}$, with $t=-1/2$ and $a=s+i$ for the former, and $a=s-i$ for the latter. But that doesn't work, since we get $-1/2!$, which is not defined.
A solution proposed in an exam is to use the transform for $t^{-1/2}=(\frac{\pi}{s})^{\frac{1}{2}}$. But how can that work, when the expression is clearly not only $t^{-1/2}$ but a product of $t^{-1/2}$ and $e^{(s\pm i)t}$?
This transform seems inconclusive from the given expressions in the table.
Any ideas?
Thanks
UPDATE: Using the correct result of $(-1/2)!=\Gamma(1/2)$ we get:
$t^ne^{at}=\frac{\Gamma(1/2)}{(-s-s+i)^{n+1}}$
this gives:
\begin{equation} \frac{1}{2i}\int_0^\infty t^{-\frac{1}{2}}e^{(-s+i)t} dt=\frac{1}{2i}\frac{\Gamma(-1/2)}{(-s-s+i)^{n+1}}=\frac{\Gamma(-1/2)}{(i-2s)^{1/2}} \end{equation}
for the second integral
\begin{equation} -\frac{1}{2i}\int_0^\infty t^{-\frac{1}{2}}e^{(-s-i)t} dt=-\frac{1}{2i}\frac{\Gamma(-1/2)}{(-s-s-i)^{n+1}}=-\frac{\Gamma(-1/2)}{(-i-2s)^{1/2}} \end{equation}