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 $n\in\N$:

\(\ds \int_0^{2\pi n}\frac {\sin x} {x} \rd x\)

\(=\)

\(\ds \sum_{k \mathop = 0}^{2n \mathop -1}\int_{\pi k}^{\pi \paren {k+1} }\frac {\sin x} {x} \rd x\)

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

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:

\(\ds \int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2\alpha x} \rd x\)

\(\to_{\alpha \mathop \to 0}\)

\(\ds \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x\)

\(\ds \)

\(=\)

\(\ds \int_0^\infty \frac {\sin x} x \rd x\)

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:

\(\ds 2\alpha \int_0^\infty \frac {\sin^2 x} x e^{-2\alpha x} \rd x\)

\(=\)

\(\ds 2\alpha \int_0^\infty \frac {\sin^2 \frac x \alpha} x e^{-2x} \rd x\)

\(\ds \)

\(\leq\)

\(\ds 2\alpha \paren {\int_0^\alpha \frac {\frac {x^2} {\alpha^2} } x \rd x + \int_\alpha^1 \frac 1 x \rd x + \int_1^\infty e^{-2x} \rd x }\)

\(\ds \)

\(=\)

\(\ds 2\alpha \paren {\frac 1 2 - \ln \alpha + \frac 1 {2 e^2} }\)

\(\ds \)

\(\to_{\alpha\mathop\to 0}\)

\(\ds 0\)

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:

\(\ds \int_0^\infty \frac {\sin x} x \rd x\)

\(=\)

\(\ds \int_0^\infty \frac {\sin 2x} x \rd x\)

\(\ds \)

\(=\)

\(\ds 2 \left[ \frac {\sin^2 x} x\right]_0^\infty - 2 \int_0^\infty \sin x \paren {\frac {\cos x} x - \frac {\sin x} {x^2} } \rd x\)

\(\ds \)

\(=\)

\(\ds - \int_0^\infty \frac {\sin 2x} x \rd x + 2\int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x\)

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:

\(\ds 0\)

\(\le\)

\(\ds \int_0^\infty {\paren {\frac {\sin x} x} }^2 \paren {1 - e^{-2\alpha x} }\rd x\)

\(\ds \)

\(=\)

\(\ds \int_0^N {\paren {\frac {\sin x} x} }^2 \paren {1 - e^{-2\alpha x} }\rd x + \int_N^\infty {\paren {\frac {\sin x} x} }^2 \paren {1 - e^{-2\alpha x} }\rd x\)

\(\ds \)

\(\le\)

\(\ds \paren {1 - e^{-2\alpha N} } \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x + \int_N^\infty {\paren {\frac {\sin x} x} }^2 \rd x\)

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

Sure, but what is the point of not using Lebesgue (which cuts out a bit of work, and makes the proof shorter and easier to grasp), when it is a proof that uses some other high end theorem of integration like Improper Integral of Partial Derivative (I believe the only time I got a proof of this was when I was introduced to Lebesque's integral) ? I see that this theorem does not have a page yet, so I'm adding it, even though I have no time to find and write a proof today (the Wikipedia page looks good though !).

Palimpseste (talk) 12:38, 12 November 2022 (UTC)

This proof looks elementary for me, say. for the first year's student. I think you do even not need the derivative of the improper integral, but (if any then) only the derivative of the integral over a closed interval. But, sorry for the distraction, please go ahead. I could add an alternative proof, afterwards. --Usagiop (talk) 10:20, 12 November 2022 (UTC)

I would be very interested in seeing a version of this proof that does not involve differentiating under the integral sign for improper integrals. It seems to me that it is the heart of the proof here (and of Feynman's method in general (speaking of which, I believe it would be good to mention that this proof is an example of applying Feynman's method, do you know where we could say this ?)), otherwise, you cannot compute a closed form for $I(\alpha)$, which (modulo a few technicalities...) gives the identity. But as long as we we need to use such a powerful tool from measure theory, I believe we are justified to also use Lebesgue's theorem to avoid getting our hands dirty every time we need to prove some convergence. It also allows to make the proof somehow clearer (this integral converges to $X$, so we have ...) by not interrupting the flow of the argument every time we need to check that something converges.

It's just a matter of taste and style, though, so I think people more akin than me with ProofWiki should have the last word. (That's why I am currently the proof by trying no to use Lebesgue's theorem.)

Palimpseste (talk) 12:38, 12 November 2022 (UTC)

By the way, I used your proof for getting rid of Lebesgue's theorem for showing that $\ds \int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2\alpha x} \rd x \to_{\alpha \to 0} \int_0^\infty {\paren {\frac {sin x} x} }^2 \rd x$ :)

And I also realized that I forgot to sign my two last messages, so there is going to be a bit of time travel unfortunately...

Palimpseste (talk) 12:38, 12 November 2022 (UTC)

I added Dirichlet Integral/Proof 5 which is my interpretation of the original Proof 2. --Usagiop (talk) 23:41, 12 November 2022 (UTC)