let $t>0$. consider the functions
$$F(t)=\int_0^{\infty} e^{-tx^2}cos(x^2)\, dx,\quad G(t)=\int_0^{\infty} e^{-tx^2}sin(x^2)\, dx.$$
i want to show that
$$F(t)^2-G(t)^2=\frac{\pi}{4}\frac{t}{1+t^2}=2tF(t)G(t).$$
as regards the first equality, i get
\begin{split} F(t)^2-G(t)^2 & =\int_0^{\infty} e^{-tx^2}cos(x^2)\, dx \int_0^{\infty} e^{-ty^2}cos(y^2)\, dy - \\ & \quad \int_0^{\infty} e^{-tx^2}sin(x^2)\, dx \int_0^{\infty} e^{-ty^2}sin(y^2)\, dy \\ & = \iint_0^{\infty} e^{-t(x^2+y^2)} \left(cos(x^2)cos(y^2) -sin(x^2)sin(y^2)\right)\, d\lambda(x,y) \\ & = \iint_0^{\infty} e^{-t(x^2+y^2)} cos(x^2+y^2)\, d\lambda(x,y) \\ & = \int_{]0,\infty[ \times ]0,\pi/2[} r e^{-tr^2}cos(r^2)\, d\lambda(r,\phi) \\ & = \frac{\pi}{2} \int_0^{\infty}r e^{-tr^2}cos(r^2)\, dr \\ & = \frac{\pi}{4} \frac{t}{t^2+1}. \end{split}
however, i don't know how to show the second equality. neither am i able to explicitly evaluate $F(t)$ nor can i proceed with
$$F(t)G(t)=\iint_0^{\infty}e^{-t(x^2+y^2)}cos(x^2)sin(y^2)\, d\lambda(x,y).$$
can i somehow get rid of this trigonometric stuff? merci!
Hint:
Use the fact that $2\cos\alpha\sin\beta=\sin(\alpha+\beta)-\sin(\alpha-\beta)$, and thus break the integral into two parts. One of them is zero; the other is easy.