Dirichlet Integral/Proof 2

Proof
Define:
 * $\displaystyle I \left({\alpha}\right) = \int_0^\infty \frac {e^{-\alpha x} \sin x} x \, \mathrm d x$

Using differentiation under the integral sign:

Integrating with respect to $\alpha$:

To determine $C$, take $\alpha \to \infty$:

Hence:


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

Setting $\alpha = 0$ yields:


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