Dirichlet Integral/Proof 2

== Theorem ==
== Proof ==
$\ds \int_0^\infty \frac {\sin x} x \rd x$ is convergent as an improper integral.
Indeed, for all $n \in \N$:
But:
so that:
:$\ds \int_0^\pi \frac {\sin x} {1 + \frac x {k \pi} } \rd x \to_{k \mathop \to \infty} 2$
Hence:
:$\ds \int_0^{2\pi n }\frac {\sin x} x \rd x = \sum_{k \mathop = 0}^{n \mathop -1} \frac 1 {2 \pi k} \int_0^\pi \frac {\sin x} {1 + \frac x {2 \pi k} } \rd x - \frac 1 {\pi \paren {2 k + 1} } \int_0^\pi \frac {\sin x} {1 + \frac x {\pi \paren {2 k + 1} } } \rd x$
can be expressed as a series whose general term is equivalent to:
:$\dfrac 2 \pi \times \dfrac 1 {2 k \paren {2 k + 1} }$
which is the term of an absolutely convergent series.
By Modulus of Sine of x Less Than or Equal To Absolute Value of x:
:$\ds \size {\frac {e^{-\alpha x} \sin x} x} \le e^{-\alpha x}$
From Laplace Transform of Real Power: 
:$\ds \int_0^\infty e^{-\alpha x} \rd x = \frac 1 \alpha$
Hence by Comparison Test for Improper Integral:
:$\ds \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$
converges whenever $\alpha > 0$.
So, we can define a real function $I : \openint 0 \infty \to \R$ by: 
:$\ds \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$
for each $\alpha \in \openint 0 \infty$.
Using Improper Integral of Partial Derivative on segments included in $\openint 0 \infty$:
Therefore, by Derivative of Arctangent Function
:$\map I \alpha = -\arctan \alpha + K$
for some $K \in \R$.
We also have: 
so:
:$\ds \lim_{\alpha \mathop \to \infty} \size {\map I \alpha} = 0$
That is: 
:$\ds \lim_{\alpha \mathop \to \infty} \map I \alpha = 0$
Therefore:
:$\map I \alpha = \dfrac \pi 2 - \arctan \alpha$
since $\ds \arctan \alpha \to_{\alpha \mathop \to \infty} \frac \pi 2$.
Note that we have:
:$\ds \map I \alpha \to_{\alpha \mathop \to 0} \frac \pi 2$ 
We now need to show that:
:$\ds \map I \alpha \to_{\alpha \mathop \to 0} \int_0^\infty \frac {\sin x} x \rd x$
Observe for this purpose that:
where all the improper integrals appearing here are convergent by Comparison Test for Improper Integral, as used above for defining $\map I \alpha$.
Therefore:
:$\ds \map I \alpha = \alpha \int_0^\infty \frac {\sin^2 x} x e^{-2 \alpha x} \rd x + \int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2 \alpha x} \rd x$
We also have:
where the improper integrals on the right hand side are convergent because the first one identifies with $\ds \int_0^\infty \frac {\sin x} x \rd x$ and the second one because $\dfrac {\sin^2 x} {x^2}$ is integrable on $\openint 0 \infty$, since it has a finite limit at $0$ and is smaller than $\frac 1 {x^2}$ at $\infty$.
Hence:
:$\ds \int_0^\infty \frac {\sin x} x \rd x = \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x$
where the second integral is absolutely convergent.
Moreover:
whenever $\alpha \le 1$.
Also:
:$\ds \int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2\alpha x} \rd x \to_{\alpha\to 0} \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x$
This is because, for any positive $R$ and $\alpha$:
because $\dfrac {\sin^2 x} {x^2}$ is integrable on $\openint 0 \infty$.
Finally, we have:
as well as:
So that, by uniqueness of limits:
:$\ds \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
Category:Dirichlet Integral