Dirichlet Integral/Proof 2

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

Using differentiation under the integral sign,

Integrating with respect to $\alpha$,

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

Hence,


 * $\displaystyle \int_0^\infty \frac{e^{-\alpha x}\sin\left(x\right)} x \mathrm dx =\frac \pi 2 - \arctan\left(\alpha\right)$

Setting $\alpha = 0$ yields,


 * $\displaystyle \int_0^\infty \frac{\sin\left(x\right)} x \mathrm dx = \frac \pi 2$