Find the Laplace Transform of $\sin\sqrt{t}$ To find the Laplace Transform of $\sin\sqrt{t}$, I use the general formula $F(s)=\int_0^\infty e^{-st}f(t)\,dt$ and I get that: $$\mathcal{L}[\sin\sqrt{t}]=\int_0^\infty e^{-st}\sin\sqrt{t}\,dt=\int_0^\infty e^{-st}\sin\frac{\sqrt{st}}{\sqrt{s}}\,dt$$
Now I make the substitution $st=u \implies dt=\frac{du}{s}$
$$\mathcal{L}[\sin\sqrt{t}]=\int_0^\infty \frac{e^{-u}}{s} \sin\left(\sqrt{\frac{u}{s}}\right)\,du=\frac{1}{s}\int_0^\infty e^{-u}\sin\left(\sqrt{\frac{u}{s}}\right)\,du=\frac{1}{s}\int_0^\infty e^{-u}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\frac{\sqrt u \cdot u^n}{\sqrt s \cdot s^n}\,du=\frac{1}{s\sqrt s}\int_0^\infty e^{-u}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\frac{\sqrt u \cdot u^n}{s^n}\,du$$
How to keep solving this integral?
 A: Using Euler's formula, $\mathcal{L}\{\sin\sqrt{t}\}(s)=\operatorname{Im}\int_0^\infty e^{i\sqrt{t}-st}\, dt$. If $I(s,t)=\int e^{i\sqrt{t}-st}\, dt$, then using the substitution $u=\sqrt{t}$ yields
$$I(s,t)=\frac{1}{s}\int (2su-i)e^{iu-su^2}\, du+\frac{i}{s}\int e^{iu-su^2}\, du=\frac{1}{s}J+\frac{i}{s}K.$$
For $J$, substitute $v=iu-su^2.$
Therefore
$$J=-\int e^v\, dv=-e^{i\sqrt{t}-st}+C_1.$$
For $K$, complete the square:
$$K=\int e^{-\left(u\sqrt{s}-\frac{i}{2\sqrt{s}}\right)^2-\frac{1}{4s}}\, du.$$
Then substitute $u=\frac{2w\sqrt{s}+i}{2s}$ to get $$\begin{align}K&=\frac{1}{\sqrt{s}}\int e^{-w^2-\frac{1}{4s}}\, dw\\&=\frac{\sqrt{\pi}e^{-\frac{1}{4s}}}{2\sqrt{s}}\operatorname{erf}w+C_2\\&=\frac{\sqrt{\pi}e^{-\frac{1}{4s}}}{2\sqrt{s}}\operatorname{erf}\frac{2s\sqrt{t}-i}{2\sqrt{s}}+C_2.\end{align}$$
So we get
$$I(s,t)=\frac{i\sqrt{\pi}e^{-\frac{1}{4s}}}{2s^{\frac{3}{2}}}\operatorname{erf}\frac{2s\sqrt{t}-i}{2\sqrt{s}}-\frac{e^{i\sqrt{t}-st}}{s}+C.$$
Now, evaluate the limits:
$$\int_0^\infty e^{i\sqrt{t}-st}\, dt=\lim_{t\to\infty}I(s,t)-\lim_{t\to 0}I(s,t)$$
which is just
$$\frac{i\sqrt{\pi}e^{-\frac{1}{4s}}}{2s^{\frac{3}{2}}}+\frac{i\sqrt{\pi}e^{-\frac{1}{4s}}}{2s^{\frac{3}{2}}}\operatorname{erf}\frac{i}{2\sqrt{s}}+\frac{1}{s}.$$
Since $\operatorname{erf}\frac{i}{2\sqrt{s}}\in i\mathbb{R}$ (under the assumption that $s\gt 0$), taking the imaginary part of the above expression gives
$$\bbox[5px,border:2px solid red]{\mathcal{L}\{\sin\sqrt{t}\}(s)=\frac{\sqrt{\pi}e^{-\frac{1}{4s}}}{2s^{\frac{3}{2}}},\, s\gt 0.}$$
