Theorem. $\int_0^\infty \sin x \phantom. dx/x = \pi/2$.

Poof. For $x>0$ write $1/x = \int_0^\infty e^{-xt} \phantom. dt$, and deduce that $\int_0^\infty \sin x \phantom. dx/x$ is $$ \int_0^\infty \sin x \int_0^\infty e^{-xt} \phantom. dt \phantom. dx = \int_0^\infty \left( \int_0^\infty e^{-tx} \sin x \phantom. dx \right) \phantom. dt = \int_0^\infty \frac{dt}{t^2+1}, $$ which is the arctangent integral for $\pi/2$, QED.

The theorem is correct, and usually obtained as an application of contour integration, or of Fourier inversion ($\sin x / x$ is a multiple of the Fourier transform of the characteristic function of an interval). The poof, which is the first one I saw (given in a footnote in an introductory textbook on quantum physics), is not correct, because the integral does not converge absolutely. One can rescue it by writing $\int_0^M \sin x \phantom. dx/x$ as a double integral in the same way, obtaining $$ \int_0^M \sin x \frac{dx}{x} = \int_0^\infty \frac{dt}{t^2+1} - \int_0^\infty e^{-Mt} (\cos M + t \cdot \sin M) \frac{dt}{t^2+1} $$ and showing that the second integral approaches $0$ as $M \rightarrow \infty$; but this detour makes for a much less appealing alternative to the usual proof by complex or Fourier analysis.

Still the double-integral trick can be used legitimately to evaluate $\int_0^\infty \sin^m x \phantom. dx/x^n$ for integers $m,n$ such that the integral converges absolutely (that is, with $2 \leq n \leq m$; NB unlike the contour or Fourier approach this technique applies also when $m \not\equiv n \bmod 2$). Write $(n-1)!/x^n = \int_0^\infty t^{n-1} e^{-xt} \phantom. dt$ to obtain $$ \int_0^\infty \sin^m x \frac{dx}{x^n} = \frac1{(n-1)!} \int_0^\infty t^{n-1} \left( \int_0^\infty e^{-tx} \sin^m x \phantom. dx \right) \phantom. dt, $$ in which the inner integral is a rational function of $t$, and then the integral with respect to $t$ is elementary. For example, when $m=n=2$ we find $$ \int_0^\infty \sin^2 x \frac{dx}{x^2} = \int_0^\infty t \frac2{t^3+4t} dt = 2 \int_0^\infty \frac{dt}{t^2+4} = \frac\pi2. $$ As a bonus, we recover a correct proof of our starting theorem by integration by parts:

$$ \frac\pi2 = \int_0^\infty \sin^2 x \frac{dx}{x^2} = \int_0^\infty \sin^2 x \phantom. d(-1/x) = \int_0^\infty \frac1x d(\sin^2 x) = \int_0^\infty 2 \sin x \cos x \frac{dx}{x}; $$ since $2 \sin x \cos x = \sin 2x$, the desired $\int_0^\infty \sin x \phantom. dx/x = \pi/2$ follows by a linear change of variable.

Exercise Use this technique to prove that $\int_0^\infty \sin^3 x \phantom. dx/x^2 = \frac34 \log 3$, and more generally $$ \int_0^\infty \sin^3 x \frac{dx}{x^\nu} = \frac{3-3^{\nu-1}}{4} \cos \frac{\nu\pi}{2} \Gamma(1-\nu) $$ when the integral converges. [Both are in Gradshteyn and Ryzhik, page 449, formula 3.827; the $\nu=2$ case is 3.827#3, credited to D. Bierens de Haan, Nouvelles tables d'intégrales définies, Amsterdam 1867; the general case is 3.827#1, from Gröbner and Hofreiter's Integraltafel II, Springer: Vienna and Innsbruck 1958.]