find the Laplace transform of cos(sqrt(t))/sqrt(t) $$ L\{ \frac{cos(\sqrt{t})}{\sqrt{t}}\}  $$
Laplace-transform
 A: using the laplace transform
$$
\int^{\infty}_{0}\mathrm{e}^{-st}f(t)\mathrm{dt} = \int^{\infty}_{0}\mathrm{e}^{-st}\frac{\mathrm{e}^{i\sqrt{t}}+\mathrm{e}^{-i\sqrt{t}}}{2\sqrt{t}}\mathrm{dt}
$$
Here I used the exponential form of cosine.
Changing variables to $v=\sqrt{t}$, we transform the integral to 
$$
\int^{\infty}_{0}\mathrm{e}^{-sv^{2}}\left(\mathrm{e}^{iv}+\mathrm{e}^{-iv}\right)\mathrm{dv}
$$
Now we can rearrange the integral and using completeing the square to yield
$$
\mathrm{e}^{-\frac{1}{4s}}\int^{\infty}_{0}\mathrm{e}^{-s\left(v-\frac{i}{2s}\right)^{2}}+\mathrm{e}^{-s\left(v+\frac{i}{2s}\right)^{2}}\mathrm{dv}
$$
Now, this next step is more of a trick i.e. not the most robust step you will ever see! but changing the variables once again $x = \sqrt{s}\left(v-\frac{i}{2s}\right)$ and $y = \sqrt{s}\left(v+\frac{i}{2s}\right)$
$$
\frac{1}{\sqrt{s}}\mathrm{e}^{-\frac{1}{4s}}\left[\int^{\infty}_{0}\mathrm{e}^{-x^{2}}\mathrm{dx}+\int^{\infty}_{0}\mathrm{e}^{-y^{2}}\mathrm{dy}\right]
$$
Both upper limits hold here as well. So now you can see that we have the standard normal integrals which is equal to $\sqrt{\pi}$, but in this case we only integrated the positive domain, which leads both these integrals to equate to $\frac{\sqrt{\pi}}{2}$.
Subbing in the results we finally arrive at the above solution from maple as
$$
\sqrt{\frac{\pi}{s}}\mathrm{e}^{-\frac{1}{4s}}.
$$
(We could of done this with more rigour by using complex analysis, but for this case it is easily shown by these steps)
Rob
