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.

Note that we can complete the square and get the following integral:
$$\frac12 \Re{\left [\int_{-\infty}^{\infty} ds \, e^{-a^2 s^2} \, e^{i 2 b s} \right ]} = \frac12 e^{-b^2/a^2} \Re{\left [\int_{-\infty}^{\infty} ds \, e^{-a^2 (s-i b/a^2)^2} \right ]}$$
We can prove that the integral on the RHS is simply $\sqrt{\pi}/a$ as you may expect (and hope) by using Cauchy's theorem.  Consider the integral
$$\oint_C dz \, e^{-a z^2}$$
where $C$ is the rectangle in the complex plane having vertices at $-R-i b/a^2$, $R-i b/a^2$, $R$, and $-R$ (in that order).  This contour integral is equal to
$$\int_{-R}^R dx \, e^{-a^2 (x-i b/a^2)^2} + i \int_{-b/a^2}^0 dy \, e^{-a^2 (R+i y)^2}\\+\int_R^{-R} dx \, e^{-a^2 x^2} + i \int_0^{-b/a^2} dy \, e^{-a^2 (-R+i y)^2}$$
By Cauchy's theorem, the contour integral is zero.  On the other hand, in the limit as $R\to\infty$, the second and fourth integrals above vanish.  Thus,
$$\int_{-\infty}^{\infty} dx \, e^{-a^2 (x-i b/a^2)^2} = \int_{-\infty}^{\infty} dx \, e^{-a^2 x^2} = \frac{\sqrt{\pi}}{a}$$
The result follows.
