$$ $$

$ \text{ Let I(a) = } \int_0^\infty e^{-bx^2} cos2ax dx \\ $

$ \frac{dI}{da} = \int_0^\infty \frac{\partial}{\partial a} e^{-bx^2} cos2ax dx \\ $

$ = \int_0^\infty e^{-bx^2} 2x (-sin2ax) dx \\ $

$ = \int_0^\infty ( e^{-bx^2}(-2x) ) sin2ax dx \\ $

$ \text{Consider } \int_0^\infty e^{-bx^2} (-2x) dx \\ $

$ \text {put } -bx^2 = t \Longrightarrow -2bxdx = dt \\ $

$ \int_0^\infty e^t \frac{dt}{b} = \frac{1}{b} e^ {-bx^2} \\ $

$ \text {Using integration by parts in (1) } \\ $

$ \frac{dI}{da} =\left ( sin2ax \int e^{-bx^2} (-2x) \right )_0^\infty - \int_0^\infty 2a cos2ax \int e^{-bx^2} (-2x) dx \\ $

$ = \left ( sin2ax * \frac{1}{b} e^{-b x^2} \right )_0^\infty - \int_0^\infty 2a cos2ax \frac { e^{-bx^2}}{b} dx \\ $

$ = - \frac{2a}{b} I \\ $

$ \therefore \frac{dI}{I} = \frac{-2a}{b} da \\ $

$ \text{ Integrating both sides } \\ $

$ log I = \frac{-a^2}{b} + c \\ $

$ I = e^{\frac{-a^2}{b} + c } \\ $

$ \text { put } a = 0, I(0) = e^c \\ $

$ I (0) = \int_0^\infty e^{-bx^2} dx \\ $

$ bx^2 = t \Longrightarrow x^2 = \frac{t}{b} \Longrightarrow x = \frac{t^{1}{2}}{\sqrt b} \Longrightarrow dx = \frac{\frac{1}{2}t^\frac{-1}{2}}{\sqrt b} dt \\ $

$ =\int_0^\infty e^{-t} \frac{1}{2} \frac{ t^{-1}{2}} { \sqrt b } dt \\ $

$ = \frac {1}{2 \sqrt b } \int_0^\infty e^{-t} t^ \frac{-1}{2} dt \\ $

$ = \frac{\Gamma \frac{1}{2} } {2\sqrt b} \\ $

$ = \frac {\sqrt (\pi) } { 2 \sqrt b } \\ $

$ e^c = \frac {\sqrt (\pi) } { 2 \sqrt b } \\ $

$ I = e^\frac{-a^2}{b} * e^c = e^\frac{-a^2} {b} \frac {\sqrt (\pi) } { 2 \sqrt b } \\ $