Dirichlet Integral/Proof 2

Proof
Let $M$ be a positive real number.

We have:

Since:


 * $\ds \int_0^M \size {\frac {e^{-\alpha x} \sin x} x} \rd x$

is increasing and bounded in $M$, we have that:


 * $\ds \int_0^\infty \size {\frac {e^{-\alpha x} \sin x} x} \rd x$

converges.

Then, from Absolutely Convergent Definite Integral is Convergent:


 * $\ds \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$

converges.

So, we can define a real function $I : \hointr 0 \infty \to \R$ by:


 * $\ds \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$

for each $\alpha \in \hointr 0 \infty$.

Using Definite Integral of Partial Derivative:

We also have:

so:


 * $\ds \lim_{\alpha \to \infty} \size {\map I \alpha} = 0$

That is:


 * $\ds \lim_{\alpha \to \infty} \map I \alpha = 0$

Note that we have:


 * $\ds \map I 0 = \int_0^\infty \frac {\sin x} x \rd x$

So we have: