How Richard Feynman would evaluate this monster log integral
HTML-код
- Опубликовано: 25 июл 2024
- Slaying yet another beast of an integral using Feynman's integration trick. The solution is surprisingly elegant and the satisfying result makes it all the more epic!
At the 20:25 mark I forgot the modulus operator on the the argument of the natural logarithm. However, it didn't affect the solution as we end up multiplying complex conjugates anyway. However, I should not have omitted it as it leaves a hole in the solution development.
The modulus operator will remain and on adding I(i) and I(-i) the moduli of two complex conjugate numbers will be multiplied (due to the logarithms) giving us exactly the same result.
I believe that your trig sub was overly complicated. Because you are in the complex plane, you can reduce as follows (1 - t^0.5) / (1-t) = 1 / (1 + t^0.5) and then solve your integral with the much simpler u sub u = 1 + t^0.5.
@@danielkanewske8473 yes I agree
@@maths_505 At 20:25, where you are referencing, I noticed that the | sqrt(i)+1 | and | sqrt(-i)+1 | terms each can change depending on which root of i or -i you take. If you were to calculate each term separately, and then multiply them, rather than combing the term into a single expression then foiling, you could get a wrong answer if you take the wrong root of i or -i. what is the reason for this?
yo creo que te podemos perdonar jeje...
How is this over powered?
This technique was developed by Leibniz, one of the inventors of calculus (whose notation we still use today). It's silly to call it the Feynman technique when the inventor of calculus used it.
He’s probably talking about how it’s the less common method of integration that Feynman was taught (and used to frequently solve complex integrals that gave others trouble). In “Surely You’re Joking” Feynman refers to it as “integrating under the curve” and explains how it is an example of why having a diverse “toolbox” of skills helps you approach problems differently and come to novel conclusions that other may have overlooked.
Yes it’s not his method but I think Feynman gives it a nice story, whereas “Leibniz” method is just a bit dry and doesn’t have the same connotations.
@@TheScreamingFedora When possible we try to name things after their originators. We don't do a good job and there are tons of exceptions, but it just doesn't make sense to do so in this case just because of the Feynman fan club, since this technique is quite old and used to be pretty ubiquitous. Another bubble to burst: Julian Schwinger has as good of a contribution to QED as Feynman, but was a not a press-hungry "curious character". Feynman got all of popular coverage (which he actively sought out) and thus is more widely known, while Schwinger modestly curated a reputation as a master amongst serious researchers.
If I'm not mistaking, in France we call it Leibniz' technique
@@drillsargentadog yet no one nowadays takes inspiration from him. So... who cares?
@@planomathandscience you’re obviously not studying physics, so your opinion is likely not going to be shared by people that are studying it
The antiderivatve in 14:58 can be done easier by noticing that (1-t) can be written as (1+sqrt(t))(1-sqrt(t)), and this last one cancels with the numerator, leaving us with the integral of dt/[sqrt(t)·(1+sqrt(t))]
Now perform a substitution making u = sqrt(t), du=dt/2sqrt(t) => int of dt/[sqrt(t)·(1+sqrt(t))] = int of 2·du/(1+u) = 2·ln(1+u) + C = 2·ln(1+sqrt(t)) + C
Deciding between contour and Feynman's is liek deciding between nuking and nuking harder...
Its actually fun trying both
Normally one can "sense" which technique would be more efficient and try that....and then there's this integral....so its actually pretty satisfying to solve it both ways and see which technique drops colder
@@maths_505 Imma try it using contour integration, and see if we can use some techniques to make it simpler, WE MUST FIND A WAY TO NERF FEYNMANN'S TECHNIQUE! IT HAS GONE FAR ENOUGH!
@@manstuckinabox3679 once you go Feynman....there ain't no turnin back!
Jeez man. Relax
U liek mudkipz?
I used log(x^4+t^4) instead to avoid dealing with complex values of t. Got the same value.
this was incredibly satisfying to watch. awesome video!
Just amazing and rigorous! I like how you used complex analysis, Euler’s formula and trigonometric substitution to arrive at the result. Thanks for sharing your knowledge and skills. I find it interesting how the argument of ln is the irrational constant pi. It seems e is the shadow of pi. Pi and e are transcendental numbers.
Feynman's Technique: Knowing the answer to everything
beautiful solution. keep rocking the integrals.
Cool! Enjoyed it from start to finish
Your channel is criminally underrated. I hope you’ll get the subscribers and views you deserve.
By the way, amazing video as always. Kudos!
criminally ? 💀
The flow of the solution was awesome and stimulating. Good work kamaal 👌
Gloriously pleasing. Chapeau!
I was eating dinner when I found this video. Now my dinner is cold, but I just found a new magical technique!
Nice solution! I mentioned a solution of mine using Feynman's trick and only real analytic methods in the comments of qncubed3's video. No complex numbers needed! Here it is: first, factor x^4 + 1 into (x^2 + sqrt(2)*x + 1)(x^2 - sqrt(2)*x + 1), then use log(ab) = log(a) + log(b) to split the integral into two separate, but very similar integrals. Under the substitution u = -x, it becomes clear that these two integrals are equivalent, leaving only one integral to solve. From there, you can use Feynman's trick to evaluate the integral of the parameterized function log(x^2 + tx + 1)/(x^2 + 1) w.r.t. x from -inf to inf, then evaluate I(sqrt(2)). After taking the partial derivative of the integrand w.r.t. t, what follows is just simple calculus integration techniques. To find the initial condition, set t = 0 in the parameterized integral, and employ the substitution x = tan(u). Here we run into the integral from 0 to pi/2 of log(cos(u))du, which is a famous integral, commonly solved using the symmetry of the integrand.
That's absolutely amazing!!!
I'll upload another video on this integral using Feynman's technique using your approach. Just let me know how to pronounce your name so I can properly credit it to you in the video.
@@maths_505 thank you so much! My username is a a blending of "violin" and "integral" since playing violin and math are my two favorite things. It's pronounced violin-tegral or equivalently viol-integral since the "in" in violin and integral are the same sound.
@@maths_505 also, have you attempted any of the 2022 MIT Integration Bee integrals? They are quite difficult and could make for some very interesting videos. I've only seen a few of them solved on other channels.
I solved a few of the fun ones on the qualifying round but I haven't seen the integrals from the competition yet
@@maths_505 the quarterfinal round has some really nasty limits of integrals which I have yet to see any solutions for
That was awesome! When you brought up complex numbers, I knew where this was going. I love it when you can step into the world of imaginary numbers only as a means of getting back to a real solution - stepping back into real numbers. It's like playing off-board chess. You can jump off the board briefly -- as long as you jump right back onto the board. That's how I envision it anyways. Cheers! Great video!
The solution actually assumed implicitly that t was a pure imaginary number. All the calculations performed are valid for complex numbers so we basically never left
Wow, thank you for the fun ride!
I personally found that most problems with complex numbers involved often end up being a massive algebraic extravaganza in order to simplify at the end. It's not the most exciting thing in the world to go through that process, but you end up with a beautiful answer afterwards, which is the only hope we have before delving right into simplifying!
Given how many contour integrals I've done recently, I can only say that π is following me around like no other before. It's everywhere!!! I'm starting to think that most Calculus problems I've solved had π in them because whoever developed the Math behind the concepts just sneakily hard-wired it in!!
It's a little conspiracy that has proven itself to me time and time again. But until we find the culprit, let's just all enjoy the Math. 😂😂
It's those sneaky egyptians! we knew they went irrelevent after the dawn of the 1st century...
This reads like a bot wrote this, surpirsed about complex numbers and pi lol
@@osamaattallah6956 What is next, that pesky "e" number? Incredible!!!
@@urosmarjanovic663 That damned Oily Macaroni Constant that likes to jump scare at random times. 0.577215664901532860606512090...
WTF is that all about???
As a rule of thumb, when you see that there is a lot of simplification right before the answer, it usually means that there was a faster way to do that.
In this case, I'(t) could be integrated (with respect to t) in a much shorter way by substituting t=u^2, while the last part could have been faster without recurring to the Euler's formula (simply multiplying the complex exponentials).
Thanks for showing this .
This just looks like a specific application of a more general approach called the Continuation Method (also sometimes invariant imbedding) to solving all sorts of problems, from root finding to nonlinear differential equations. Wasserstrom 1973 is a nice review of it. Didn't know about any attribution to Feynmen in its development. Very nice video!
it's attributed to Leibnitz
@@lolilollolilol7773 True, but Leibnitz's method also pertains only to integration and so it is also just a specific application of a much more general method.
follow your explanation is like to listen to a detective story, great!
thank you for your interesting content you make math seems to like very simple
Beautiful
Fantastic class
The solution is so beatiful😮
Just subbed, great channel!!
That indeed was awesome 👌
I kept thinking you were done but you simplified it even further 😂
I love how this piece went from very easy to hard to harder to almost impossible
Put this channel in the teaching portion of your CV bro you've earned it.
Already done
Can you show us an example of Feynman's technique solving fractional derivative of a spherical special function such as the Bessel function?
I haven't looked at integrals since calc 2 in college almost 15 years ago so i don't understand anything beyond the first 2 minutes but the final answer is truly elegant
Actually you can also just figure it out by knowing what ln(sin(x)) integrated over (0, pi/2] is
Edit: I wrote all real numbers instead of (0, pi/2] by mistake
thank you sir❤
Can please someone explain to me why at 8:32 we can jiust subsitute u=x when we earlier substituted x= 1/u. I cant make sense of this
Wow 👌 👏, thank you 👍
An interesting thing I found is if you do this same integral with ln(x^2+1) instead of ln(x^4+1) you get 2pi*ln2, meaning there's probably some sort of general formula for integrals like this
I believe you can derive a formula for integrals of the form ln(x^n +1)/(x^2+1) through the use of complex analysis, namely contour integration. Might be difficult, as you'll have as many of what are known as branch cuts as your power of n, which may be a bit of a pain to go through (since you'll have to compute I believe around 6+4n integrals, that's a rough estimate. However, most of them go to zero anyway), but I believe it's doable.
Sounds like residue theorem to me as there u always have 2pi*i * res(z)
Cheery cheery cheery color, and voice is a service
super neat.
for antiderivative at 15:00, you could use difference of squares to get 1/(1+sqrt(t)) * 1/sqrt(t). Then you can make U = sqrt(t). Good video though, I liked how you integrated ln(x)/(1+x^2), I'm not used to using techniques like that, even though I've seen it used a few times. Any tips on how to spot stuff like that?
It's mostly hit and trial but it works pretty well with logarithmic bois especially on this interval.
nothing is more overpowered than guessing the solution
integral at 15:24 can be done by substituting sqrt(t) as u after simplifying by canceling the 1-sqrt(t).
I loved watching this tour de force.
However, I found myself wondering, what was Feynman's technique? What was special or different about it? I heard some discussion about other techniques at the beginning but I'm still not getting what makes this unique or special.
I remember learning this in my applied mathematics 1 class
Very entertaining delivery! I would enjoy watching you solve this using contours.
But I wouldn't enjoy solving it😂
Check out qncubed3. He solved it using complex analysis
At 21:23 dont you need to consider both square roots of i?
My favourite part is when he said binomial expansion time and binomial expansioned all over the place. Truly, one of the maths of all time.
I had flashbacks to QM2 - not PTSD quality but slightly stressful. We effed around with this stuff for half a year non-stop.
I managed to do most of the exercises - and in the end I had developed a perverse liking to it - but lots of trees lost theirs lives in the process.
I got my BS in mathematics and just wanted to say be proud of your knowledge of mathematics. To me, this is well above and beyond other scientific fields. I truly believe that there is such thing as math brain and not everyone finds this stuff prideful or interesting. The few that do are the ones that are carrying the progress of the future. Fascinating stuff.
that's Feynman's technique™ !
The trouble with using contour for this problem is that ln(1+z^4) has a singularity at z=+/- sqrt(i) that's non-removable. It can still be done but, as you said, not easy.
A question.
Can't we use Infinite Geometric Series for the 1/(x^2+1) and convert it to a sum series- Integration?
We may then use by parts formula and then simplify it perhaps?
not helpful. that series is va;id when IxI
Prof Fred Adams, "If you use it once its a trick, if you use it twice its a technique"
Great video! What drawing app are you using?
Could you do a video covering when you can differentiate under the integral ? i.e., what does it mean for the integrand to converge therefor allow for the partial derivative inside the integral?
Search up Dirichlet's convergence theorem for integrals....that'll help you decide on convergence and switch up of limits.
Hm I had done this factoring differently. Instead of factoring into complex variables I factored this into x^2 +/- sqrt2 x +1, and used a parameter on the sqrt2
That's the other video on this integral 😂
What does it looks like ?
every example I have ever seen of using Liebniz' Integration Rule, was an example of solving an DEFINITE integral. Yes, sometimes an improper definite integral, but always a definite integral. Does anyone have an example of using the technique to solve an indefinite integral, that is when one limit of integration is a variable?
Video: "We can try solving this integral with the Feyman technique"
Me, sat on the sofa eating chips and having no idea what that means: "....Go on"
why do we replace -1 with i when i = -1^1/2 ???
Man this video is nostalgic
How so?
take Residue theorem into consideration and expand the integration core?
Awesome! I'm currently studying physics (2nd year) and was just curious of the Feynmann's method, I have just one question though.
Doesn't nullifying the factors one by one at 11:38 create any kind of problem? Didn't we obtain the equation 1=A(...) + B(,,,) from the fraction 1/(...)(,,,)? Shouldn't it mean that we're dividing by zero?
The equation 1=A( )+B( ) is valid for all x so there's no harm in extracting the values of A and B from that equation. In fact, any arbitrary values of the variable will work. You can try it out and believe me you'll like the results.
At 8:40 when you change back to the x world from u, you didn't change the du into dx which would give you du=-1/x^2 so the assumption that I(0)=-I(0) being equal to zero was not a true assumption, right? Not sure if there is something I'm missing here. Granted I'm a new calc 3 student so this integral is not something I've worked on before...
In terms of definite integrals, the u's and x's are just dummy variables; meaning you can name them whatever you want. All you have to do is rename the du to dx. It's not a substitution back into something.
What matters here is structure: if the functions involved and the limits look exactly the same (only difference being the name of the variables) the integrals are the same. You can find this in literally any cal2 book.
In case of the indefinite integral (antiderivative), the variables are no longer dummy variables and yes you would've had to substitute back the relationship for the final answer.
14:58 this could have been done easily if you factorise the 1-t into (1+sqrt(t))(1-sqrt(t))
Then integral become 1/sqrt(t)(1+sqrt(t)
This can be easily solved by putting 1+sqrt(t) as u
Did you get the idea for this integral from qncubed3? I often take integrals (including this one😉) from his channel and solve them on mine using Feynman integration.
Yup. I saw his video using contour integration which was pretty cool but I wanted to try this using Feynman's technique and the result is indeed marvelous!
Feynman's technique definitely beats contour integration for this beast of an integral!
Just checked out your video.
It's the exact same line of thought which is awesome!
I skipped most of the video though obviously cuz I knew what would happen but I found the explanation quite nice
What app are you using?
The Residue Theorem is clearly more overpowered, since you brought up complex numbers.
What app is he using?
The thumbnail is like “if Feynman was a Platinum-record selling rapper”…. Lmaooo
I'm pretty sure Feynman would treat every video on this technique as a diss track towards contour integration 😂
Random, which app do you use to record the lecture ?
lost my shit laughing when you pulled the pi into the exponent
21:30 Doesn’t the square root of i have 2 roots (they are offset by 180 degrees).
don't you need the convergence of the original integral ensured to split it on the sums of integrals?
How about using residue theorem? This would be simpler... but I'm not sure...
To think that I once thought long division was complicated
What tablet / software do you use for your videos?
At 8:29 you eliminated u = x at last of integral but few times before you have let x = 1/u => u =1/x how you eliminated u=x? how can we do that please explain me
This was awesome!!!!!!!!
Why is there a parental advisory sticker😂
Why not😂😂😂
Feynman's technique? This is in fact the Leibniz technique!
PI just has to show it self everywhere
using I for both the final solution and the integral function is potentially very confusing
Nice
For the trig sub, a much easier solution is to see that (1-sqrt(t))/(1-t) = 1/(1+sqrt(t)) and then sub u=1+sqrt(t).
Man i still have problems visualising the countours before integrating
Integrand is even so we can integrate it only from 0..infinity and double the result
ln(x^4+1) = Int(4x^4t^3/((xt)^4+1),t=0..1)
so we have
Int(1/(x^2+1)*Int(4x^4t^3/(x^4t^4+1),t=0..1),x=0..infinity)
Int(Int(4x^4t^3/((x^2+1)(x^4t^4+1)),x=0..infinity),t=0..1)
Is it correct or we may choose better our parameter
As we can see this approach is similar to the Leibnitz's differentiation under integral sign
Int(4x^4t^3/((x^2+1)(x^4t^4+1)),x=0..infinity)
u=xt
du=tdx
dx=dt/t
Int(4u^4/t*1/((u^2/t^2+1)(u^4+1))*1/t,u=0..infinity) , t>0
Int(4u^4*1/(t^2(u^2/t^2+1)(u^4+1)),u=0..infinity)
Int(4u^4/((u^2+t^2)(u^4+1)),u=0..infinity)
Int(4(u^4+1-1)/((u^2+t^2)(u^4+1)),u=0..infinity)
4Int(1/(u^2+t^2),u=0..infinity)-4Int(1/((u^2+t^2)(u^4+1)),u=0..infinity)
(u^4+1) - (u^2 + t^2)(u^2 - t^2) = (u^4+1) - (u^4 - t^4)
(u^4+1) - (u^2 + t^2)(u^2 - t^2) = 1+t^4
4Int(1/(u^2+t^2),u=0..infinity)-4/(1+t^4)Int(((u^4+1) - (u^2 + t^2)(u^2 - t^2))/((u^2+t^2)(u^4+1)),u=0..infinity)
4Int(1/(u^2+t^2),u=0..infinity)-4/(1+t^4)Int(1/(u^2+t^2),u=0..infinity)+4/(1+t^4)Int((u^2-t^2)/(u^4+1),u=0..infinity)
(4 - 4/(1+t^4))Int(1/(u^2+t^2),u=0..infinity)+4/(1+t^4)Int((u^2-t^2)/(u^4+1),u=0..infinity)
4t^4/(1+t^4)Int(1/(u^2+t^2),u=0..infinity) + 4/(1+t^4)Int(u^2/(u^4+1),u=0..infinity)-4t^2/(1+t^4)Int(1/(u^4+1),u=0..infinity)
4t^3/(1+t^4)Int(1/t*1/(1+(u/t)^2),u=0..infinity) + 4/(1+t^4)Int(u^2/(u^4+1),u=0..infinity)-4t^2/(1+t^4)Int(1/(u^4+1),u=0..infinity)
Int(u^2/(u^4+1),u=0..infinity)
u=1/w
du = -1/w^2dw
Int(1/w^2/(1/w^4+1)(-1/w^2),w=infinity..0)
Int(1/w^2/(1/w^2+w^2),w=0..infinity)
Int(1/(1+w^4),w=0..infinity)
Int(u^2/(u^4+1),u=0..infinity) = Int(1/(1+w^4),w=0..infinity)
4t^3/(1+t^4)Int(1/t*1/(1+(u/t)^2),u=0..infinity) + 4(1-t^2)/(1+t^4)Int(u^2/(u^4+1),u=0..infinity)
Int(u^2/(u^4+1),u=0..infinity) = 1/2Int((1+u^2)/(u^4+1),u=0..infinity)
1/2Int((1+1/u^2)/(u^2+1/u^2),u=0..infinity)
1/2Int((1+1/u^2)/((u-1/u)^2+2),u=0..infinity)
u-1/u = sqrt(2)y
(1+1/u^2)du= sqrt(2)dy
sqrt(2)/2Int(1/(2y^2+2),y=-infinity..infinity)
sqrt(2)/4Int(1/(y^2+1),y=-infinity..infinity)
sqrt(2)/4π
4t^3/(1+t^4)*π/2+4sqrt(2)/4π(1-t^2)/(1+t^4)
2πt^3/(1+t^4)+sqrt(2)π(1-t^2)/(1+t^4)
π/2Int(4t^3/(1+t^4),t=0..1) - sqrt(2)πInt((t^2-1)/(t^4+1),t=0..1)
π/2Int(4t^3/(1+t^4),t=0..1) - sqrt(2)πInt((1-1/t^2)/(t^2-1/t^2),t=0..1)
π/2ln(1+t^4)|_{0}^{1} - sqrt(2)πInt((1-1/t^2)/((t+1/t)^2-2),t=0..1)
π/2ln(2) - sqrt(2)πInt((1-1/t^2)/((t+1/t)^2-2),t=0..1)
t+1/t=sqrt(2)y
(1-1/t^2)dt=sqrt(2)dy
π/2ln(2) - 2πInt(1/(2y^2-2),y=infinity..sqrt(2))
π/2ln(2)+ πInt(1/(y^2-1),y=sqrt(2)..infinity)
π/2ln(2)+ π/2Int(2/(y^2-1),y=sqrt(2)..infinity)
π/2ln(2)+ π/2Int(((y+1)-(y-1))/((y-1)(y+1)),y=sqrt(2)..infinity)
π/2ln(2)+ π/2(Int(1/(y-1),y=sqrt(2)..infinity)-Int(1/(y+1),y=sqrt(2)..infinity))
π/2ln(2)+ π/2ln((y-1)/(y+1))|_{sqrt(2)}^{infinity}
π/2ln(2)+ π/2(0-ln((sqrt(2)-1)/(sqrt(2)+1)))
π/2ln(2) - π/2ln((sqrt(2)-1)/(sqrt(2)+1))
π/2ln(2(sqrt(2)+1)/(sqrt(2)-1))
π/2ln(2(sqrt(2)+1)^2)
π/2ln(2(3+2sqrt(2)))
π/2ln(6+4sqrt(2))
Int(ln(x^4+1)/(x^2+1),x=-infinity..infinity) = π ln(6+4sqrt(2))
Int(ln(x^4+1)/(x^2+1),x=-infinity..infinity) = π ln(4+2*2*sqrt(2)+2)
Int(ln(x^4+1)/(x^2+1),x=-infinity..infinity) = π ln((2+sqrt(2))^2)
Int(ln(x^4+1)/(x^2+1),x=-infinity..infinity) = 2π ln(2+sqrt(2))
Great work, much appreciated. No complex function used. The use of Fubini's Theorem and dominated convergence is crucial. (Using complex functions will need to use the diffrentiation under the integral sign for complex valued function and path integral whose proof is much harder.)
15:50 Is replacement t=sin^2 phi correct? It means that 0
Some of us would argue it is the Risch algorithm.
Doing t=sin(o)^2 there is a problem: while "t" can vary from minus infinite to infinite, this relationship can´t. It is a domain problem.
The form I=2 pi ln(2+sqrt(2)) reflects the periphery of a circle with radius ln(2+sqrt(2)). :-)
A little irrelevant but what device are you writing on in this video? I am curious cause i want to get used to doing math on a tablet/iPad
What happens if the log is not defined? Don't you have to be more precise bevor coming along with log?
There's nothing wrong with the behaviour of the log in this case
My brain combusted everytime he used "easy" in any form to describe a step he just completed
😂😂😂
Great work.... just 1 over sq root of 2 is sq of 2 over 2
could you please explain why at 8:40 you can just switch u to x, this means u=x, however earlier we set x to be reciprocal of u?
For definite integrals it doesn't matter
@maths_505 thank you, but why?
12:41 i just realised i went mad for math. I laughed so hard when he forgot he putted a there
At 18:29, you show the integral of sec x to be ln (sec x + tan x) and the integral of tan x to be ln (sec x). However, according to CRC, these should be log (sec x + tan x) and log (sec x), respectively. What that means is that if you carry down log instead of ln through to your final equation, the answer would be pi * log (6 + 4 sqrt (2)) rather than pi * ln (6 + 4 sqrt (2)). When you plug in the numbers, you get 3.35 instead of 7.71543. Right?
Log is ln in the context of mathematics, log id assumed base e not base 10 as it normally would be in physics for example.
Hey what do you use to draw?
It's an s pen
Interesting. However, I could not stop wondering whether you should have tracked if your substitutions were valid for all t values you plugged in. The sin \phi = t substitution requires e.g. t in (-1,1). Not sure what happens for imaginary t to be frank, never substituted in the context of complex integrals.