Cauchy-Goursat II

int{0}{infty}{e^{-x^2}cos 2bx dx}={sqrt{pi}/2}e^{-b^2}

In order to derive the integration formula in question, we integrate the function e^{-z^2} around the closed rectangular path shown in the figure

C_1 ({-}a right a)

z=x

e^{-z^2}=e^{-x^2}

int{-a}{a}{e^{-x^2}dx}=2int{0}{a}{e^{-x^2}dx}

C_2 (a right -a)

z=x+ib

z^2=x^2+2ixb-b^2

int{a}{-a}{e^{-(x+ib)^2}dx}=-int{-a}{a}{e^{-(x+ib)^2}dx}

{=} -int{-a}{a}{e^{-x^2}e^{b^2}(cos2xb-isin2xb)dx}

{=}-{e^{b^2}}int{-a}{a}{e^{-x^2}(cos2xb)dx}+{ie^{b^2}}int{-a}{a}{e^{-x^2}(sin2xb)dx}

C_3 (0<=y<=b)

z=a+iy

dz=idy

int{0}{b}{e^{-(a+iy)^2}idy}=int{0}{b}{e^{-a^2}e^{-2ayi}e^{y^2}idy}

{=}{e^{-a^2}}int{0}{b}{e^{y^2}(icos2ay+sin2ay)dy}

{=}{ie^{-a^2}}int{0}{b}{e^{y^2}(cos2ay)dy}+{e^{-a^2}}int{0}{b}{e^{y^2}(sin2ay)dy}

C_4 (b<=y<=0)

z=-a+iy

dz=idy

int{b}{0}{e^{-(-a+iy)^2}idy}={-}int{0}{b}{e^{-a^2}e^{2ayi}e^{y^2}idy}

{=}{-ie^{-a^2}}int{0}{b}{e^{y^2}(cos2ay+isin2ay)dy}

{=}{-ie^{-a^2}}int{0}{b}{e^{y^2}(cos2ay)dy}+{e^{-a^2}}int{0}{b}{e^{y^2}(sin2ay)dy}

According to the Cauchy-Goursat theorem, then

C_1+C_2+C_3+C_4=0

{2e^{b^2}}int{0}{a}{e^{-x^2}(cos2xb)dx}=2int{0}{a}{e^{-x^2}dx}+{2e^{-a^2}}int{0}{b}{e^{y^2}(sin2ay)dy}

int{0}{a}{e^{-x^2}(cos2xb)dx}={e^{-b^2}}int{0}{a}{e^{-x^2}dx}+{e^{-(a^2+b^2)}}int{0}{b}{e^{y^2}(sin2ay)dy}

We now let a right infty, using the known integration formula

int{0}{infty}{e^{-x^2}dx}=sqrt{pi}/2

as e^{-(a^2+b^2)} right 0

int{o}{infty}{e^{-x^2}cos2xbdx}={e^{-b^2}}{sqrt{pi}/2}