Prove $F(t)^2-G(t)^2=2tF(t)G(t)$

Question

Let $F, G:]0,\infty[ \to \mathbb R$ be $$F(t):=\int_{[0,\infty[}e^{-tx^2}\cos{x^2}d\lambda(x)$$ and $$G(t):=\int_{[0,\infty[}e^{-tx^2}\sin{x^2}d\lambda(x)\,.$$ Prove that
$$F(t)^2-G(t)^2=2tF(t)G(t)\,.$$

Note:

$F(t)^2-G(t)^2=(\int_{[0,\infty[}e^{-tx^2}\cos{x^2}d\lambda(x))^2-(\int_{[0,\infty[}e^{-tx^2}\sin{x^2}d\lambda(x))^2$

Specifically looking at

$(\int_{[0,\infty[}e^{-tx^2}\cos{x^2}d\lambda(x))^2=\int_{[0,\infty[}e^{-tx^2}\cos{x^2}d\lambda(x)\times \int_{[0,\infty[}e^{-tx^2}\cos{x^2}d\lambda(x)$

am I allowed to state: $\int_{[0,\infty[}e^{-tx^2}\cos{x^2}d\lambda(x)\times \int_{[0,\infty[}e^{-tx^2}\cos{x^2}d\lambda(x)=\int_{[0,\infty[^2}e^{-tx^2}\cos{x^2}d\lambda^{2}(x)$? And then I assume I should substitute $y = x^2$

$\int_{[0,\infty[}2x(\int_{[0,\infty[}e^{-ty}\cos{y}dy)dx$

I can integrate $e^{-y}\cos{y}$ tediously by parts, however, I am not sure what to do with the parameter $t$ in $e^{-ty}\cos{y}$, as in the above case.

Any support is greatly appreciated.

Answer 1

You can compute $F$ and $G$ directly, since $$ H(t) = F(t) + i G(t) = \int_0^\infty e^{-tx^2} \left( \cos x^2 + i \sin x^2 \right) dx = \int_0^\infty e^{(i - t)x^2} \, dx = \frac{1}{\sqrt{t - i}} \frac{\sqrt{\pi}}{2} \, . $$ Then $$ F^2(t) - G^2(t) = \Re H^2(t) = \Re \frac{\pi}{4(t-i)} = \frac{\pi}{4} \frac{t}{t^2 + 1} $$ and $$ 2 F(t) G(t) = \Im H^2(t) = \Im \frac{\pi}{4(t-i)} = \frac{\pi}{4} \frac{1}{t^2 + 1} \, . $$ It follows that $F^2(t) - G^2(t) = 2tF(t)G(t)$.