Dini's magical integral
Вставка
- Опубліковано 9 кві 2024
- 🌟Support the channel🌟
Patreon: / michaelpennmath
Channel Membership: / @michaelpennmath
Merch: teespring.com/stores/michael-...
My amazon shop: www.amazon.com/shop/michaelpenn
🟢 Discord: / discord
🌟my other channels🌟
mathmajor: / @mathmajor
pennpav podcast: / @thepennpavpodcast7878
🌟My Links🌟
Personal Website: www.michael-penn.net
Instagram: / melp2718
Twitter: / michaelpennmath
Randolph College Math: www.randolphcollege.edu/mathem...
Research Gate profile: www.researchgate.net/profile/...
Google Scholar profile: scholar.google.com/citations?...
🌟How I make Thumbnails🌟
Canva: partner.canva.com/c/3036853/6...
Color Pallet: coolors.co/?ref=61d217df7d705...
🌟Suggest a problem🌟
forms.gle/ea7Pw7HcKePGB4my5
So I(α) approaches the same limit from both sides of 1... It can be regularized into a continuous function
Indeed, it's a potential of a charged ring in 2D space, potentials are continuous (unless there is a charge bilayer = dipole density somewhere in the system). In fact, the constant value at α1 is just the fact that from the outside, a ring and a point are indistinguishable, just like you can use Newton's law of gravity for a point or a sphere in the same form.
Beautiful problem and an elegant solution
The thumbnail.......
The thumbnail has the solution
=|pi/a^2-1|
Talk about a spoiler amiright
@@rafaelfreitas6159 my bad
@@devilspattern6164 no, lol relax I was jokingly referring to the integral: it clearly suggests Feynman trick (which's kinda obvious to be honest).
@@rafaelfreitas6159 I solved it with substitution
I'm not sure you can make that conclusion at 14:30. You know that I'(α) = 0 on (0,1) so it's constant on (0,1). But knowing I(0) = 0 doesn't help you since it's not on the interval.
However, you do know that cos(x) belongs to [-1 , 1] so 1-2αcos(x)+α^2 belongs to [1-2α+α^2 , 1+2α+α^2] = [(1-α)^2 , (1+α)^2]. ln() is a strictly increasing continuous function, so ln(1-2αcos(x)+α^2) belongs to [2ln(1-α) , 2ln(1+α)]. So our integral, I(α) belongs to [2πln(1-α) , 2πln(1+α)].
If we assume I(α) equals some positive ε, we could simply shrink our α down to the point where our interval no longer contains it, but we're still on (0 , 1). So clearly no positive ε could satisfy the condition that we're in the interval [2πln(1-α) , 2πln(1+α)], while also holding I(α) constant. Same idea for any negative ε. Thus, the only possible number left that I(α) can take is 0.
I assume the conclusion works if there's also some (not covered here, but probably easily demonstrated) assumption that the antiderivative has to be smooth everywhere besides the pole at 1. So the only way it can be smooth is by being equal to 0.
Just ordered "Inside Interesting Integrals." Thank you for the reference Michael!
There's a "deep-dive" into integrals of these form, with respect t ox and with respect to alpha, in the book "a treatise of the integral calculus - second volume", which is free on the internet archive.
Just as an example, if m,n,r,v,u are integers such that m=ru and n=rv, then:
integral 1/(1-2tcos(mx)+t^2)*1/(1-2scos(nx)+s^2)dx from 0 to pi = pi/(1-t^2)*1/(1-s^2)*(1+t^v*s^u)/(1-t^v*s^u)
BTW, isn't that the generating function of something like legendre polynomials or some such?
@@archismanrudra9336 at a power of negative one half I believe.
Here is a solution without doing any calculations:
You can have 2π in the upper limit if you add a square root inside the log.
As a physicist, I see this as 2D potential (e.g. electrostatic) of a ring with unit radius, at distance α, because the argument is just the cosine law (abs value of vector distance squared).
As a consequence, we can use the symmetry to see that the result will still be a logarithmic potential of a point "particle" (just like a sphere and a point mass have the same potential on the outside), while on the inside, you have a Faraday cage effect, so the potential will be a constant (no electric field inside). We know this constant because we can evaluate it at α=0, which is trivially I=0. On the outside, the result will be 2*π*ln(α), where the 2π is the "mass/charge" of the ring (its length). We know there is no constant added because the potential must be continuous between inside and outside.
One cute thing you can do is expand it as a power series in alpha and that will give you an infinite sequence of simpler integrals as coefficients.
I had a differentiation under the integral sign, but simplified the ln term to ln ((1 - alpha e^{i x}) (1 - alpha e^{-ix})), eventually simplifying the derivative wrt alpha to i times integral of 1/(1 - alpha z) around a unit circle counter-clockwise. Since the pole is at 1/alpha, this makes clear that if alpha is between -1 and 1, the derivative is 0, and you can evaluate the residue in the other case quite simply
(getting the same answer, of course)
The formula for any alpha is really nice : max(0, 2π ln |alpha|)
while trying to work on it, I prooved that this integral equals 1/2**n times the integral from 0 to pi of ln(a**(2**n+1) + 2a**(2**n)cost + 1) dt for all natrual number n then I used the case of a in 0,1 to then generalize it and when n goes to the infinity and it gave me the same answer
Expnading the denominator as geometric series after dividing out by 1+a^2 gives you some series representations relating to the catalan numbers.
Edit: bringing this to its conclusion you get I'(a)=pi/a((a^2-1)/abs(a^2-1)+1)
For α = 1, the integral also exists and is equal to zero.
Is that the same Dini as in Dini derivative?
I'm not saying people should investigate the Internet Archive if they want to peer Inside Interesting Integrals, but you should. Yes, specifically you.
thanks for the shoutout
Mike do a video with Dini derivatives.
I love 1 - two tacos
16:55
How fucking fast are you, Jesus, it's one minute after he posted
In the computation you divide by alpha, you have to assume alpha is not 0.
...which is why he pointed out that you can compute I(0) from the original equation before getting into the simplifying calculations for alpha0, no 0/0 step used. The longer calculations are for all the other cases.
@@BridgeBum I know but it's better in mathematics to be rigourous to avoid to tell bullshit.
Poisson❤❤
What has this to do with Poisson?
@@bjornfeuerbacher5514after differentiating we have the poisson kernel
@@bjornfeuerbacher5514after differentiation its basically integral of the poisson kernel
my first thought as well
Fourier analysis 😅
What does this have to do with Fourier analysis?
I believe that you can say from the beginning why alpha cannot be one. If alpha is one you will get ln(0) when we include the upper bound.
Yeah but I think that divergence can be tamed in a principal value sense. If I'm not mistaken, a similar if not identical integral appeared in my research and evaluates to pilog(2) or something lol. I have it inside one of my envelopes...
Why can't alpha be negative?
It actually can, but it produces the same values. Basically, I(α) is an even function.
@@ShaunakDesaiPiano thanks!
When alpha is 1 the integral is equal to zero
-How so? At 0 the integrand goes like ln(x) (it's ln(1 - cos(x)) + const. So, in effect ln(x^2) = 2 ln(x)).
Not integrable.-
nvm
@@beatn2473 you seem to be mistaken in the fact that lnx is not integrable around 0 when it is. Its antiderivative is x(lnx-1). Near zero this approaches 0 since linear terms dominate logarithmic ones
@@skylardeslypere9909 you are right of course. I must have been drunk or something.
Nice but skipping some passages it has become somehow difficult to follow
Noice
the integral of an analytic function is piecewise continuous? Why does calculus have to be so mean😂
sorry but you are not doing math, you are doing physics, there's no rigor in your arguments
There is always 1
@@williammartin4416 ??
The unrigour cancels out in the end
Noice