Laplace Transformations of $left(frac{cos sqrt t}{sqrt t}right)$ and $left(sin sqrt tright)$

Find the Laplace transforms of the following:

1.$left(dfrac{cos sqrt t}{sqrt t}right)$,

2.$left(sin sqrt tright)$.

Well, in the first, I used the cosine series expansion and solved the problem.
But in the next one, I used the same logic, but didn’t get the answer as expected.

Please explain. Can question 2 be obtained from question 1?

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Batominovski’s edit:

Here is how to solve question 1. Using $cos(theta)=sum_{k=0}^inftyfrac{(-1)^k}{(2k)!}theta^{2k}$, we have
$$cos(sqrt{t})=sum_{k=0}^inftyfrac{(-1)^k}{(2k)!} t^k,$$
so
$$frac{cos(sqrt{t})}{sqrt{t}}=sum_{k=0}^inftyfrac{(-1)^k}{(2k)!}t^{k-frac12}.$$
Therefore
$$f(s)=int_0^infty frac{cos(sqrt{t})}{sqrt{t}}e^{-st}dt=sum_{k=0}^inftyint_0^inftyfrac{(-1)^k}{(2k)!}t^{k-frac12}e^{-st} dt.$$
Thus, for $s>0$, by setting $u=st$, we have
$$f(s)=int_0^inftyfrac{(-1)^k}{(2k)!}frac{1}{s^{k+frac12}}int_0^infty u^{left(k+frac12right)-1}e^{-u}du.$$
Therefore,
$$f(s)=frac1{sqrt{s}}sum_{k=0}^inftyfrac{(-1)^k}{(2k)!}left(frac{1}{s}right)^kGammaleft(k+frac12right).$$
But $Gammaleft(k+frac12right)=frac{(2k-1)!!}{2^k}sqrt{pi}$, so
$$f(s)=sqrt{frac{pi}{s}}sum_{k=0}^infty(-1)^kfrac{(2k-1)!!}{(2k!)}left(frac{1}{2s}right)^k,$$
so
$$f(s)=sqrt{frac{pi}{s}}sum_{k=0}^inftyfrac{(-1)^k}{k!}left(frac{1}{4s}right)^k=sqrt{frac{pi}{s}}e^{-frac1{4s}}.$$