I know nothing about it but I became interested in it rn when I saw him use it on an integral that I thought could only cleanly be done w/Feynman’s technique. Would you say if I were to fully learn all concepts used in this video(and ofc be able to replicate em in other problems), i would have learned the essence of complex analysis? Also, what would you describe the point of complex analysis now that you’ve become well-versed in it 😅 based on how it sounds, is it like a deep dive into the utility of the complex plane for solving problems?
*Happy first contour integral with chalk board* Yeah, I watched a whiteboard version of it before, but with some difficulty. But this one is great, in all aspects. And ... Please when you are busy, at least make short videos. Thank you so much dear *QN³* ❤️
Thanks for this great video and explanation. Just a question, at 7:16, should there be two or three poles in the lower right quadrant (positive Real, negative Imaginary)?
It doesn't matter since this is only a rough sketch of where the poles could be. Depending on the value of n, the number and position of the poles will be entirely changed. The only pole that we are concerned about is the first one.
@@qncubed3 It is not at all obvious, though nonetheless true, that the integral value remains the same if the pizza slice includes the first two nth-roots of (-1), or the first three, .... or in fact all of them, in which case you have the whole pizza minus a slice with no roots in its interior. Proving that the integral does not depend on how many residues you trap inside your region of integration would be a great exercise. I did it for the first two, and the result is far from obvious until the very end, when a magical simplification comes to save you in the nick of time! Will try to find the general answer tomorrow.
A much simpler aproach, about how i solved it. Substitute x=t^1/n This makes dx=t^((1/n)-1)dt The given function resolves to the form of beta function. Which later simplifies into eulers reflection formula
I'm not sure if contour integration can be used to evaluate indefinite integrals. However, here's a related post I found :) math.stackexchange.com/questions/1999869/evaluate-int-frac11xndx-for-n-in-mathbb-r
@@Pommes736 If you want to gain intuition for things like this, use software to numerically solve your integrals for different bounds; this will inform you of the answer immediately, although to prove that they are equivalent may be difficult.
@@davidraveh5966 Oh you didn't understand my question. I can solve these integrals without problem, my question was if I can use THIS METHOD for INDEFINITE integrals.
What if we replaced n by 5, how will the integfation be, and what will the answer be?? Do i just replace n by 5 in all the steps of the solution and in the final answer or what??
@@harisserdarevic4913 try it for n=5 using this method and try it using decompoaition and if you reached same answer then it is correct. I think for odd power, it has another way of solving
But why are you allowed to choose a contour that's only around a single pole? Why not choose a contour that encloses 2 poles? How diff would the answer be?
Note: Typo at 3:55 should be an element symbol instead of equality ... silly me
Without any doubt:
You're *The King Of Complex analysis* on RUclips.
Please continue this playlist.
Thank you 💖
math505 is cool too
HE JUST WANTED TO GIVE AN EXERCISE IN COMPLEX ANANYSIS FOR THERE ARE MUCH MORE SIMPLER WAYS TO SOLVE ruclips.net/video/bshl5HqiAYA/видео.html
He sorta looks like Jacob Collier...
When I first did this integral and got the right answer, I knew finally that I really understood complex analysis.
I know nothing about it but I became interested in it rn when I saw him use it on an integral that I thought could only cleanly be done w/Feynman’s technique. Would you say if I were to fully learn all concepts used in this video(and ofc be able to replicate em in other problems), i would have learned the essence of complex analysis? Also, what would you describe the point of complex analysis now that you’ve become well-versed in it 😅 based on how it sounds, is it like a deep dive into the utility of the complex plane for solving problems?
this channel is so f underrated ! .. the best on complex analysis thank you
underrated by whom please?
*Happy first contour integral with chalk board*
Yeah, I watched a whiteboard version of it before, but with some difficulty. But this one is great, in all aspects.
And ...
Please when you are busy, at least make short videos.
Thank you so much dear *QN³* ❤️
Blackboard videos are noice.
Thanks for this great video and explanation. Just a question, at 7:16, should there be two or three poles in the lower right quadrant (positive Real, negative Imaginary)?
It doesn't matter since this is only a rough sketch of where the poles could be. Depending on the value of n, the number and position of the poles will be entirely changed. The only pole that we are concerned about is the first one.
@@qncubed3 ah ok, thank you!
@@qncubed3 It is not at all obvious, though nonetheless true, that the integral value remains the same if the pizza slice includes the first two nth-roots of (-1), or the first three, .... or in fact all of them, in which case you have the whole pizza minus a slice with no roots in its interior. Proving that the integral does not depend on how many residues you trap inside your region of integration would be a great exercise. I did it for the first two, and the result is far from obvious until the very end, when a magical simplification comes to save you in the nick of time! Will try to find the general answer tomorrow.
A question: residue method can only be used to calculate definite or improper integrals but not for indefinite in order to obtain only the primitive?
A much simpler aproach, about how i solved it. Substitute x=t^1/n
This makes dx=t^((1/n)-1)dt
The given function resolves to the form of beta function. Which later simplifies into eulers reflection formula
Can you make a video explaining contour integrals?
great job!
wow this is great!
you saved me thank you
An interesting, alternative form for the final answer:
I = (1/n) * Γ(1/n) * Γ(1-1/n)
I = Γ(1+1/n) * Γ(1-1/n)
I = B(1+1/n, 1-1/n)
nice one
We are waiting .....
🧐 It's me, looking at screen, for your notification 🧐
Videos coming back by the end of this week :)
Finalllllly ......
Interesting the result looks like the reflection formula for the gamma function but with 1/n
Greattttttt video! However, can n be non-integer?
Is there a way to compute the indefinite integral of this with complex analysis or do you have to have bounds?
I'm not sure if contour integration can be used to evaluate indefinite integrals. However, here's a related post I found :)
math.stackexchange.com/questions/1999869/evaluate-int-frac11xndx-for-n-in-mathbb-r
@@qncubed3 I'm not interested in this school integral per se. I wanna know if it's possible in general for any function without any bounds.
@@Pommes736 If you want to gain intuition for things like this, use software to numerically solve your integrals for different bounds; this will inform you of the answer immediately, although to prove that they are equivalent may be difficult.
@@davidraveh5966 Oh you didn't understand my question. I can solve these integrals without problem, my question was if I can use THIS METHOD for INDEFINITE integrals.
That's pretty nice. Would it be possible to generalize this result for R=1?
There’s actually a better method divide the denominator and numerator with x^n and then apply partial fraction and then resolve the contour
Thank you for that wonderful piece of delivery, pls, can you help when n=5 , I.e f(x) = 1/ x^5 + 1
all u need is the beta function then put into gamma form and use euler's reflection formula
this video looks like asian flammable maths
:O
Bad boy you don't clean up your own mess 😤
What if we replaced n by 5, how will the integfation be, and what will the answer be??
Do i just replace n by 5 in all the steps of the solution and in the final answer or what??
uh yeah thats what it means to solve something for a general variable n. it holds for any n>1 so you don't have to redo any work
@@harisserdarevic4913 try it for n=5 using this method and try it using decompoaition and if you reached same answer then it is correct. I think for odd power, it has another way of solving
I would probably calculate it with Beta function then change it to Gamma function , finally i would finish it with reflection formula for Gamma
After a such long time.......
:D
But why are you allowed to choose a contour that's only around a single pole? Why not choose a contour that encloses 2 poles? How diff would the answer be?
It is possible, but then you would have to calculate two residues
lovely accent as well!
cheers, from straya mate
lovely country
I thought you were english or something by the way you dressed lol
I wish you were my classmate!
Good
P I Z Z A
Blackboard videos
Bêta fonction mène à la même résultat
Je crois que bêta mene a la même résultat