Dirichlet Integral/Proof 5

Proof
Let $M \in \R_{>0}$.

Define a real function $I_M : \R \to \R$ by:
 * $\ds \map {I_M} \alpha := \int_0^M \dfrac {\sin x} x e^{-\alpha x} \rd x$

Then:

Observe:

On the other hand:

Thus:

Therefore:

As:
 * $\ds \map {I_M} 0 = \int_0^M \dfrac {\sin x} x \rd x$

we have shown:
 * $\ds \forall M \in \R_{>0} : \size {\int_0^M \dfrac {\sin x} x \rd x - \dfrac \pi 2} \le \dfrac 2 M$

In particular: