Laplace transform of $\sin(\sqrt t)$ How can I use this differential equation
$$4tf''(t) +2 f'(t) +  a^2 f(t)=0$$
 to show that 
$$L(\sin(\sqrt{t}))=\frac{1}{2}\sqrt{\pi}\,\frac{1}{s^{\frac{3}{2}}}\,e^{\frac{-1}{4s}}$$
 A: HINTS:
Note that $f(t)=\sin(\sqrt{t})$ is a solution to the differential equation
$$4tf''(t)+2f'(t)+f(t)=0$$
Let $F(s)$ represent the Laplace Transform of $f(t)$. Then note that Laplace Transform of $tf''(t)$ is given by 
$$\mathscr{L}\{tf''(t)\}(s)=-s^2\frac{dF(s)}{ds}-2sF(s)+f(0)$$
Now solve a simple ODE for $F(s)$.
NOTE:
Solution to the ODE without prescribed conditions (e.g., initial conditions) has the general form $f(t)=A\sin(\sqrt{t})+B\cos(\sqrt{t})$.  Therefore, if one is restricted to the ODE only to find the Laplace Transform of $\sin(\sqrt{t})$, then all we can say is that 
$$\mathscr{L}\{\sin(\sqrt{t})\}(s)=Ae^{-1/(4s)}s^{-3/2}$$  
A: I think it is faster to use the series definition of $\sin t$ and the Laplace transform of $t^{n+\frac{1}{2}}$, or just the definition of the Laplace transform:
$$ \int_{0}^{+\infty}e^{-st}\sin\sqrt{t}\,dt = \int_{0}^{+\infty} 2t\sin(t)e^{-st^2}\,dt = \text{Im} \int_{0}^{+\infty} 2t\, e^{it-st^2}\,dt. $$
The last integral can be computed by setting $t=\frac{u}{\sqrt{s}}$ and completing the square.
