Proving that $\int_0^\infty e^{-a^2x^2} \cos (bx) dx = \frac {\sqrt \pi}{2a}e^{\frac{-b^2}{4a^2}}$ Any clue regarding this:
$$\int_0^\infty e^{-a^2x^2} \cos (bx)  dx = \frac {\sqrt \pi}{2a}e^{\frac{-b^2}{4a^2}}$$

Let 
$$I(b) = \int^\infty_0 e^{-a^2 x^2} \cos (b x) \, dx, \quad a > 0.$$
Integrating by parts gives
$$I(b) = \frac{2a^2}{b} \int^\infty_0 x e^{-a^2 x^2} \sin (bx) \, dx.$$
Also, differentiating $I(b)$ with respect to the parameter $b$ we have
$$I'(b) = - \int^\infty_0 x e^{-a^2 x^2} \sin (bx) \, dx = - \frac{b}{2a^2} I(b).$$
Solving this first-order differential equation yields
$$I(b) = K e^{-\frac{b^2}{4a^2}},$$
where $K$ is a constant to be determined. To find this constant, setting $b = 0$ one has
$$I(0) = \int^\infty_0 e^{-(ax)^2} \, dx = \frac{\sqrt{\pi}}{2a} \cdot \frac{2}{\sqrt{\pi}} \int^\infty_0 e^{-u^2} \, du = \frac{\sqrt{\pi}}{2a} \cdot \text{erf} (\infty) = \frac{\sqrt{\pi}}{2a}.$$
So $K = \sqrt{\pi}/(2a)$ and we have
$$I(b) = \int^\infty_0 e^{-a^2 x^2} \cos (b x) \, dx = \frac{\sqrt{\pi}}{2a} e^{-\frac{b^2}{4a^2}}, \quad a > 0,$$
as required.
