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.

Try this standard trick:
$$
I(a,b)^2=\left[\int_0^\infty e^{-a^2 x^2} \cos(2 b x) \,\mathrm dx\right]\,\left[\int_0^\infty e^{-a^2 y^2} \cos(2 b y) \,\mathrm dy\right]
$$
Use Eulers formula to convert this all to an exponential integration and use polar coordinates in the double integral
$$
I(a,b)^2= \int_0^\infty \int_0^\infty  e^{-a^2 (x^2+y^2)} \cos(2 b x) \cos(2 b y) \,\mathrm dx\,\mathrm dy$$
