Find the Laplace transforms of the following:
1.$\left(\dfrac{\cos \sqrt t}{\sqrt t}\right)$,
2.$\left(\sin \sqrt t\right)$.
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?
Batominovski's edit:
Here is how to solve question 1. Using $\cos(\theta)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\theta^{2k}$, we have
$$\cos(\sqrt{t})=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!} t^k,$$
so
$$\frac{\cos(\sqrt{t})}{\sqrt{t}}=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}t^{k-\frac12}.$$
Therefore
$$f(s)=\int_0^\infty \frac{\cos(\sqrt{t})}{\sqrt{t}}e^{-st}dt=\sum_{k=0}^\infty\int_0^\infty\frac{(-1)^k}{(2k)!}t^{k-\frac12}e^{-st} dt.$$
Thus, for $s>0$, by setting $u=st$, we have
$$f(s)=\int_0^\infty\frac{(-1)^k}{(2k)!}\frac{1}{s^{k+\frac12}}\int_0^\infty u^{\left(k+\frac12\right)-1}e^{-u}du.$$
Therefore,
$$f(s)=\frac1{\sqrt{s}}\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\left(\frac{1}{s}\right)^k\Gamma\left(k+\frac12\right).$$
But $\Gamma\left(k+\frac12\right)=\frac{(2k-1)!!}{2^k}\sqrt{\pi}$, so
$$f(s)=\sqrt{\frac{\pi}{s}}\sum_{k=0}^\infty(-1)^k\frac{(2k-1)!!}{(2k!)}\left(\frac{1}{2s}\right)^k,$$
so
$$f(s)=\sqrt{\frac{\pi}{s}}\sum_{k=0}^\infty\frac{(-1)^k}{k!}\left(\frac{1}{4s}\right)^k=\sqrt{\frac{\pi}{s}}e^{-\frac1{4s}}.$$
Best Answer
Note that you have : $$( \sin ( \sqrt t))'=\frac 1 2\frac {\cos ( \sqrt t)}{ \sqrt t}$$ And $$\mathscr {L} \{f'(t)\}(s)=s\,\mathscr {L} \{f(t)\}(s)-f(0)$$ Therefore: $$\mathscr {L} \{(\sin ( \sqrt t))'\}(s)=s\,\mathscr {L} \{\sin ( \sqrt t)\}(s)$$ $$\mathscr {L} \left \{\frac 1 2\frac {\cos ( \sqrt t)}{ \sqrt t} \right \}(s)=s\,\mathscr {L} \{\sin ( \sqrt t)\}(s)$$ $$\boxed {\mathscr {L} \left \{\frac {\cos ( \sqrt t)}{ \sqrt t} \right \}(s)=2s\,\mathscr {L} \{\sin ( \sqrt t)\}(s)}$$