Dirichlet Integral/Proof 2

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:
 * $\ds \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$