Compute laplace transform of $\frac{\cos\sqrt t}{\sqrt t}$? What is the Laplace transform of $\frac{\cos\sqrt t}{\sqrt t}$?
 A: I give you some hints along the way.
First let $u=\sqrt{t}$. This transforms your integral to
$$
\frac{1}{2}\int_0^{+\infty}e^{-su^2}\cos u\,du=\int_{-\infty}^{+\infty}e^{-su^2}\cos u\,du,
$$
where we in the second step used that the function is even. Now,
$$
\cos u=\frac{1}{2}\bigl(e^{iu}+e^{-iu}\bigr).
$$
We get
$$
\frac{1}{2}\int_{-\infty}^{+\infty} e^{-su^2}\bigl(e^{iu}+e^{-iu}\bigr)\,du
=\frac{1}{2}\Bigl(\mathcal{F}(e^{-su^2})(1)+\mathcal{F}(e^{-su^2})(-1)\Bigr)
$$
where $\mathcal F$ denotes the Fourier transform. But, you can find almost everywhere, the Fourier transform of the Gaussian (for example here),
$$
\mathcal{F}(e^{-su^2})(w)=\sqrt{\frac{\pi}{s}}e^{-w^2/(4s)}.
$$
I guess that is it.
A: We have
$$ \int_0^{\infty} e^{-st} \frac{\cos{\sqrt{t}}}{\sqrt{t}} \, dt. $$
Clearly we need $s>0$ for this to converge. Set $u^2=st$, and then $2/\sqrt{s} du = dt/\sqrt{t}$, so the integral becomes
$$ \frac{2}{\sqrt{s}}\int_0^{\infty} e^{-u^2} \cos{\left( \frac{u}{s} \right)} \, du. $$
To do this integral, define
$$ I(a) = 2\int_0^{\infty} e^{-u^2} \cos{au} \, du. $$
Then
$$ I'(a) = -\int_0^{\infty} 2ue^{-u^2} \sin{au} \, du = \left[ e^{-u^2}\sin{au} \right]_0^{\infty} - a \int_0^{\infty} e^{-u^2} \cos{au} \, du = 0-\frac{a}{2}I(a). $$
Hence $I(a)=I(0)e^{-a^2/4}$, by the usual method of integrating factors. It is well-known that $I(0)=2\int_0^{\infty} e^{-u^2} \, du = \sqrt{\pi} $, so the Laplace transform is
$$ \frac{1}{\sqrt{s}}I(1/s) = \frac{\sqrt{\pi}e^{-1/(4s)}}{\sqrt{s}} .$$
