Using Fubini's theorem for Riemann integrals I'm wondering what the argument is for reversing the order of integration of those Riemann integrals:

$$ \int_{0}^{\infty}\int_{0}^{\infty}e^{-xt}\sin{x} dt dx = \int_{0}^{\infty}\int_{0}^{\infty}e^{-xt}\sin{x} dx dt$$

Is there somehow a way of using Fubini's theorem for Lebesgue integrals?
EDIT: Because $e^{-xt}\sin{x}\notin L^1([0,\infty)^2)$ as Lorenzo Q. has shown, you can't use Fubini's theorem. But is there another way of showing the equality above? 
 A: As mentioned, the integrand is not absolutely integrable.  However, we can justify changing the order of integration by a combination of uniform and dominated convergence.
Since the integrand is continuous the interchange is permitted on the bounded rectangle:
$$\int_{a}^{b} \left(\int_{c}^{d}e^{-xt} \sin x \, dx\right)  \, dt=  \int_{c}^{d} \left(\int_{a}^{b}e^{-xt} \sin x \, dt \right) \, dx.$$
The inner integral on the RHS is uniformly convergent for $x \in [c,d]$ by the Weierstrass test as $a \to 0$ and $b \to \infty$, and   it follows that
$$\tag{*}\int_{0}^{\infty} \left(\int_{c}^{d}e^{-xt} \sin x \, dx \right) \, dt=  \lim_{a \to 0, \, b\to \infty}\int_{c}^{d} \left(\int_{a}^{b}e^{-xt} \sin x \, dy\right)  \, dx \\ = \int_{c}^{d} \lim_{a \to 0, \, b \to \infty}\left(\int_{a}^{b}e^{-xt} \sin x \, dt \right) \, dx \\ =  \int_{c}^{d} \left(\int_{0}^{\infty}e^{-xt} \sin x \, dt\right)  \, dx. $$
If we  can show that the limit of the LHS of (*) as $c \to 0$ and $d \to \infty$ can be passed under the integral then we are done.
Note that, using integration by parts,
$$\left|\int_{c}^{d} e^{-xt} \sin x \, dx \right|  = \left| \frac{-e^{-xt} (t\sin x + \cos x)}{1+t^2}\right|_{x= c}^{x = d}\\ \leqslant \left|\frac{e^{-ct} \cos c}{1+t^2}\right| + \left|\frac{e^{-ct} t\sin c}{1+t^2}\right| + \left|\frac{e^{-dt} \cos d}{1+t^2}\right| + \left|\frac{e^{-dt} t\sin d}{1+t^2}\right| \\ \leqslant \frac{A}{1+t^2},$$
where $A$ is a constant.
This bound holds because
$$\left| \frac{e^{-xt} \cos x}{1+t^2}\right| = \frac{e^{-xt}|\cos x|}{1 +t^2} \leqslant \frac{1}{1+t^2},$$
and $f(x) = te^{-xt} \sin x $ has a maximum for $x \in [0,\infty)$ at $x^{*} = \arctan(1/t)$ where 
$$f(x^{*}) = t e^{-t \arctan(1/t)} \sin \arctan(1/t) = te^{-t \arctan(1/t)}\frac {1/t}{\sqrt{1 + (1/t)^2}} \leqslant 1 $$
Since $1/(1+t^2)$ is integrable we can apply DCT in taking the limit of both sides of (*) to obtain 
$$\int_{0}^{\infty} \left(\int_{0}^{\infty}e^{-xt} \sin x \, dx\right)  \, dt=  \int_{0}^{\infty} \left(\int_{0}^{\infty}e^{-xt} \sin x \, dt \right) \, dx.$$
A: Fubini's theorem cannot be applied because the integral isn't absolutely convergent:
$$\int_0^{\infty}\int_0^{\infty}|e^{-xt}\sin x |dtdx\geq \int_0^{\infty}\left|\int_0^{\infty}e^{-xt}\sin xdt\right|dx = \int_0^{\infty}\left|\frac{\sin x}{x}\right|dx= \infty$$
Either way, you can manually check that the integrals are equal:
$$\int_0^{\infty}\left(\int_0^{\infty}e^{-xt}\sin x dt\right)dx=\int_0^{\infty}\frac{\sin x}{x}dx= \frac{\pi}{2} $$
\begin{gather*}\int_0^{\infty}\left(\int_0^{\infty}e^{-xt}\sin x dx\right)dt=\int_0^{\infty}-\left[\frac{e^{-tx}(t\sin x+\cos x)}{t^2+1}\right]_0^{\infty}dt=\int_0^{\infty}\frac{1}{t^2+1}dt=\\ 
=\left[\arctan t\right]_0^{\infty}=\frac{\pi}{2} \end{gather*}
