Laplace transform. How to derive I have this integral related to a Laplace transform and I was wondering if anyone knows of a clever way to derive it. I know we usually look these up in a table, but this form is not in a table I have.
The transform is:
$$\mathcal{L}\left\{ \frac{\sin 1/t}{\sqrt{t}} \right\}(s)=\int\limits_{0}^{\infty}\frac{\sin(1/t)}{\sqrt{t}}e^{-st}\,\mathrm dt, \;\ s>0.$$
I ran it through Maple and it gave me $$\sqrt{\frac{\pi}{s}}\sin(\sqrt{2s})e^{-\sqrt{2s}}.$$
A rather simplistic looking solution, but how to derive?
I did notice that the $$\frac{1}{t^{\frac{1}{2}}}=\sqrt{\frac{\pi}{s}},$$ 
since the Laplace of $$t^{k}=\frac{\Gamma(k+1)}{s^{k+1}}, \;\ k>-1.$$  This would mean 
$$t^{\frac{-1}{2}}=\frac{\Gamma(\tfrac12)}{\sqrt{s}}=\sqrt{\frac{\pi}{s}}.$$
Perhaps it is too difficult to do by hand. I just thought it was interesting how to derive this if possible. 
Thanks very much for your time, interest, and expertise.
 A: The Key Lemma: For all $a\in \mathbb{C}$ with $\Re(a)>0$ we have $$ \int^{\infty}_0 \frac{1}{\sqrt{t} } e^{- a (t+ 1/t)^2 } dt = e^{-2a} \sqrt{ \frac{\pi}{a} } .$$
Proof: Let the integral be denoted by $I$ and $t=u^2$ so that $$ I = 2 \int^{\infty}_0 e^{-a (u^2+1/u^2) } du= 2 e^{-2a} \int^{\infty}_0 e^{-a (u-1/u)} du. $$
By letting $ u=1/t$ we see that $$ \int^{\infty}_0 \frac{1}{u^2} e^{-a (u-1/u)^2} du = \int^{\infty}_0 e^{-a (t-1/t)^2 } dt $$
so then $$ I = e^{-2a} \int^{\infty}_0 \left(1 + \frac{1}{u^2} \right) e^{-a (u-1/u)^2 } du. $$
Now letting $ x= u-1/u $ gives $$ I = e^{-2a} \int^{\infty}_{-\infty} e^{-ax^2} = e^{-2a} \sqrt{ \frac{\pi}{a} } .$$

Idea for the rest of the proof: (I will come back soon and complete it.)
Prove that $$\int^{\infty}_0 \frac{1}{\sqrt{t} } e^{-st + i/t } dt = \sqrt{\frac{\pi}{s}} e^{-(1-i)\sqrt{2s} } dt $$
from which the result follows by taking imaginary parts of both sides. To do this, let $ t= \frac{1-i}{\sqrt{2s} } x$ (This "substitution" can be made valid by integrating along a suitable sector in the complex plane). Then the resulting integral becomes of the form of our lemma (up to a constant multiple) and then we are basically done.
A: Let's set $t:=x^2$ then :
$$I=2\int_0^{\infty} \sin(\frac 1{x^2}) e^{-sx^2}dx=i\int_0^{\infty} \left(e^{-\frac i{x^2}}-e^{\frac i{x^2}}\right) e^{-sx^2}dx$$
At this point you may use following formula :
$$J(a,s)=\int_0^{\infty} e^{-\frac a{x^2}-s x^2} dx=\sqrt{\frac{\pi}{4s}}e^{-2\sqrt{as}}$$
(right and left terms both verify $J(a,s)=-\frac{\partial J(s,a)}{\partial s}$, $\frac{\partial^2 J}{\partial a \partial s}=J$ and $J(0,s)=\sqrt{\frac{\pi}{4s}}$, this formula is usually provided with $a\gt 0$, $s\gt 0$ but may be extended by analytic continuation)
For a more serious justification of this formula of P. Laplace you may search it in Gradshteyn and Ryzhik's famous "Table of integrals, series and products" getting (3.325) and use Moll and other's heroic effort to justify them all! In this case see the chapter 5 of this paper of Albano, Amdeberhan, Beyerstedt and Moll.
so that we get :
$$I=i\sqrt{\frac{\pi}{4s}}\left(e^{-2\sqrt{is}}-e^{-2\sqrt{-is}}\right)$$
Use $\sqrt{i}=\frac{1+i}2$ and $\sqrt{-i}=\frac{1-i}2$ to get :
$$I=i\sqrt{\frac{\pi}{4s}}\left(e^{-(1+i)\sqrt{2s}}-e^{-(1-i)\sqrt{2s}}\right)$$
$$I=\sqrt{\frac{\pi}{s}}\sin(\sqrt{2s})e^{-\sqrt{2s}}$$
Simple looking things are sometimes not so simple and some of the previous steps  are not so easy to justify!
(if you remove the $\sqrt{t}$ at the denominator or put a $t$ or something else there you won't get a simple answer!)
