It's rather unexpected

First we let

$F(a) = \displaystyle \int_0^{\infty} \int_0^{\infty} \dfrac{ \sin x \sin t \sin (a(x+t))}{xt(x+t)}$

Now, $\dfrac{ \mathrm{d}}{\mathrm{d}a} \sin (a(x+t)) = (x+t) \cos (a(x+t))$ . That will simplify $F'(a)$ into

$F'(a) = \displaystyle \int_0^{\infty} \int_0^{\infty} \dfrac{ \sin x \sin t ( \cos ax \cos at - \sin ax \sin at)}{xt}$

while this may seem a little more complicated, note that

$\displaystyle\int_0^{\infty} \dfrac{ \sin x \cos ax }{x} \mathrm{d}x = \frac{\pi}{2}$

we'll let this integral $I_A$ . Meanwhile, note too that

$\displaystyle\int_0^{\infty} \dfrac{ \sin x \sin ax }{x} \mathrm{d}x = \frac{1}{2} \ln \bigg( \frac{1-a}{1+a} \bigg)$

we'll let this integral $I_B$ .

This will further simplify $F'(a)$ .

$F'(a) = (I_A)^2 - (I_B)^2$

$F'(a) = \dfrac{\pi^2}{4} - \frac{1}{4} \bigg( \ln^2 \frac {1-a}{1+a}\bigg)$

integrating this from $0$ to $1$ , we will get

$F(1) = \dfrac{\pi^2}{4} - \int_0^1 \ln^2 \bigg( \frac{1-a}{1+a} \bigg) \mathrm{d}a$

Note again, that $\displaystyle \int_0^1 \ln^2 \bigg( \frac{1-x}{1+x} \bigg) \mathrm{d}x = \frac{\pi^2}{3}$ .

This will now let us find our answer.

$F(1) = \frac{ \pi^2 }{4} - \frac{1}{4} \cdot \frac{\pi^2}{3} = \boxed{ \frac{\pi^2}{6}}$

and there, that enables us to find $\lfloor 10^6 A \rfloor$ .