Dirichlet Integral

Theorem

 * $\displaystyle \int_0^\infty \frac {\sin x} {x} \mathrm d x = \frac \pi 2$

Proof
By Fubini's Theorem:


 * $\displaystyle \int_0^\infty \left({\int_0^\infty e^{- x y} \sin x \ \mathrm d y}\right) \ \mathrm d x = \int_0^\infty \left({\int_0^\infty e^{- x y} \sin x \ \mathrm d x}\right) \ \mathrm d y$

Hence: