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/ayoungervoice/r...ke7afae85e25c 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: $\int_{0}^{\pi} \ln(22\cos{x}) dx$ 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. 
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 $I = 0$. 
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.
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  
September 26th, 2018, 04:55 PM  #6  
Senior Member Joined: Feb 2016 From: Australia Posts: 1,765 Thanks: 623 Math Focus: Yet to find out.  Quote:
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 nontextbook.. 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  
Senior Member Joined: Sep 2016 From: USA Posts: 562 Thanks: 325 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
 
September 26th, 2018, 06:04 PM  #8 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,765 Thanks: 623 Math Focus: Yet to find out.  
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!

September 26th, 2018, 06:33 PM  #10  
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.  Quote:
Last edited by ProofOfALifetime; September 26th, 2018 at 06:37 PM. Reason: Just being more specific  

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