Observe the fact that $$\frac{1}{x} = \int_0^{\infty} e^{-xy}dy.$$
Apply the fact above and Fubini Theorem with $$\int_0^b\int_0^\infty e^{-xy}\sin(x)dydx$$ to deduce $$\lim_{b \rightarrow \infty} \int_0^b \frac{\sin x}{x} = \frac{\pi}{2}.$$
My approach :
Since $f(x,y) = e^{-xy}\sin(x)$ is continuous on $\mathbb{R}^2$, $f$ is two-dimensional Lebesgue measurable.
We have $$\int_0^b\int_0^\infty |f| dydx \leq \int_0^b\int_0^\infty xe^{-xy}dydx = \int_0^b -e^{-xy}\Big|_{y=0}^{y = \infty} dx = b < \infty.$$
So $f$ is $(m \times m)$-integrable.
Now we can apply Fubini, $$\int_0^b \frac{\sin(x)}{x}dx = \int_0^\infty\int_0^b e^{-xy}\sin(x)dxdy= \int_0^\infty \frac{e^{-by}(y \sin(-b) - \cos(-b))}{1+y^2} dy + \int_0^\infty \frac{1}{y^2+1} dy.$$
So $$\lim_{b \rightarrow \infty} \int_0^b \frac{\sin(x)}{x}dx = \lim_{b \rightarrow \infty} \int_0^\infty \frac{e^{-by}(y \sin(-b) - \cos(-b))}{1+y^2} dy + \arctan(y)\Big|_{y=0}^{y=\infty}.$$ It turn out that I have to prove that $$\lim_{b \rightarrow \infty} \int_0^\infty \frac{e^{-by}(y \sin(-b) - \cos(-b))}{1+y^2} dy = 0.$$ I try to approximate it using $$\Big|\lim_{b \rightarrow \infty} \int_0^\infty \frac{e^{-by}(y \sin(-b) - \cos(-b))}{1+y^2} dy\Big| \leq \lim_{b \rightarrow \infty} \int_0^\infty \Big|\frac{e^{-by}(y \sin(-b) - \cos(-b))}{1+y^2} \Big|dy$$ but I cannot find a good function $g$ satisfying $$\lim_{b \rightarrow \infty} \int_0^\infty \Big|\frac{e^{-by}(y \sin(-b) - \cos(-b))}{1+y^2}\Big| dy \leq \lim_{b \rightarrow \infty} \int_0^\infty g(x) dy = 0.$$ Any suggestion about $g$ ? Or any suggestion to a more simpler way to show $$\lim_{b \rightarrow \infty} \int_0^\infty \frac{e^{-by}(y \sin(-b) - \cos(-b))}{1+y^2} dy = 0.$$
HINT:
For $b\ge 1$ and $y>0$, we have
$$e^{-by}\le \frac{1}{1+by}<\frac{1}{1+y}$$
and therefore
$$\left|\frac{e^{-by}\left(y\sin(b)+\cos (b)\right)}{1+y^2}\right|\le\frac{1+y}{(1+y)(1+y^2)}=\frac{1}{1+y^2}$$
which is integrable. Then, apply the Dominated Convergence Theorem.