Switching Integral Order A way to prove $\int_0^\infty \sin(x) / x dx = \frac{\pi}{2}$ is to switch integrals as in
$$\tag{1}\int_0^\infty \int_0^\infty e^{-xy} \sin(x) dx dy= \int_0^\infty \int_0^\infty e^{-xy} \sin(x) dy dx $$
because $\int_0^\infty e^{-xy} \sin(x) dy = \sin(x)/x$ and twice integration by parts gives $\int e^{-xy}\sin(x) dx = C - e^{-xy}(y \sin(x) + \cos(x))/(1 + y^2)$, etc.
My question here is how to justify switching the integrals in equation 1 without computing?
Fubini theorem is not helpful since $e^{-xy}\sin(x)$ is not nonnegative and not absolutely integrable on $\mathbb{R}^2$.
 A: The integrand is definitely not absolutely integrable.  You can justify changing the order of integration by a combination of uniform and dominated convergence.
Note that by the Weierstrass test, the improper integral
$$\tag{1} F(x) = \int_0^\infty e^{-xy} \sin x \, dy$$
is uniformly convergent for $x$ in any compact interval.
Take sequences $a_n,c_m \to 0$  and $b_n,d_m \to +\infty$. Since the integrand is continuous the interchange is permitted on the bounded rectangle:
$$\int_{a_n}^{b_n} \int_{c_m}^{d_m}e^{-xy} \sin x \, dx  \, dy=  \int_{c_m}^{d_m} \int_{a_n}^{b_n}e^{-xy} \sin x \, dy  \, dx.$$
By the uniform convergence of the improper integral (1),  it follows that
$$\tag{2}\int_{0}^{\infty} \int_{c_m}^{d_m}e^{-xy} \sin x \, dx  \, dy=  \lim_{n \to \infty}\int_{c_m}^{d_m} \int_{a_n}^{b_n}e^{-xy} \sin x \, dy  \, dx \\ = \int_{c_m}^{d_m} \lim_{n \to \infty}\int_{a_n}^{b_n}e^{-xy} \sin x \, dy  \, dx \\ =  \int_{c_m}^{d_m} \int_{0}^{\infty}e^{-xy} \sin x \, dy  \, dx. $$
If we  can show that the limit of the LHS of (2) as $m \to \infty$ can be passed under the integral then we are done.
Note that
$$\left|\int_{c_m}^{d_m} e^{-xy} \sin x \, dx \right|  = \left| \left.\frac{-e^{-xy} (y\sin x + \cos x)}{1+y^2}\right|_{x= c_m}^{x = d_m} \right| \leqslant \frac{C}{1+y^2}.$$
This bound holds because
$$\left| \frac{e^{-xy} \cos x}{1+y^2}\right| = \frac{e^{-xy}|\cos x|}{1 +y^2} \leqslant \frac{1}{1+y^2},$$
and $f(x) = ye^{-xy} \sin x $ has a maximum for $x \in [0,\infty)$ at $x^{*} = \arctan(1/y)$ where 
$$f(x^{*}) = y e^{-y \arctan(1/y)} \sin \arctan(1/y) = ye^{-y \arctan(1/y)}\frac{1/y}{\sqrt{1 + (1/y)^2}} \leqslant 1 $$
Since $1/(1+y^2)$ is integrable we can apply DCT in taking the limit of both sides of (2) to obtain the desired result.
