TITLE: Laplace transform. How to derive
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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.