Prove that $\int_0^\infty e^{-a^2 s^2} \cos(2 b s) \,\mathrm ds=\frac{\sqrt{\pi}}{2a}e^{-b^2/a^2} $ Prove that 
$$I(a,b)=\int_0^\infty e^{-a^2 s^2} \cos(2 b s) \,\mathrm ds=\frac{\sqrt{\pi}}{2a}e^{-b^2/a^2}\quad a>0 $$
I can prove it by differential-equations technique(taking the derivative with respect to $b$ to become a first-order equation) but I wanna prove it directly.

Since the integrand is an even function in $s$, then we can write the integral as

$$ I(a,b)=\int_0^\infty e^{-a^2 s^2} \cos(2 b s) \,\mathrm ds = \frac{1}{2}\int_{-\infty}^\infty e^{-a^2 s^2} \cos(2 b s) \,\mathrm ds.$$

Now, Follow the steps 
i) use the identity $ \cos(t)=\frac{e^{it}+e^{-it}}{2}.$
ii) complete the square of the argument of the exponential.
iii) use Gaussian integrals.
