Are all conditions met for Fubini? Needed to work out $\lim_{R\to \infty}\int_{[0,R]}\frac{\sin{x}}{x}d \lambda (x)$ As a background I have proven that $\int_{[0,\infty[}e^{-xt}dt=\frac{1}{x}$
I am required to use Fubini's Theorem to calculate:
$\lim_{R\to \infty}\int_{[0,R]}\frac{\sin{x}}{x}d \lambda (x)$
I believe I know the manner in which to do this, but I believe I do not have the correct conditions to apply Fubini's theorem. 
My ideas: 
$\lim_{R\to \infty}\int_{[0,R]}\frac{\sin{x}}{x}d \lambda (x)=\lim_{R\to \infty}\int_{[0,R]}\sin{x}\int_{[0,\infty[}e^{-xt}d\lambda (t)d \lambda (x)$
$=\lim_{R\to \infty}\int_{[0,R]}\int_{[0,\infty[}\sin{x}e^{-xt}d\lambda (t)d \lambda (x)$
and then if I were able to apply Fubini:
$=\int_{[0,\infty[} \lim_{R\to \infty}\int_{[0,R]}\sin{x}e^{-xt}d\lambda (x)d \lambda (t)$
and since $[0,R]$ is a compact interval: Riemann-Integral = Lebesgue integral, therefore 
$\int_{[0,\infty[} (\lim_{R\to \infty}\int_{[0,R]}\sin{x}e^{-xt}dx)d \lambda (t)$ etc. 
as previously stated, my biggest problem is proving the condition necessary for Fubini's theorem, which would mean: 
$\lim_{R\to \infty}\int_{[0,R]}\int_{[0,\infty[}|\sin{x}e^{-xt}|d\lambda (t)d \lambda (x)<\infty$, or $\lim_{R\to \infty}\int_{[0,R]}|\frac{\sin{x}}{x}|d \lambda (x)<\infty$
and I cannot prove any of those. 
 A: Unfortunately the function $(x,t) \mapsto e^{-xt} \sin x$ is not absolutely integrable and Fubini's theorem does not apply.  If it were then were then we could say
$$\int_0^\infty \int_0^\infty e^{-xt} | \sin x| \, dx \, dt = \int_0^\infty \int_0^\infty e^{-xt} | \sin x| \, dt \, dx $$
However,
$$\int_0^\infty e^{-xt} | \sin x| \, dt= \frac{|\sin x|}{x} $$
and this is neither Lebesgue nor improperly Riemann integrable over  $[0,\infty]$.
You can justify
$$\int_0^\infty \int_0^\infty e^{-xt}  \sin x \, dx \, dt = \int_0^\infty \int_0^\infty e^{-xt}  \sin x \, dt \, dx $$
in terms of iterated (conditionally convergent) improper integrals by a combination of uniform and dominated convergence.  See this answer for details.
A: Though $(x,t)\in (0,\infty)^2 \mapsto e^{-xt} \sin x$ is not absolutely integrable on the whole domain, we know that it is integrable on $[0,R]\times (0,\infty)$. Thus Fubini's theorem implies that for all $R>0$, it holds that
$$\begin{eqnarray}
\int_0^R\frac{\sin{x}}{x}d x &=& \int_0^R\sin{x}\int_0^\infty e^{-xt}dtdx\\
&=&\int_0^\infty\left(\int_0^R\sin{x} e^{-xt}dx\right)dt\\
&=&\int_0^\infty\frac{1-t\sin (R)e^{-Rt}-\cos (R)e^{-Rt}}{1+t^2}dt\\
&=&\frac{\pi}{2} - \int_0^\infty\frac{(t\sin (R)+\cos (R))e^{-Rt}}{1+t^2}dt.
\end{eqnarray}$$ Note that the error term
$$
|\int_0^\infty\frac{(t\sin (R)+\cos (R))e^{-Rt}}{1+t^2}dt|\leq C\int_0^\infty e^{-Rt}dt\to 0
$$as $R\to \infty.$ Thus by taking $R\to \infty$, we get
$$
\lim_{R\to\infty}\int_0^R\frac{\sin{x}}{x}d x =\frac{\pi}{2}.
$$
