Talk:Dirichlet integral

Confusing proof
The proof is very confusing, there might be minor mistakes. Also, I don't think you need to bring in the second variable (beta). On the Bosnian version of the page the proof is in one variable and it is much clearer. The beta variable just makes it more general.

There is a step where the imaginary part of a complex exponential is made to include the whole integral, is that justified? —Preceding unsigned comment added by 188.62.241.118 (talk) 20:04, 17 April 2010 (UTC)

Using Complex Analysis
Can this be proven using Complex Analysis? --Hirak 99 (talk) 09:30, 12 August 2010 (UTC)
 * Yes, indeed. It can be evaluated using complex analysis. This is a standard example for using contour integration. Nerd271 (talk) 21:05, 26 February 2019 (UTC)

Complex Integration
The method here seems unnecessarily difficult. In particular I think it is preferable not to have to refer to another page (Sokhotski-Plemelj theorem) - especially when that theorem is not even proved on that page!

The obvious alternative would be to integrate $$f(z)=\frac{e^{iz}}{z}$$ around a composite path parameterised by &epsilon; &rarr; 0 and R &rarr; &infin;, consisting of the real line segments [-R,-&epsilon;] and [&epsilon;,R], the semicircle diameter R centred about zero in the upper half-plane in the positive direction, and the semicircle diameter &epsilon; centered about zero in the upper half-plane in the negative direction. The whole integral is zero, the integral around the large semicircle tends to zero and the integral around the small semicircle tends to -i&pi; (via Taylor expansion of the exponent function), so the integral on the real line segments of sin(x)/x is &pi;. Finally the function is even, so the integral over the positive real line is &pi;/2.

It might also be worth justifying the statement that the integral around the large semicircle tends to zero.

I think that would make the page accessible to the average second-year maths undergraduate.

BobHatt (talk) 19:04, 23 January 2013 (UTC)

Another Proof?
Using the function $$f(a)=\int_{\infty}^{\infty} \frac{sin(ax)}{x} dx $$ and differentiating wrt to a, under the integral sign, the result can be shown very easily(ultimately using the elementary property of delta functions). But I'm not sure if after differentiating, it is correct to write $$f'(a)=\int_{\infty}^{\infty} cos(ax) dx =2\pi\delta(a)$$. Should that go into this page? Aritrop (talk) 03:22, 25 April 2015 (UTC)
 * It's just the g(1) value of the bottom of section 1.2. Cuzkatzimhut (talk) 20:21, 22 January 2016 (UTC)

Missing explanation for step in Laplace transform argument
I don't understand the very first equation in the Laplace transform argument, namely

\int_{0}^{\infty} \frac{\sin t}{t} \, dt = \lim_{s \rightarrow 0} \int_{0}^{\infty} e^{-st}\frac{\sin t}{t} \, dt. $$ Why is it okay to exchange the limit with the integration in this case (as well as with the implied limit for the upper end of the integration domain)? Standard arguments like dominated convergence don't apply here as the integrant is not integrable. — Preceding unsigned comment added by 81.104.255.29 (talk) 11:28, 19 March 2021 (UTC)


 * Not sure what you're talking about. This is a single rather than double integral. No limits are being exchanged. We simply rewrite the integrand and the lower limit of integration. Nerd271 (talk) 15:21, 19 March 2021 (UTC)


 * Hey! Thanks for your quick response. I asked my math PhD friends before changing the article, so I feel rather strongly this should be clarified.


 * In the equation

\int_{0}^{\infty} \frac{\sin t}{t} \, dt = \lim_{s \rightarrow 0} \int_{0}^{\infty} e^{-st}\frac{\sin t}{t} \, dt $$
 * the integral on the left-hand side is $$\int_{0}^{\infty}\lim_{s \rightarrow 0}e^{-st}\frac{\sin t}{t} \, dt$$. On the right-hand side the $$\lim_{s \rightarrow 0}$$ and the integral have changed places. I don't understand why this is correct in this case (I do not doubt it is, but I feel it requires an argument). — Preceding unsigned comment added by 81.104.255.29 (talk) 17:12, 19 March 2021 (UTC)


 * You may be over thinking it. Who said the limit has to be inside the integral? Why not just apply it outside and rewrite the boundary and the integrand accordingly? Nerd271 (talk) 17:33, 19 March 2021 (UTC)


 * I'm asking why the equation $$\int_{0}^{\infty} \frac{\sin t}{t} \, dt = \lim_{s \rightarrow 0} \int_{0}^{\infty} e^{-st}\frac{\sin t}{t} \, dt$$, which is in the article without explanation, holds. I'm not sure what you mean by "boundary and the integrand" -- if you could elaborate your argument as to why this equality holds I (and I suspect, other readers of the article) would much appreciate it. (I don't want to just bitch: I realize you worked on this article for a long time, so let me take this opportunity to thank you for your work which has already helped me a lot!) — Preceding unsigned comment added by 81.104.255.29 (talk • contribs)


 * Does it really need an explanation? Isn't inspection sufficient? This is precisely what I learned when I was introduced to the Laplace transform. Clearly, in that limit, the exponential factor tends to one, and so does not affect the integrand. This is one of the most frequently used tricks in mathematics, namely, multiplying by one. The 'lim' operator acts on the whole thing. Don't stick it in the middle then confuse yourself about whether or not you can pull it out. Keep it outside the whole time. (Glad you find this useful!) Nerd271 (talk) 18:41, 19 March 2021 (UTC)


 * I'm afraid it needs an explanation. I'm afraid yours doesn't work. In the limit, the exponential factor tends to one, but for that to work the limit needs to be inside the integral . The statement you need here is called a Tauberian theorem and is highly non-trivial: It actually implies the Prime Number Theorem! See Theorem 6.2 in this manuscript for details. As is, the article is not proving what it claims to prove. You should add the clarification tag back in. (The other arguments, e.g., the double integration one, also have problems as Fubini does not apply in this case, but I won't bother you with the details there just now.)  81.104.255.29 (talk) 18:53, 19 March 2021 (UTC)

No, it works just fine. Don't make it more complicated than it needs to be. As a professor of mine put it, "Analysis should not be used to impress." If you want to add a note clarifying the "interchange of limits," go ahead. Otherwise, the article is fine the way it is. Nerd271 (talk) 19:14, 19 March 2021 (UTC)


 * It's really not fine. Let me try another way: By your logic, this should work:

\int_0^\infty \sin(x)\, dx = \lim_{s \rightarrow 0} \int_{0}^{\infty} e^{-st}\sin(t)\, dt = \lim_{s \rightarrow 0} \frac{1}{1 + s^2} = 1. $$
 * Don't you think something is a bit weird about that equation? Perhaps ask your professor about it too. — Preceding unsigned comment added by 81.104.255.29 (talk) 19:22, 19 March 2021 (UTC)


 * The integral on the left-hand side does not exist. Therefore, it does not work. Nerd271 (talk) 19:26, 19 March 2021 (UTC)


 * (1) So why does $$\int_{0}^{\infty} \frac{\sin t}{t} \, dt$$ exist?
 * (2) The function $$f_n(x) := n\chi_{(0,1/n)}(x)$$ where $$\chi$$ is the indicator function satisfies $$\lim_{n\to\infty}f_n(x) = 0$$ for all $$x\in\mathbb{R}$$, but $$\int_0^1 \lim_{n\to\infty} f_n(x)\,dx = 0 \neq 1 = \lim_{n\to\infty}\int_0^1 f_n(x)\,dx$$. (This is the example in the Wikipedia article on dominated convergence.) — Preceding unsigned comment added by 81.104.255.29 (talk) 19:48, 19 March 2021 (UTC)


 * Because it has been proven to converge. If you start with a false premise, you can prove even the most absurd of claims. If you want to add a note to satisfy hair-splitters and people who want complications when none is necessary, then do so. If you cannot (because you are new here), write it here and I will add it for you. Otherwise, no change is necessary. Nerd271 (talk) 19:57, 19 March 2021 (UTC)


 * In the context of the article, the equation is used to prove that it converges. In the same sense my (incorrect) computation for $$

\int_0^\infty \sin(x)\, dx$$ could be claimed to "prove" that that integral converges.


 * Anyway, thanks for suggesting I could add a note. I would modify the sentence after the equation block to read "where the first equation is due to a Tauberian theorem for Laplace transforms and the third is because $$\mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1}$$ is the Laplace transform of the function $$\sin t$$." Let me know if that is okay with you? — Preceding unsigned comment added by 81.104.255.29 (talk) 20:10, 19 March 2021 (UTC)


 * No, that "proof" is like one of those for 1 = 2. You make use of faulty arguments, in fact, you start with one, so it is no surprise that your conclusion is incorrect.
 * In case you missed it, the Laplace transform for the sine function is evaluated in full in the section that uses the Leibniz rule for differentiating under the integral sign. We do not need to repeat ourselves. Nerd271 (talk) 20:23, 19 March 2021 (UTC)


 * I might have been unclear. Currently, the article has the following after the equation block: "because $$\mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1}$$ is the Laplace transform of the function $$\sin t$$. (See the section 'Differentiating under the integral sign' for a derivation.)". I was proposing replacing the first sentence with the above.


 * As for $$\int_0^\infty \frac{\sin t}{t} \, dt$$, the article attempts to prove this integral converges by a calculation, the first step of which is to insert the $$e^{-st}$$. If that's always a valid step in such a proof (it isn't), my calculation for $$\int_0^\infty \sin(x)\, dx$$ would be fine too. If it only works if the integral we start with converges, the article should better establish that convergence first.  — Preceding unsigned comment added by 81.104.255.29 (talk) 21:35, 19 March 2021 (UTC)

Convergence is settled in the introduction. Therefore, we may proceed to evaluate the integral. 21:54, 19 March 2021 (UTC)

Okay, so I thought about this some more. You were actually right, Nerd271, and what I proposed was too complicated. In fact, it was quite complicated (a Tauberian theorem) but it doesn't even apply: Theorem 6.2 in the manuscript I linked requires the Laplace transform to be analytic on an open set containing the closed half-plane. That doesn't work for $$\sin$$ because its Laplace transform $$s\mapsto\frac{1}{1 + s^2}$$ has poles in $$\pm i$$, but it doesn't work for $$\sin(x)/x$$ either as its Laplace transform $$s\mapsto\pi/2 - \arctan(s)$$ also has singularities for $$\pm i$$.

However, after some thought I found an easier solution: The statement we need here is a consequence of the final value theorem for the Laplace transform plus an integration by parts: If $$\int_0^\infty f(x)\, dx$$ exists (possibly as an improper integral), then $$\lim_{s\downarrow 0}\int_0^\infty e^{-sx} f(x)\, dx$$ converges to it. I added this statement and its proof to the final value theorem article. As you assumed, the reason $$\lim_{L\to\infty}\int_0^L\sin(x) \, dx$$ doesn't work is that this integral doesn't exist (interestingly, it varies between 0 and 2, so the Laplace transform at 0 gives the correct "mean" value).

This means one can indeed interchange the limit and the integral here, provided one establishes that the integral converges first. You say this is done in the introduction, but I couldn't find in there (it's just asserted with no explanation why). To see that the integral converges, Dirchlet's test can be used (this is basically a version of the Leibniz criterion for integrals).

So I would now propose: Let's add a reference to Dirichlet's test to the introduction to establish convergence of the integral and replace the reference to Tauberian theorems to a link to the subsection of the final value theorem.

BTW, this does not mean it's fine to interchange limits and integrals in all situations, as you seem to believe (see the counterexample above). Such an interchange is also very much what is happening here, so the idea of "just leave the limit outside" is faulty. Orthogonalist (talk) 13:24, 22 March 2021 (UTC)


 * Thank you for enlightening me on something my professors and textbooks already told me. I already know it is not always permissible to interchange limits. I already pointed out why your "counterexample" is faulty because the premise is false. Nerd271 (talk) 13:49, 23 March 2021 (UTC)


 * Nerd271: The OP is perfectly right saying that :$$

\int_{0}^{\infty} \frac{\sin t}{t} \, dt = \lim_{s \rightarrow 0} \int_{0}^{\infty} e^{-st}\frac{\sin t}{t} \, dt. $$ needs an explanation; indeed, it is a key point of the whole computation. pm a 21:02, 30 March 2021 (UTC)


 * So explain properly without throwing arguments you later withdraw and without confusing students who are new to this material. If you think you can impress people with your mathematical terminology, think again. Nerd271 (talk) 00:13, 2 April 2021 (UTC)

Missing explanation
Here is the simplest and most elementary argument I can see to show that the function $$f:[0,\infty)\to\R$$ defined by the integral
 * $$f(s)=\int_0^\infty \frac{e^{-st}}t \sin t \, dt$$

is continuous at $$s=0$$. Simply write, for $$s>0$$
 * $$f(0)-f(s)=s\int_0^\infty \frac{1-e^{-st}}{st} \sin t \, dt=s\sum_{k=0}^\infty \int_{k\pi}^{(k+1)\pi} \frac{1-e^{-st}}{st} \sin t \, dt=s\sum_{k=0}^\infty (-1)^k\int_0^\pi \frac{1-e^{-s(t+k\pi)}}{s(t+k\pi)}   \sin t  \, dt.$$

Now recall that the function $$x\mapsto \frac{1-e^{-x}}x $$ is decreasing with limits at $$0$$ and $$+\infty$$ resp. $$1$$ and $$0$$ (e.g. by convexity: it's (minus) an incremental ratio of the exponential function). Therefore for any $$s>0$$ the integral $$\int_0^\pi \frac{1-e^{-s(t+k\pi)}}{s(t+k\pi)}\sin t\, dt$$ is decreasing wrto $$k$$ (and of course tends to zero for $$k\to+\infty$$ because the integrand goes to $$0$$ uniformly). So by the alternating series test the sum of the latter series has the sign of its first term, and it is bounded by it. We conclude
 * $$0\le f(0)-f(s)\le s\int_0^\pi \frac{1-e^{-st}}{st}  \sin t  \, dt\le s\int_0^\pi \sin t  \, dt=2s$$,

so $$f(s)=f(0)+o(1)$$ as $$s\to0$$ pm a 21:03, 30 March 2021 (UTC)


 * Yes, this works. As I tried to indicate in the "Laplace transform" section, the continuity at 0 is also a consequence of the final value theorem for the Laplace transform (I added the derivation at Final_value_theorem; it's basically the same integration by parts argument you mentioned in the article). I feel the article for Abel's theorem could be improved to include that statement directly.
 * The alternating series test (in its integral form, called Dirchlet's test for improper integrals) is also the reason the Dirichlet integral exists in the first place, which wasn't mentioned in the article until relatively recently. Orthogonalist (talk) 11:09, 1 April 2021 (UTC)
 * P.S.: As I mentioned above, this article has some other formal correctness issues, e.g., the reversal of the order of integration in the "double integration" section: Fubini/Tonelli needs integrable or positive integrands, neither of which is the case here.
 * Tangentially related, similar arguments to the ones here happen in Laplace_transform. The comment there ("Even when the interchange cannot be justified the calculation can be suggestive.") confuses me a bit since Abel's theorem does indeed show that if the improper integral exists/converges, the Laplace transform is right-continuous at 0. Orthogonalist (talk) 11:41, 1 April 2021 (UTC)


 * The function $$\sin x / x$$ is one of exponential order because $$\sin x$$ is already a function of exponential order. The integrand is clearly continuous in the interval $$[0, \infty[$$. Therefore, the function is Riemann-integrable in that interval and its Laplace transform exists. Nerd271 (talk) 13:55, 2 April 2021 (UTC)

This is getting silly. The (proper) Riemann integral is only defined on compact intervals. Moreover, the existence of the Laplace transform isn't the issue at hand. Orthogonalist (talk) 21:44, 2 April 2021 (UTC)


 * Except the definition of a proper Riemann integral can be extended to include improper integrals. Laplace transforms are by definition improper integrals. I am aware that the original definition of the Riemann integral applies only to finite domains, which is why I mentioned it. Let's not forget differentiability implies continuity. Nerd271 (talk) 23:07, 2 April 2021 (UTC)