integration - $lim_{A to infty} int_0^{A} int_0^{infty} sin(x) e^{-xt}dtdx$

I would like to compute the following integral:

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty} \sin(x) e^{-xt}dtdx \qquad (1)$$

I would like to swap the order of integration because then the integral becomes easier (I could evaluate it by taking its imaginary part as was done Here or i could do a partial integration two times to get that:

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty} \sin(x) e^{-xt}dxdt = \lim_{A \to \infty}\int_0^A\dfrac{1}{1+t^2}dt=\frac{\pi}{2}$$

In order to do that i need to apply Fubini-Lebesgue, because the function $\sin(x)e^{-xt}$ takes also negative values, but i can't show that

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty} |\sin(x)| e^{-xt}dtdx < \infty$$

I tried to approximate the function in this way:

$$|\sin(x)e^{-xt}|\le e^{-xt}$$

but

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty}e^{-xt}dtdx = \infty$$

How can i use Fubini-Lebesuge to evaluate $(1)$? Any hint would be really appreciated, thank you!

I also tried to bound

$$\lim_{A \to \infty} \int_0^{A} \int_0^{\infty} |\sin(x)| e^{-xt}dtdx \le \lim_{A\to \infty}A$$

but $\lim_{A\to \infty} = \infty$ hence again this argument does not work

Answer

How about, since $x\geq 0$ the following holds for all $t\geq 0$,

$$|\sin(x)e^{-tx}|\leq xe^{-tx}$$
then,
$$\int_0^A\int_0^{\infty}xe^{-tx}dtdx=\int_0^Ax\cdot\frac{1}{x}dx=A$$