Too busy these days!

In my calculus classes, I do this as follows: $\int_{0}^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}\int_{0}^{\infty} e^{-xy}\sin x {dy}{dx} = \int_{0}^{\infty}\int_{0}^{\infty} e^{-xy}\sin x dx dy=\int_{0}^{\infty}\frac{1}{1+y^2}dy=\frac{\pi}{2}$ At the second equality, we use an extension of Fubini's theorem that you can find in a good calculus text; the other steps are basic calculus.

Hint: Call the integral $I(a)=\int_{0}^{\infty} \exp(-ax) \frac{\sin x}{x} dx$ . This integral exists and is finite for all $a>0$ . Taking derivative with respect to $a$ and differentiating under the integral sign , we have $\frac{dI(a)}{da} = -\int_{0}^{\infty} \exp(-ax) \sin x dx=-\frac{1}{1+a^2}$ Hence, $I(a)= -\tan^{-1}(a) + C$ Also, as $a \to \infty$ , $I(a) \to 0$ (which follows from Dominated Convergence Theorem (DCT)). Thus, we have $I(a)=\frac{\pi}{2}-\tan^{-1}(a)$ Invoking DCT once again, this time letting $a\to 0$ , we conclude that $\int_{0}^{\infty} \frac{\sin x}{x} dx=\frac{\pi}{2}$