What a journey man. 45 minutes of insanity but also fun. The most tricky part would be coming up with f(z) on your own. I wouldn't have come up with it and prob just used the given f(x) and ended up crying in the corner xD
its actually not that hard to come up with if you do it ‘forwards’ rather than ‘backwards’ as he showed it off for purposes of the video. You can easily notice that a cos^2 term would complete the square in the denominator, so simply rewrite 1 as sin^2 + cos^2 and you end up with a sum of squares which can be factored with complex numbers and then rewrite with euler’s identity to get whats essentially f(z)
Here is a little remark: Calling J(a) the integral, an immediate calculation shows that J(1/a) = a^2 J(a) for every non zero a. In order to get your final result you can therefore restrict yourself to the case a>1 and then deduce the other : If a>1 then J(a) = pi/a * log(1+1/a). Thus if 0
The function g(x)=xsin(x) /[2(1-cos(x)] from the interval [0, pi] to R (define it as you like in 0) is perfectly Riemann integrable, because the lim x-->0 of g(x) is 1 (in fact xsin(x) is asymptotic to x^2 and the same goes for 2(1-cos(x)) and it's defined in pi, so the integral in this case is not even an improper one: the function differs from a continuous one only on the set {0}, which has measure 0.
Hello Quan! Learning from mistakes is always good ;-) In fact I didn't look closely and confounded the exercise with the integral of sin(x)/(1+a^2-2acos(x)) from 0 to 2pi (which is improper when a=1 ). Now the good news is that this leads us to a another, maybe more natural way, to solve your exercise. Recall that some of your viewers asked how you invented the auxiliary meromorphic function; you had to admit that it was supplied by the exercise statement. Yesterday I took some time to redo your exercise with the same standard procedure that one would apply to sin(x)/(1+a^2-2acos(x)). You certainly know this procedure, it gives you in a natural way the auxiliary function to use. In our case the factor x in front of the sin creates some obstacles. But they can be overcome, and I'm grateful to you that this problem made me think about some interesting technical points (feel free to contact me for details). Of course, with the standard procedure I found the same result as you in the video. It was not easier, but it has the advantage that one comes up with the auxiliary function by oneself ;-)
Yesterday I challenged a friend with this integral. He sat down at his desk and finished within less than 15 minutes! His method is completely different, it uses only first year calculus. Wow ;-) (My friend teaches math in a french "classe préparatoire", i.e., he trains 18 years old students to do this kind of tricky exercises.)
Acompanho você a mais de 1 ano aqui do Brasil, já assistir todos seus vídeos da lista “complex analysis”, seus vídeos me ajudaram muito a me desenvolver, parabéns pelo trabalho.
42:55 Wouldn't we need to take a semi-circular epsilon contour around the pole and show its integral tends to zero in the limit, or is it actually legit to just take the limit as a goes to 1 of those two answers?
To justify taking the limit as I did, you'll need to use the dominated convergence theorem which works out fine in this case, probably should have mentioned that. In general, you wouldn't be able to interchange limits and integrals (which is what I essentially did). And yes, you could probably take a=1 as a special case, so an extra semicircle at the origin would be needed.
@@qncubed3 Do you have a guide or video on how to choose integration contours and f(z) functions for different types of integral? again thank you for sharing, you do very high quality work
What a journey man. 45 minutes of insanity but also fun. The most tricky part would be coming up with f(z) on your own. I wouldn't have come up with it and prob just used the given f(x) and ended up crying in the corner xD
its actually not that hard to come up with if you do it ‘forwards’ rather than ‘backwards’ as he showed it off for purposes of the video.
You can easily notice that a cos^2 term would complete the square in the denominator, so simply rewrite 1 as sin^2 + cos^2 and you end up with a sum of squares which can be factored with complex numbers and then rewrite with euler’s identity to get whats essentially f(z)
Here is a little remark: Calling J(a) the integral, an immediate calculation shows that J(1/a) = a^2 J(a) for every non zero a.
In order to get your final result you can therefore restrict yourself to the case a>1 and then deduce the other :
If a>1 then J(a) = pi/a * log(1+1/a).
Thus if 0
The function g(x)=xsin(x) /[2(1-cos(x)] from the interval [0, pi] to R (define it as you like in 0) is perfectly Riemann integrable, because the lim x-->0 of g(x) is 1 (in fact xsin(x) is asymptotic to x^2 and the same goes for 2(1-cos(x)) and it's defined in pi, so the integral in this case is not even an improper one: the function differs from a continuous one only on the set {0}, which has measure 0.
@@guidotoschi7284 You are right and I'm wrong! So I will have to try to understand in another way why the function a --> J(a) is not analytic in a=1.
Hello Quan!
Learning from mistakes is always good ;-) In fact I didn't look closely and confounded the exercise with the integral of sin(x)/(1+a^2-2acos(x)) from 0 to 2pi (which is improper when a=1 ).
Now the good news is that this leads us to a another, maybe more natural way, to solve your exercise. Recall that some of your viewers asked how you invented the auxiliary meromorphic function; you had to admit that it was supplied by the exercise statement. Yesterday I took some time to redo your exercise with the same standard procedure that one would apply to sin(x)/(1+a^2-2acos(x)). You certainly know this procedure, it gives you in a natural way the auxiliary function to use. In our case the factor x in front of the sin creates some obstacles. But they can be overcome, and I'm grateful to you that this problem made me think about some interesting technical points (feel free to contact me for details).
Of course, with the standard procedure I found the same result as you in the video. It was not easier, but it has the advantage that one comes up with the auxiliary function by oneself ;-)
Yesterday I challenged a friend with this integral. He sat down at his desk and finished within less than 15 minutes! His method is completely different, it uses only first year calculus. Wow ;-)
(My friend teaches math in a french "classe préparatoire", i.e., he trains 18 years old students to do this kind of tricky exercises.)
How did he do that? Have tried but can't get anything other than contour integration to work out nicely.
@@Decrupt If you contact me via email I will sent it to you.
@@BernhardElsner Thanks a lot for your time. Where can I find your email?
@@Decrupt On the internet ;-)
@@Decruptdo you need it?? I can solve it without complex analysis. Do you want me to write here?
im just in calc 2 but i dream with being able to solve integrals like these 😢
I have a contour integration playlist ordered by difficulty which hopefully helps if you're trying to learn this.
@@qncubed3 thank you so much!
Parabéns, sou apaixonado por análise complexa e sempre acompanho seu trabalho. Parabéns!
Acompanho você a mais de 1 ano aqui do Brasil, já assistir todos seus vídeos da lista “complex analysis”, seus vídeos me ajudaram muito a me desenvolver, parabéns pelo trabalho.
that was cool!
42:55 Wouldn't we need to take a semi-circular epsilon contour around the pole and show its integral tends to zero in the limit, or is it actually legit to just take the limit as a goes to 1 of those two answers?
To justify taking the limit as I did, you'll need to use the dominated convergence theorem which works out fine in this case, probably should have mentioned that. In general, you wouldn't be able to interchange limits and integrals (which is what I essentially did). And yes, you could probably take a=1 as a special case, so an extra semicircle at the origin would be needed.
You are a complex analysis guru 😊🥳
When I see the integral, I felt somewhat familiar with the integrand, and when I figured it out, it took me five minutes to solve it!!!
😳
It s incredible
15:20 For Xi I normally pronounce it like “Ksi”emphasising the k for psi I just say “si” like you do
👍
Greetings from UoM. Great thing, thanks for UA-cam algorithm to remind me complex analysis exam is coming up.
This was also filmed at UoM :D good luck for your exam!
@@qncubed3 wait whatttttttttt
quan goes to unimelb
@@firstname9845 correct
In my experience Xi is pronounced "Kai"
Thank you very much for sharing, why did you choose this function f(z)?
It was the function provided by the question I got
@@qncubed3 Do you have a guide or video on how to choose integration contours and f(z) functions for different types of integral? again thank you for sharing, you do very high quality work
Vrom wanna email 📧U 🥺🥺🥺