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.
@@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?
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.
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!
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
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.
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!
@@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.
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.
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 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
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
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. 😂😂
@@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).
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
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.
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
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.
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.
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.)
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.
Not sure why people use such advanced methods for integrals like at 15:21. When the most "challenging" part of an integral is a simple root, the easiest solution always seems to be basic u substitution. In this case u = sqrt(t) so t = u^2, dt= 2u dt. That integral is of 2u(1-u)/(u(1-u^2)) du. Trivially this is of 2(1-u)/(1-u^2) du, which factors out via long division or basic inspection as 2/(1+u) du. The fact the inverse of the substitution of a simple root of t is a simple polynomial of u makes the change of coordinates very convenient to apply to the integral.
There is a much more sstraightforward way of calculating this define g(x,a) = log(i(a^2+x^2)(1+a^2x^2)/(1+x^2) then g(x,0) = log(ix^2)/(1+x^2) and g(x,i exp(i*pi/4)) = log(1+x^4)/(1+x^2) Now int(log(i)/(1+x^2),x-infinity..infnity)=I/2Pi^2 and int(log(x^2)/(1+x^2),x-infinity..infnity)=0 so int(g(z,0)=i/2Pi^2 Now dg/da= 2*a/(1-a^2)*[1/(a^2+x^2)+1/(1+a^2*x^2)-2/(1+x^2)) so int(dg/da dx) = 4 pi/(a-1) ( Take care here the sign of Re(a)) Finally we need to integrate this from 0 to get 4 pi log(a-1) and adding the value for int(g(x,0)dx) we get i/2 pi^2 + -ipi^2 + 4 pi log(i exp(i*pi/4)-1) = 2 pi log(2 + sqrt(2)) The only integral need is int(1/(1+x^2) dx , -infinity..infinity) = pi/2
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?
@7:30 "…integral from infinity to zero, which is quite weird. Twice that, so it's twice as weird…" @24:00 "…we're not going to evaluate this using our calculators, we'll use the binomial theorem. Why? Because we're sickos, obviously."
I don't know if I believe every passage, but it was nice. I think that all the trigonometric part was a little useless though, you could have factorized 1-t=(1-\sqrt(t))(1+\sqrt(t)) simplify and substitute t=u^2 (If I'm correct, the integral is rather trivially the log you find this way)
At 13:47, you plug x=infinity in the first part and state that it is equal to pi/2 but here we have arctan(x/sqrt(t)) where clearly the denominator could clearly be a complex number (as at the end we need to replace t by i or -i). So arctan(infinity/sqrt(t)) is slightly more challenging to calculate in that case…
We want to evaluate the integral functions at i and -i so we want the t variable in the denominator to be a purely imaginary number. In that case, the limit does evaluate to pi/2. You can try to evaluate the integral using brute force; all you'll when you get the arctan function is its logarithmic definition from complex analysis
Did you mean purely imaginary? Well, if so, sqrt(i) and sqrt(-i) evaluate to + or -exp(i.pi/4) and + or -exp (-i.pi/4) which have both real and imaginary parts, all of which non zero and positive and negative real parts so that the arctan could end end being equal to -pi/2. I think something more convincing is needed: for me there is still a problem in that calculation, at that precise point of the derivation
@@marcfreydefont7520 yes ofcourse Purely imaginary As far as the positive and negative values of the square root of i are concerned, it's quite a common practice to take just the positive square while considering principal branches. However this is an interesting proposition but I think it will check out once we multiply the two complex arguments which are conjugates
It's also worth noting that the development of this technique had nothing to do with Feynman; it was known by Leibniz and expanded upon by other. Feynman was just famous
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
I dislike where you put the exponent especially at the end, I believe it should be in the brackets as to not confuse (ln(6+4√2))^π with ln((6+4√2)^π). Great video
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?
6 months late, but ... @15:36, making the substitution t = sin(phi)^2 seems unnecessarily complicated. The substitution u = sqrt(t) leads to dt = 2*u*du, and the integrand becomes 2*u*(1-u)/(u*(1-u^2)) du, which simplifies to 2 du/(1 + u). You wind up with the same antiderivative in the end, so I don't suppose it matters all that much.
Interesting... but of course, *one must know _when and where_ it is true that _each step is valid_ if one is to apply the technique more generally. Which makes me wonder: How would 3Blue1Brown explain this?
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?
exp(2 pi i ) = exp( pi i) * exp( pi i), but 1 = exp( 2 pi i) so log( exp(2 pi i))=0. However log(exp( pi i)) = pi i, and 0 neq 2 pi i. The problem is that log( u v) is not necessarily = log u + log v for complex numbers.
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
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?
Feynman's Technique: Knowing the answer to everything
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
I used log(x^4+t^4) instead to avoid dealing with complex values of t. Got the same value.
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
I was eating dinner when I found this video. Now my dinner is cold, but I just found a new magical technique!
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 👌
this was incredibly satisfying to watch. awesome video!
Put this channel in the teaching portion of your CV bro you've earned it.
Already done
Gloriously pleasing. Chapeau!
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.
I love how this piece went from very easy to hard to harder to almost impossible
I kept thinking you were done but you simplified it even further 😂
Cool! Enjoyed it from start to finish
beautiful solution. keep rocking the integrals.
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.
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)
follow your explanation is like to listen to a detective story, great!
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 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
The solution is so beatiful😮
Prof Fred Adams, "If you use it once its a trick, if you use it twice its a technique"
Wow, thank you for the fun ride!
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"
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
Cheery cheery cheery color, and voice is a service
The Residue Theorem is clearly more overpowered, since you brought up complex numbers.
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 😂
integral at 15:24 can be done by substituting sqrt(t) as u after simplifying by canceling the 1-sqrt(t).
Beautiful
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).
nothing is more overpowered than guessing the solution
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
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).
On 15:41, if t equals to sin^2 φ, then sin φ =+-√t. To avoid this, you could have defined sin φ as t in the first place
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.
My brain combusted everytime he used "easy" in any form to describe a step he just completed
😂😂😂
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
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.
Why is there a parental advisory sticker😂
Why not😂😂😂
To think that I once thought long division was complicated
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.
lost my shit laughing when you pulled the pi into the exponent
Some of us would argue it is the Risch algorithm.
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
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.)
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.
Not sure why people use such advanced methods for integrals like at 15:21. When the most "challenging" part of an integral is a simple root, the easiest solution always seems to be basic u substitution. In this case u = sqrt(t) so t = u^2, dt= 2u dt. That integral is of 2u(1-u)/(u(1-u^2)) du. Trivially this is of 2(1-u)/(1-u^2) du, which factors out via long division or basic inspection as 2/(1+u) du.
The fact the inverse of the substitution of a simple root of t is a simple polynomial of u makes the change of coordinates very convenient to apply to the integral.
There is a much more sstraightforward way of calculating this
define g(x,a) = log(i(a^2+x^2)(1+a^2x^2)/(1+x^2)
then g(x,0) = log(ix^2)/(1+x^2) and
g(x,i exp(i*pi/4)) = log(1+x^4)/(1+x^2)
Now int(log(i)/(1+x^2),x-infinity..infnity)=I/2Pi^2
and int(log(x^2)/(1+x^2),x-infinity..infnity)=0
so int(g(z,0)=i/2Pi^2
Now
dg/da= 2*a/(1-a^2)*[1/(a^2+x^2)+1/(1+a^2*x^2)-2/(1+x^2))
so int(dg/da dx) = 4 pi/(a-1) ( Take care here the sign of Re(a))
Finally we need to integrate this from 0 to get 4 pi log(a-1)
and adding the value for int(g(x,0)dx) we get
i/2 pi^2 + -ipi^2 + 4 pi log(i exp(i*pi/4)-1) = 2 pi log(2 + sqrt(2))
The only integral need is int(1/(1+x^2) dx , -infinity..infinity) = pi/2
Fantastic class
It's also called Leibnitz' technique, so I guess Leibnitz discovered it first.
Feynman didn't discover it at all it was already a famous technique. He was just more famous than everyone else who used it
PI just has to show it self everywhere
The form I=2 pi ln(2+sqrt(2)) reflects the periphery of a circle with radius ln(2+sqrt(2)). :-)
thank you for your interesting content you make math seems to like very simple
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
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
I remember learning this in my applied mathematics 1 class
@7:30 "…integral from infinity to zero, which is quite weird. Twice that, so it's twice as weird…"
@24:00 "…we're not going to evaluate this using our calculators, we'll use the binomial theorem. Why? Because we're sickos, obviously."
*Feynman's technique is NOT the single most overpowered integration technique in existence!*
OpenAI CHAT GPT4 is!
I don't know if I believe every passage, but it was nice. I think that all the trigonometric part was a little useless though, you could have factorized 1-t=(1-\sqrt(t))(1+\sqrt(t)) simplify and substitute t=u^2 (If I'm correct, the integral is rather trivially the log you find this way)
At 13:47, you plug x=infinity in the first part and state that it is equal to pi/2 but here we have arctan(x/sqrt(t)) where clearly the denominator could clearly be a complex number (as at the end we need to replace t by i or -i). So arctan(infinity/sqrt(t)) is slightly more challenging to calculate in that case…
We want to evaluate the integral functions at i and -i so we want the t variable in the denominator to be a purely imaginary number. In that case, the limit does evaluate to pi/2.
You can try to evaluate the integral using brute force; all you'll when you get the arctan function is its logarithmic definition from complex analysis
Did you mean purely imaginary? Well, if so, sqrt(i) and sqrt(-i) evaluate to + or -exp(i.pi/4) and + or -exp (-i.pi/4) which have both real and imaginary parts, all of which non zero and positive and negative real parts so that the arctan could end end being equal to -pi/2. I think something more convincing is needed: for me there is still a problem in that calculation, at that precise point of the derivation
@@marcfreydefont7520 yes ofcourse
Purely imaginary
As far as the positive and negative values of the square root of i are concerned, it's quite a common practice to take just the positive square while considering principal branches. However this is an interesting proposition but I think it will check out once we multiply the two complex arguments which are conjugates
damn bro as someone who has not done integration in over 6 years I followed along just well. Wolfram asks you for solutions to their website right?
I've been leavin em on read 😂
thank you sir❤
It's also worth noting that the development of this technique had nothing to do with Feynman; it was known by Leibniz and expanded upon by other. Feynman was just famous
Just subbed, great channel!!
23:52 this is just the formula
(A+1)²=A²+2A+1
Great work.... just 1 over sq root of 2 is sq of 2 over 2
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 😂
quite delighted to see Samsung Notes in this video
12:41 i just realised i went mad for math. I laughed so hard when he forgot he putted a there
take a shot everytime dude says imply in the video
why do we replace -1 with i when i = -1^1/2 ???
super neat.
I dislike where you put the exponent especially at the end, I believe it should be in the brackets as to not confuse (ln(6+4√2))^π with ln((6+4√2)^π). Great video
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.
Great video! What drawing app are you using?
Navy Nuke School has about a 48% percent failure rate, and those rejects can already DO calculus going in. holy fuck!
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.
6 months late, but ... @15:36, making the substitution t = sin(phi)^2 seems unnecessarily complicated. The substitution u = sqrt(t) leads to dt = 2*u*du, and the integrand becomes 2*u*(1-u)/(u*(1-u^2)) du, which simplifies to 2 du/(1 + u). You wind up with the same antiderivative in the end, so I don't suppose it matters all that much.
The answer is pi * ln(6+4*sqrt(2))
What app are you using?
"what he said"
This was enjoyable, the way you solved it. But I want a cigarette now, and don't smoke.
Wow 👌 👏, thank you 👍
Interesting... but of course, *one must know _when and where_ it is true that _each step is valid_ if one is to apply the technique more generally. Which makes me wonder: How would 3Blue1Brown explain this?
In the integral to determine I(t), just substitute s=1-sqrt(t). You’ll get the solution after 1 step!
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.
How about using residue theorem? This would be simpler... but I'm not sure...
exp(2 pi i ) = exp( pi i) * exp( pi i), but 1 = exp( 2 pi i) so log( exp(2 pi i))=0. However log(exp( pi i)) = pi i, and 0 neq 2 pi i. The problem is that log( u v) is not necessarily = log u + log v for complex numbers.
Man i still have problems visualising the countours before integrating
I'd remove the plus and minus infinity as it makes no sense to me. But what do i know!
12:46 the 2 different t’s is driving me insane
We learn all of this in the school
When I feel that a extremely hard Integral requires Complex Analysis...😂
using I for both the final solution and the integral function is potentially very confusing
Feynman takes too much credit for this. It was standard in math before he was around doing lazy tricks with poor groundings.
I think Leibniz was using this technique quite often.