piskolforji Posted March 8, 2020 Share Posted March 8, 2020 hi. how to solve the integral that i attached in this post? i tried to solve it but seems its too hard for me 😕 plz help me Link to comment Share on other sites More sharing options...

taeto Posted March 8, 2020 Share Posted March 8, 2020 You could begin by substituting \(r = a\cdot s,\) that surely helps. What is the upper integration bound for \(\varphi,\) it looks like \(\gamma_\pi?\) Link to comment Share on other sites More sharing options...

piskolforji Posted March 8, 2020 Author Share Posted March 8, 2020 8 minutes ago, taeto said: You could begin by substituting r=a⋅s, that surely helps. What is the upper integration bound for φ, it looks like γπ? The upper integration bound for fi is 2 pi Link to comment Share on other sites More sharing options...

taeto Posted March 8, 2020 Share Posted March 8, 2020 10 minutes ago, piskolforji said: The upper integration bound for fi is 2 pi Great. After substituting \(r\) do you see \(1 + \cos \varphi\) somewhere, and does that remind you of something, like as if it could be another expression squared? Link to comment Share on other sites More sharing options...

piskolforji Posted March 8, 2020 Author Share Posted March 8, 2020 (edited) 16 minutes ago, taeto said: Great. After substituting r do you see 1+cosφ somewhere, and does that remind you of something, like as if it could be another expression squared? 1+cosfi= 2 (cosfi/2)^2 But the denominator is a+r cosfi Edited March 8, 2020 by piskolforji Link to comment Share on other sites More sharing options...

taeto Posted March 9, 2020 Share Posted March 9, 2020 12 hours ago, piskolforji said: 1+cosfi= 2 (cosfi/2)^2 But the denominator is a+r cosfi Indeed. I misread the integrand, sorry about that. But re-reading it now, I wonder if you copied it correctly from its source. The differentials \(dr\) and \(d \varphi\) are in the wrong order with respect to the order of the integration symbols. You have to know which integral is the "outer" and which is the "inner", since in this special case you are not free to swap them around as you please. The integrand is not defined at \(r = a\) and \(\varphi = \pi,\) and that makes it harder than usual. Link to comment Share on other sites More sharing options...

piskolforji Posted March 9, 2020 Author Share Posted March 9, 2020 i could solve it by Mathematica but i need to know how to solve it.☹️ Link to comment Share on other sites More sharing options...

taeto Posted March 9, 2020 Share Posted March 9, 2020 (edited) Good, that helps. Now the integral over \(r\) is of standard form \(\int \frac{r}{a+br}dr,\) with \(b=\cos \varphi.\) An antiderivative is \( \frac{r}{b} - \frac{a}{b^2}\log | a+br|.\) So we get \(\int_0^a \frac{r}{a+br}dr = \frac{a}{b} (1-\frac{\log (1+b)}{b} ) \) when \(b \not\in \{0,-1\}.\) Agree? Edited March 9, 2020 by taeto Link to comment Share on other sites More sharing options...

piskolforji Posted March 9, 2020 Author Share Posted March 9, 2020 (edited) did u use integral table for that integral? can u give me a reference plz ? Edited March 9, 2020 by piskolforji Link to comment Share on other sites More sharing options...

taeto Posted March 9, 2020 Share Posted March 9, 2020 en.wikipedia.org/wiki/List_of_integrals_of_rational_functions Link to comment Share on other sites More sharing options...

piskolforji Posted March 9, 2020 Author Share Posted March 9, 2020 8 hours ago, taeto said: en.wikipedia.org/wiki/List_of_integrals_of_rational_functions But it gives a more difficult integral for fi Link to comment Share on other sites More sharing options...

taeto Posted March 10, 2020 Share Posted March 10, 2020 Maybe use \(\cos(2\pi - \varphi) = \cos \varphi\) to change the integral from \(0\) to \(2\pi\) to two times an integral from \(0\) to \(\pi.\) Then use \(\cos(\pi - \varphi) = -\cos \varphi\) to change this integral to an integral from \(0\) to \(\pi/2.\) Finally substitute \(s = \sin \varphi\) to get an integral from \(0\) to \(1\) without any trigonometric functions (but a \(\log s\) ). I will try to look if that can be done. Link to comment Share on other sites More sharing options...

joigus Posted May 7, 2020 Share Posted May 7, 2020 (edited) I've found a pedestrian way to do it without looking at tables. If you're not interested, don't pay attention. First you reorder your integral by taking care of the $r$ part, while factoring the angular part to the left. You integrate by parts the r factor, \[\int_{0}^{2\pi}d\varphi \frac{1}{\cos\varphi}\int_{0}^{a}dr r\frac{\partial}{\partial r}\left(\log\left(a+r\cos\varphi\right)\right)\] After some algebraic steps you get to, \[\int_{0}^{2\pi}d\varphi\frac{1}{\cos\varphi}\left(-a\cos\varphi\log\left(a\left(1+\cos\varphi\right)\right)+a\cos\varphi+a\log a\right)\] This reduces to, \[-a\int_{0}^{2\pi}d\varphi\log\left(a\left(1+\cos\varphi\right)\right)+a\int_{0}^{2\pi}d\varphi+a\log a\int_{0}^{2\pi}d\varphi\frac{1}{\cos\varphi}\] Both the first and the last integrals have singular integrands, but you can argue that they're convergent. And the value they converge to is precisely zero, as they are periodic functions of $\varphi$ between zero and $2\pi$. The only thing you're left with is, \[a\int_{0}^{2\pi}d\varphi\] Check me for mistakes, I'm sleepy. Edited May 7, 2020 by joigus bad rendering of eq. Link to comment Share on other sites More sharing options...

taeto Posted May 7, 2020 Share Posted May 7, 2020 (edited) Thanks joigus for following up on this! It is a mistake when you say "this reduces to". Because it will not be difficult to come up with examples like \[ \int_a^b f(x)+g(x) dx\] where \(f\) and \(g\) are both not defined in the entire interval \([a,b]\) and where \[\int_a^b f(x)+g(x) dx \neq \int_a^b f(x)dx + \int_a^b g(x)dx\] holds, even when all those three integrals converge. The Riemann integral is not defined unless the integrand is defined throughout the integration interval, and this is not the case here. In such a case when it is not defined, you can resort to the theorem that if the number of discontinuities is countable you can ignore them. But that only goes when the integrand is bounded in absolute value, which in our case it is not. So however much I want to believe your solution, I think we should offer yet a better proof. Edited May 7, 2020 by taeto Link to comment Share on other sites More sharing options...

joigus Posted May 7, 2020 Share Posted May 7, 2020 1 hour ago, taeto said: The Riemann integral is not defined unless the integrand is defined throughout the integration interval, and this is not the case here. There are more integrals in this world. Have you ever heard of the Lebesgue measure? I often think of integrals in terms of the Lebesgue measure. The trick is looking at the domain and the range of the function. "The range is infinite" becomes no more worrying than "the domain is infinite." Another name for it is mathematical power. No matter how you want to do it with Riemann, you're gonna have to face the singularities, because those are in the domain. If I've found them with my method, believe me, you're gonna find them too. No integral handbook in the world will get around that. That's a property of the integral, not of my method. I may take a look at how rigorously it is defined with the Lebesgue measure. Maybe Cauchy's theorem. But not now. Most integrals in QFT are Lebesgue or Cauchy complex, not Riemann. Think just a little bit outside the Riemann box, please. 1 hour ago, taeto said: It is a mistake when you say "this reduces to". I said "this reduces to" about the simple algebraic steps, not the convergence arguments. 1 hour ago, taeto said: So however much I want to believe your solution, I think we should offer yet a better proof. I honestly think I should go to sleep now, but who knows. I'm finding it difficult lately. Just now, joigus said: There are more integrals in this world. Have you ever heard of the Lebesgue measure? I often think of integrals in terms of the Lebesgue measure. The trick is looking at the domain and the range of the function. "The range is infinite" becomes no more worrying than "the domain is infinite." Another name for it is mathematical power. No matter how you want to do it with Riemann, you're gonna have to face the singularities, because those are in the domain. If I've found them with my method, believe me, you're gonna find them too. No integral handbook in the world will get around that. That's a property of the integral, not of my method. I may take a look at how rigorously it is defined with the Lebesgue measure. Maybe Cauchy's theorem. But not now. Most integrals in QFT are Lebesgue or Cauchy complex, not Riemann. Think just a little bit outside the Riemann box, please. I said "this reduces to" about the simple algebraic steps, not the convergence arguments. I honestly think I should go to sleep now, but who knows. I'm finding it difficult lately. And oh, please, believe you me: If you find an integral with singularities in the domain that has been solved in a handbook. Some mathematician has proven convergence way ahead of the handbook going to press. Same goes for Wolfram or the like. Link to comment Share on other sites More sharing options...

taeto Posted May 7, 2020 Share Posted May 7, 2020 3 minutes ago, joigus said: There are more integrals in this world. Have you ever heard of the Lebesgue measure? I often think of integrals in terms of the Lebesgue measure. The trick is looking at the domain and the range of the function. "The range is infinite" becomes no more worrying than "the domain is infinite." Another name for it is mathematical power. No matter how you want to do it with Riemann, you're gonna have to face the singularities, because those are in the domain. If I've found them with my method, believe me, you're gonna find them too. No integral handbook in the world will get around that. That's a property of the integral, not of my method. I may take a look at how rigorously it is defined with the Lebesgue measure. Maybe Cauchy's theorem. But not now. Most integrals in QFT are Lebesgue or Cauchy complex, not Riemann. Think just a little bit outside the Riemann box, please. I said "this reduces to" about the simple algebraic steps, not the convergence arguments. I honestly think I should go to sleep now, but who knows. I'm finding it difficult lately. Please do not patronize. The theorem that I mentioned specifically concerns the Lebesgue integral. Before that I already mentioned that we cannot be talking about the Riemann integral, because that is not even defined for this integrand with the given integration domain. Now you should justify your "this reduces to". Typically by appealing to the monotone convergence theorem or to the dominated convergence theorem for Lebesgue integrals. Or to something else? Link to comment Share on other sites More sharing options...

joigus Posted May 7, 2020 Share Posted May 7, 2020 2 hours ago, joigus said: I've found a pedestrian way to do it without looking at tables. If you're not interested, don't pay attention. What part of this couple of sentences you didn't understand, taeto? I wasn't talking to you when I said it, but anyway, as you seem to care so much... And sorry for 'patronizing' you. Really. Link to comment Share on other sites More sharing options...

taeto Posted May 7, 2020 Share Posted May 7, 2020 27 minutes ago, joigus said: What part of this couple of sentences you didn't understand, taeto? I wasn't talking to you when I said it, but anyway, as you seem to care so much... And sorry for 'patronizing' you. Really. You are right, I apologize for being so bitchy. I care a little bit, because I wanted to find a good argument for the integral, and I could not readily do that. Your 'pedestrian way' ought to provide the correct answer. But it does not work as a formal proof, unless you can show that the integrals converge as Lebesgue integrals. Something like the Henstock-Kurzweil integral may work even if Lebesgue does not. Link to comment Share on other sites More sharing options...

joigus Posted May 8, 2020 Share Posted May 8, 2020 9 hours ago, taeto said: You are right, I apologize for being so bitchy. I care a little bit, because I wanted to find a good argument for the integral, and I could not readily do that. Your 'pedestrian way' ought to provide the correct answer. But it does not work as a formal proof, unless you can show that the integrals converge as Lebesgue integrals. Something like the Henstock-Kurzweil integral may work even if Lebesgue does not. No offence taken. I gave you an apology too, and I owe you an explanation. I thought this was probably just about a Physics student working on a physical problem looking for mnemonic/algorithm. Looks like the magnetic moment of a given current density. The integrand comes from the geometry of the circuit. The way I see it, physical problems in terms of different variables are just useful parametrizations of something 'real.' If the integrals are not manifestly divergent and simple symmetry arguments tell me it better be zero, the maths probably are telling me I must assume those integrals not to be of physical relevance. There are two peaks in the integrand, one at \varphi = \pi and the other at \varphi = \fraq{3}{2}\pi, with opposite signs. In such a way that, if I further change the variables \cos x = u, the integral formally reduces to a \[\int_{1}^{1}\] That tells me it must be zero. I wasn't going for any kind of mathematical rigor. I'm not cut out for that. And I haven't the faintest idea what the Henstock-Kurzweil integral is. Link to comment Share on other sites More sharing options...

joigus Posted May 9, 2020 Share Posted May 9, 2020 17 hours ago, joigus said: one at \varphi = \pi pi over two. Sorry. Link to comment Share on other sites More sharing options...

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