Talk:Dirichlet Integral/Proof 2

How do you see that $I$ is actually continuous at 0 from the comparison theorem ? It seems to me that $\dfrac{\sin x}{x}$ is far from being integrable on $(0;+\infty)$...

In fact, I believe there are two things to prove in addition to what is there:

1. prove that $\ds\int_0^{+\infty}\frac{\sin x}{x} \rd x$ actually has a finite value;

2. prove that $\ds I \to_{0}\int_0^{+\infty}\frac{\sin x}{x} \rd x$.

Here is more or less a sketch of how I would show these. Perhaps there is an easier way ? This is intendend so that you just have to copy and paste the proof in the article if you agree with me.

The first point is more or less straightforward (basically the same as proving that the alternated harmonic series converge):

For any $n\in\N$:

Now, using Lebesgue's Dominated Convergence Theorem:
 * $\ds \int_0^\pi \frac {\sin x} {1 + \frac x {\pi k} } \rd x \to_{k \mathop \to \infty} 2$

so that


 * $\ds \int_0^{2\pi n }\frac {\sin x} {x} \rd x = \sum_{k \mathop = 0}^{n \mathop -1} \frac 1 {2 \pi k} \int_0^\pi \frac {\sin x} {1 + \frac x {2\pi k} } \rd x - \frac 1 {\pi \paren {2k+1} } \int_0^\pi \frac {\sin x} {1 + \frac x {\pi \paren {2k+1} } } \rd x$

can be expressed as a series whose general term is equivalent to
 * $\ds \frac 2 \pi \times \frac 1 {2k \paren {2k + 1} }$

which is the term of an absolutely convergent series.

For the second point, observe that by integration by parts:
 * $\ds \int_0^\infty \frac {\sin x} x \rd x = \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x$

In the same way, one gets
 * $\ds I(\alpha) = 2\alpha \int_0^\infty \frac {\sin^2 x} x e^{-2\alpha x} \rd x + \int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2\alpha x} \rd x$

Using Lebesgue's Dominated Convergence Theorem:

so it only remains to show that:
 * $\ds 2\alpha \int_0^\infty \frac {\sin^2 x} x e^{-2\alpha x} \rd x \to_{\alpha \mathop \to 0} 0$

This can be seen in the following way:

whenever $\alpha \leq 1$.

Palimpseste (talk) 20:53, 6 November 2022 (UTC)
 * Definitely a problem, this was all kind of rushed to just get rid of the previous version which was extremely unrigorous. Do you see any way to compute that $I'(0) = -1$? Then the derivative is indeed $-\paren {1 + \alpha^2}^{-1}$ and all is well. Of course you'll still need to show that $\int \frac {\sin x} x \rd x < \infty$ as you've done. Caliburn (talk) 21:36, 6 November 2022 (UTC)

In fact, you don't need to show that $I'(0) =-1$: it is enough to know that
 * $ \ds I'(\alpha) = -\frac 1 {1+\alpha^2}$

for $\alpha\in (0;\infty)$ to conclude that $I(\alpha) = K -\arctan\alpha$ for some $K\in\R$.

Then, taking the limit for $\alpha\to\infty$ yields $K=\frac \pi 2$.

Thus, $I\to_{0}\frac \pi 2$. My correctio above shows that $I$ converges at $0$ to the Dirichlet integral, so we get the desired identity.

Palimpseste (talk) 22:17, 6 November 2022 (UTC)
 * Oh yes you are of course correct here, it slipped my mind why you were proving continuity. Feel free to fill this in and I'll clean it up in a bit. Caliburn (talk) 23:08, 6 November 2022 (UTC)


 * I do not get:
 * 1. why $\ds \int_0^\infty \frac {\sin x} x \rd x = \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x$
 * 2. why Lebesgue's Dominated Convergence Theorem can be applied
 * With some correction, I see:
 * $\ds \lim_{\alpha \to 0} \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x = \dfrac \pi 2$
 * but I have no idea how to finish the proof from here. So excited to see how. --Usagiop (talk) 00:00, 7 November 2022 (UTC)

Yes, I'm sorry I just didn't have time to write down details for how to integrate by parts yesterday. It is not that straightforward either, but a more classical result:

Reorganizing the terms gives the identity I was mentioning. Exactly the same procedure of integrating by parts yields the identity for $I(\alpha)$.

For the dominated convergence, simply observe that $\ds {\paren {\frac {\sin x} x} }^2$ is integrable on $(0;+\infty)$ (limits to $1$ at $0$, and dominated by $\ds \frac 1 {x^2}$ at $\infty$).

Palimpseste (talk) 08:04, 7 November 2022 (UTC)


 * Thank you. Sorry, I just wanted to say "please improve the proof" and let me see the correct proof. By the way, I still think that ${\paren {\frac {\sin x} x} }^2 e^{-2\alpha x}$ is not dominated by $\ds \frac 1 {x^2}$. --Usagiop (talk) 10:23, 7 November 2022 (UTC)

Let's say it is dominated by $\dfrac 1 {x^2}$ on $(1;\infty)$, and by $1$ on $(0;1)$.

Palimpseste (talk) 12:04, 7 November 2022 (UTC)

(I was not saying it is dominated by $\dfrac 1 {x^2}$, but by $\ds \paren {\frac {\sin x} x}^2$, which is integrable on $(0;\infty)$, so I am not sure I got your answer or question right...)

Palimpseste (talk) 12:14, 7 November 2022 (UTC)


 * Sorry, I somehow overlooked $e^{-2\alpha x} \le 1$. --Usagiop (talk) 17:32, 7 November 2022 (UTC)


 * I think you do even not need Lebesgue's Dominated Convergence Theorem. You can show the convergence with elementary estimates. --Usagiop (talk) 18:43, 7 November 2022 (UTC)

Let me just note:

First, choose $N$ large enough, and then choose $\alpha$ small enough. --Usagiop (talk) 18:51, 7 November 2022 (UTC)