piskolforji 0 Posted March 8 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 0 Share this post Link to post Share on other sites

taeto 92 Posted March 8 You could begin by substituting \(r = a\cdot s,\) that surely helps. What is the upper integration bound for \(\varphi,\) it looks like \(\gamma_\pi?\) 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 8 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 0 Share this post Link to post Share on other sites

taeto 92 Posted March 8 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? 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 8 (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 by piskolforji 0 Share this post Link to post Share on other sites

taeto 92 Posted March 9 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. 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 9 i could solve it by Mathematica but i need to know how to solve it.☹️ 0 Share this post Link to post Share on other sites

taeto 92 Posted March 9 (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 by taeto 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 9 (edited) did u use integral table for that integral? can u give me a reference plz ? Edited March 9 by piskolforji 0 Share this post Link to post Share on other sites

taeto 92 Posted March 9 en.wikipedia.org/wiki/List_of_integrals_of_rational_functions 0 Share this post Link to post Share on other sites

piskolforji 0 Posted March 9 8 hours ago, taeto said: en.wikipedia.org/wiki/List_of_integrals_of_rational_functions But it gives a more difficult integral for fi 0 Share this post Link to post Share on other sites

taeto 92 Posted March 10 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. 0 Share this post Link to post Share on other sites

joigus 101 Posted May 7 (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 by joigus bad rendering of eq. 0 Share this post Link to post Share on other sites

taeto 92 Posted May 7 (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 by taeto 0 Share this post Link to post Share on other sites

joigus 101 Posted May 7 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. 0 Share this post Link to post Share on other sites

taeto 92 Posted May 7 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? 0 Share this post Link to post Share on other sites

joigus 101 Posted May 7 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. 0 Share this post Link to post Share on other sites

taeto 92 Posted May 7 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. 0 Share this post Link to post Share on other sites

joigus 101 Posted May 8 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. 0 Share this post Link to post Share on other sites

joigus 101 Posted May 9 17 hours ago, joigus said: one at \varphi = \pi pi over two. Sorry. 0 Share this post Link to post Share on other sites