My Math Forum I am lost on this example problem

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 September 25th, 2018, 04:59 PM #1 Senior Member     Joined: Oct 2016 From: Arizona Posts: 189 Thanks: 33 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. I am lost on this example problem Hi, so my question has to do with this link (the very first example problem) https://medium.com/a-younger-voice/r...k-e7afae85e25c I am learning how to integrate using the "Feynman integral trick", and I already got a little lost when they used the convenient value $\alpha =1$. They got that the integral is 0 for that case, and I was wondering if someone could quickly just mention what technique was used to get 0. They make it seem like it's a simple matter, but I'm not sure how to get that result. This is the integral: $\int_{0}^{\pi} \ln(2-2\cos{x}) dx$ and they get that this is $0$, and they state it like it's obvious. Maybe it is, but I would like to know how they got that result. Thank you! Last edited by ProofOfALifetime; September 25th, 2018 at 05:26 PM. Reason: Provided more detail
 September 26th, 2018, 01:27 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,913 Thanks: 1113 Math Focus: Elementary mathematics and beyond $$\int_0^{\pi}\log(\alpha-\alpha\cos x)\,dx=\pi\log\left(\frac{\alpha}{2}\right)$$ Follow the steps numbered 3 through 6 as they appear in the beginning of the article. Good luck finding the constant of integration mentioned in step 2 - I wasn't able to. Thanks from ProofOfALifetime
 September 26th, 2018, 06:12 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,307 Thanks: 1976 Let $\displaystyle I = \int_0^\pi \ln(2 - 2\cos x)\,dx$. By the substitution $x \to \pi - x$, $\displaystyle I = \int_0^\pi \ln(2 + 2\cos x)\,dx$. Hence $\displaystyle I = \frac12 \int_0^\pi \ln(4 - 4\cos^2\! x)\,dx = \frac12 \int_0^\pi \ln(2 - 2\cos 2x)\,dx$. By instead using the substitution $x \to 2\pi - x$, $\displaystyle I = \int_\pi^{2\pi} \ln(2 - 2\cos x)\,dx$. Hence $\displaystyle I = \frac12 \int_0^{2\pi} \ln(2 - 2\cos x)\,dx$. Now, by the substitution $x \to 2x$, $\displaystyle I = \int_0^\pi \ln(2 - 2\cos 2x)\,dx = 2I$ (by the previous result), so $I = 0$. Thanks from greg1313 and ProofOfALifetime
 September 26th, 2018, 10:51 AM #4 Senior Member     Joined: Oct 2016 From: Arizona Posts: 189 Thanks: 33 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. Thank you both so much. I have a question for skipjack! How do I find these kinds of trick integrals for practice, and how did you know to do that? What I mean is, how do I get practice doing these kinds of integrals that are not exactly, "textbook" ? I would like to get better at these. Thanks from greg1313 Last edited by ProofOfALifetime; September 26th, 2018 at 10:54 AM.
 September 26th, 2018, 03:19 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,913 Thanks: 1113 Math Focus: Elementary mathematics and beyond Thanks from ProofOfALifetime
September 26th, 2018, 04:55 PM   #6
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 Originally Posted by ProofOfALifetime Thank you both so much. I have a question for skipjack! How do I find these kinds of trick integrals for practice, and how did you know to do that? What I mean is, how do I get practice doing these kinds of integrals that are not exactly, "textbook" ? I would like to get better at these.
There are several YouTube channels which have a focus these 'non-textbook' style integrals. blackpenredpen and LetsSolveMathProblems are a couple which come to mind.

Another source might be the MIT integration bees (an event started at MIT which is a bit like a spelling bee but for integration). The 2018 bee can be found here, and solutions here. Note that the integrals increase in difficulty from 1 through 20.

edit: I guess for some the integration bee questions aren't really non-textbook.. in which case maybe check out some Putnam style integration problems.

Last edited by Joppy; September 26th, 2018 at 05:01 PM.

September 26th, 2018, 06:02 PM   #7
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 Originally Posted by Joppy Another source might be the MIT integration bees (an event started at MIT which is a bit like a spelling bee but for integration). The 2018 bee can be found here
There goes the next 6 hours of my night. Thanks

September 26th, 2018, 06:04 PM   #8
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 Originally Posted by SDK There goes the next 6 hours of my night. Thanks
My pleasure .

 September 26th, 2018, 06:32 PM #9 Senior Member     Joined: Oct 2016 From: Arizona Posts: 189 Thanks: 33 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. I am super super grateful for all the resources that have been mentioned, and I am enthusiastically checking them out! Thank you Joppy and greg1313. You guys are the best! Great to be back in touch with you! Thanks from greg1313
September 26th, 2018, 06:33 PM   #10
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Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.
Quote:
 edit: I guess for some the integration bee questions aren't really non-textbook.. in which case maybe check out some Putnam style integration problems.
That's okay, I am still grateful for them. I am actually training myself for an integration problem (more like a proof since they give you that it's $\frac{\pi}{6}$ but you have to prove it) featured in the American Mathematical Monthly! It is due in 35 days! I very much want to solve it! Hence, why I'm learning the Feynman integral trick!

Last edited by ProofOfALifetime; September 26th, 2018 at 06:37 PM. Reason: Just being more specific

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