How Richard Feynman would evaluate this monster log integral

[@maths_505]

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.

[@danielkanewske8473]

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]

@@danielkanewske8473 yes I agree

[@antoniomora1621]

​@@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?

[@pabloarmenteros]

yo creo que te podemos perdonar jeje...

[@joeboxter3635]

How is this over powered?

[@TimothyOBrien6]

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.

[@TheScreamingFedora]

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.

[@drillsargentadog]

@@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.

[@damon1588]

If I'm not mistaking, in France we call it Leibniz' technique

[@planomathandscience]

@@drillsargentadog yet no one nowadays takes inspiration from him. So... who cares?

[@lanog40]

@@planomathandscience you’re obviously not studying physics, so your opinion is likely not going to be shared by people that are studying it

[@pablosarrosanchez460]

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

[@riadsouissi]

I used log(x^4+t^4) instead to avoid dealing with complex values of t. Got the same value.

[@ahmetleventtakr7625]

Your channel is criminally underrated. I hope you’ll get the subscribers and views you deserve. By the way, amazing video as always. Kudos!

[@NaN_000]

criminally ? 💀

[@rajendramisir3530]

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.

[@violintegral]

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.

[@maths_505]

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.

[@violintegral]

@@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.

[@violintegral]

@@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.

[@maths_505]

I solved a few of the fun ones on the qualifying round but I haven't seen the integrals from the competition yet

[@violintegral]

@@maths_505 the quarterfinal round has some really nasty limits of integrals which I have yet to see any solutions for

[@amrendrasingh7140]

The flow of the solution was awesome and stimulating. Good work kamaal 👌

[@matthew.y]

I was eating dinner when I found this video. Now my dinner is cold, but I just found a new magical technique!

[@AnsisPlepis]

this was incredibly satisfying to watch. awesome video!

[@manstuckinabox3679]

Deciding between contour and Feynman's is liek deciding between nuking and nuking harder...

[@maths_505]

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

[@manstuckinabox3679]

@@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!

[@maths_505]

@@manstuckinabox3679 once you go Feynman....there ain't no turnin back!

[@anthonymichael970]

Jeez man. Relax

[@vogelvogeltje]

U liek mudkipz?

[@ublade82]

Feynman's Technique: Knowing the answer to everything

[@felixlucanus7922]

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]

it's attributed to Leibnitz

[@felixlucanus7922]

@@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.

[@nicogehren6566]

beautiful solution. keep rocking the integrals.

[@edcoad4930]

Gloriously pleasing. Chapeau!

[@DaveJ6515]

Cool! Enjoyed it from start to finish

[@drstrangecoin6050]

Put this channel in the teaching portion of your CV bro you've earned it.

[@maths_505]

Already done

[@unidentifieduser5346]

I love how this piece went from very easy to hard to harder to almost impossible

[@sorooshusa]

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.

[@paarths.5281]

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

[@newplayer3259]

integral at 15:24 can be done by substituting sqrt(t) as u after simplifying by canceling the 1-sqrt(t).

[@ryangosling239]

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

[@seegeeaye]

follow your explanation is like to listen to a detective story, great!

[@DominicProMax]

I kept thinking you were done but you simplified it even further 😂

[@ElliotUnbound]

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

[@daddy_myers]

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.

[@samssams1619]

Sounds like residue theorem to me as there u always have 2pi*i * res(z)

[@bjrnleonsrenriedel8585]

The solution is so beatiful😮

[@alanrodriguez9365]

Wow, thank you for the fun ride!

[@Aryan-ut7rl]

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

[@Spielzeit85]

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

[@holyshit922]

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))

[@tzebengng9722]

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.)

[@nathanmenezes7914]

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).

[@bardistass]

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"

[@nafaidni]

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.

[@georgesheffield1580]

Thanks for showing this .

[@wilurbean]

Prof Fred Adams, "If you use it once its a trick, if you use it twice its a technique"

[@procerpat9223]

Very entertaining delivery! I would enjoy watching you solve this using contours.

[@maths_505]

But I wouldn't enjoy solving it😂 Check out qncubed3. He solved it using complex analysis

[@psychedelictranscendental811]

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.

[@Facetime_Curvature]

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...

[@maths_505]

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.

[@wolfgangreichl3361]

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.

[@marcfreydefont7520]

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…

[@maths_505]

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

[@marcfreydefont7520]

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

[@maths_505]

@@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

[@nablahnjr.6728]

nothing is more overpowered than guessing the solution

[@aerialwinston9932]

Cheery cheery cheery color, and voice is a service

[@nitroxide17]

21:30 Doesn’t the square root of i have 2 roots (they are offset by 180 degrees).

[@ALI_S_abdalrahman]

thank you for your interesting content you make math seems to like very simple

[@Galileosays]

The form I=2 pi ln(2+sqrt(2)) reflects the periphery of a circle with radius ln(2+sqrt(2)). :-)

[@Czeckie]

i find integration with complex numbers kind of iffy. At the start you say that a contour integration would need a branch cut. Your computations also uses a branch cut, but it's hidden in not being careful enough. The crucial point is integration of the partial fractions - the mindless use of basic calculus formulas hides there's a complex logarithm behind the scene. I'm not criticizing the video, it's very nice. I just want people to appreciate the subtlety. One place to see definitely more is going on is the evaluation of arctan(x/sqrt(t)). Why is it pi/2? Arctan(i) is a pole, so there must be some argument somewhere to limit our t's in the calculation. More care needs to be taken when computing with complex functions. This is the lesson of complex analysis and the reason why we have Riemann surfaces in the first place.

[@maths_505]

We needed t to be a pure complex number and in that case, the definite integral does indeed evaluate to (1/sqrt(t))(pi/2). This can be verified by evaluating the integral of 1/(i+x²) from zero to infinity; the logarithmic definition of the arctan function comes in handy here. But this is in fact a nice idea for a follow up math snack video

[@konchady1]

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.

[@CameronTacklind]

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.

[@jimschneider799]

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.

[@hadikareem2335]

Can you show us an example of Feynman's technique solving fractional derivative of a spherical special function such as the Bessel function?

[@geraltofrivia9424]

Beautiful

[@xxthelinkxx3296]

My brain combusted everytime he used "easy" in any form to describe a step he just completed

[@maths_505]

😂😂😂

[@JohnSmith-cg3cv]

The thumbnail is like “if Feynman was a Platinum-record selling rapper”…. Lmaooo

[@maths_505]

I'm pretty sure Feynman would treat every video on this technique as a diss track towards contour integration 😂

[@Sci24]

Fantastic class

[@samssams1619]

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

[@lucaspeciale9838]

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)

[@jnm11]

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

[@captainchicky3744]

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

[@maths_505]

That's the other video on this integral 😂

[@twistedcubic]

The Residue Theorem is clearly more overpowered, since you brought up complex numbers.

[@burningtime7746]

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

[@JamesJoyce12]

Some of us would argue it is the Risch algorithm.

[@JYT256]

lost my shit laughing when you pulled the pi into the exponent

[@jieyuenlee1758]

23:52 this is just the formula (A+1)²=A²+2A+1

[@severoon]

@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."

[@gplgomes]

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.

[@jrarsenault47]

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?

[@Bruhong99]

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.

[@andreaspapoutsakis6342]

Great work.... just 1 over sq root of 2 is sq of 2 over 2

[@robertorossano6442]

that's Feynman's technique™ !

[@yusuke4964]

How about using residue theorem? This would be simpler... but I'm not sure...

[@konoveldorada5990]

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?

[@-Curved]

not helpful. that series is va;id when IxI

[@Fictionarious]

To think that I once thought long division was complicated

[@YodaWhat]

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?

[@魏義漢]

What does it looks like ?

[@sandorszabo2470]

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.

[@darksoul.0x7]

I remember learning this in my applied mathematics 1 class

[@cavesalamander6308]

15:50 Is replacement t=sin^2 phi correct? It means that 0

[@wynautvideos4263]

At 21:23 dont you need to consider both square roots of i?

[@ajskilton]

Wow 👌 👏, thank you 👍

[@silverfox1754]

Man i still have problems visualising the countours before integrating

[@himanka1roy237]

thank you sir❤

[@AmanBansal-xb8uk]

Just subbed, great channel!!

[@vickramratnam2320]

The answer is pi * ln(6+4*sqrt(2))

[@ghk27]

take Residue theorem into consideration and expand the integration core?

[@danielfrancis3660]

I'd remove the plus and minus infinity as it makes no sense to me. But what do i know!

[@Saki630]

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?

[@maths_505]

I've been leavin em on read 😂

[@bernat8331]

4:25 You need to justify that both integrals converge before using linerity of integrals 13:18 that thing doesnt converge if t=0. There is lack of rigor everywhere in this proof 13:25 You literally will use t as a complex number. A root of a complex number isnt even defined propery, as there are 0 possible solutions. It can be fixed because any of them work, but again lack of rigor 16:40 again taking square roots of complex numbers without caring 21:38 what about e^5ipi/4 Overall, it semms like you assumed t was real for the whole vídeo and them randomly generalised it for complex t with no justification As a person with strong mathematical training but not yet cursing complex integration, im unable to believe most of the steps in this video.

[@JamesSarantidis]

As an engineer, I have to admire the original solver's creativity even if he "skipped the rigorous part". I have to salute the brave mathematicians too though, who dwell in the semantics and give us all these beautiful tools. Math are both terrifying and beautiful. o7

[@jamesfrancese6091]

@@JamesSarantidisit’s useful yes, but not original to Feynman or another hard-headed problem solver - this integration technique (via analytic continuation) was discovered by mathematicians almost as soon as the residue theorem was proved, in the 1830s

[@Aerxis]

@@jamesfrancese6091 which mathrmaticians? if you don't mind sharing....

[@epicmarschmallow5049]

Most of these complaints are trivial and easily resolved with elementary analysis and don't really change the argument, just add small pointless technical steps that aren't really the point of the video

[@jorgearreola-cisneros6952]

@8:28 I don’t think the u substitution he set was reverted properly at the end. He stated that x=1/u, but then substituted x as if x=u which doesn’t make any sense. Am I tripping?

[@maths_505]

It wasnt a reverse substitution In definite integrals, the variable is just a dummy variable so all I did was rename it back to x The name of the variable doesn't matter....all that matters is structure

[@ishaneshkhanal]

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

[@zahari20]

Feynman's technique? This is in fact the Leibniz technique!

[@ulisesbussi]

don't you need the convergence of the original integral ensured to split it on the sums of integrals?

[@ahmadnoorbig5191]

What app are you using?

[@Satnam7275]

What is the pressing need to spend 25 minutes of your valuable life period in solving one mathematical question ? Will it improve our life quality or can we use it in any application affecting human beings :)?

[@epicmarschmallow5049]

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

[@funnydog7817]

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?

[@maths_505]

Search up Dirichlet's convergence theorem for integrals....that'll help you decide on convergence and switch up of limits.

[@rafvermeer6777]

In the integral to determine I(t), just substitute s=1-sqrt(t). You’ll get the solution after 1 step!

[@holyshit922]

Without Leibniz rule Integrand is even function and interval is symmetric around zero so =Int(2ln(x^4+1)/(x^2+1),x=0..infinity) x=tan(t) dx = (1+tan^2(t))dt =Int(2ln(tan^4(t)+1),t=0..Pi/2) =2Int(ln(tan^4(t)+1),t=0..Pi/2) =2Int(ln(sin^4(t)+cos^4(t))/cos^4(t),t=0..Pi/2) =2(Int(ln(sin^4(t)+cos^4(t)),t=0..Pi/2)-4Int(ln(cos(t)),t=0..Pi/2)) =2Int(ln(sin^4(t)+cos^4(t)),t=0..Pi/2)-8Int(ln(cos(t)),t=0..Pi/2) sin^4(t)+cos^4(t)=(sin(t)^2+cos(t)^2)^2-2sin(t)^2cos(t)^2=1-1/2sin^2(2t) =2Int(ln(1-1/2sin^2(2t)),t=0..Pi/2)-8Int(ln(cos(t)),t=0..Pi/2) 1=cos(t)^2+sin(t)^2 cos(2t)=cos(t)^2-sin(t)^2 1-cos(2t)=2sin(t)^2 =2Int(ln(1-1/4(1-cos(4t))),t=0..Pi/2)-8Int(ln(cos(t)),t=0..Pi/2) =2Int(ln(3/4+1/4cos(4t)),t=0..Pi/2)-8Int(ln(cos(t)),t=0..Pi/2) =2Int(ln(3+cos(4t),t=0..Pi/2))-2Int(ln(4),t=0..Pi/2)-8Int(ln(cos(t)),t=0..Pi/2) u=4t , du=4dt 1/2Int(ln(3+cos(u)),u=0..2Pi)-4Int(ln(2),t=0..Pi/2)-8Int(ln(cos(t)),t=0..Pi/2) I tried to calculate Int(ln(3+cos(u)),u=0..2Pi) using real numbers only but maybe it is good idea to express cosine as complex exponentials

[@maths_505]

Or you could just use Feynman's technique at that stage and then use the Weierstrass substitution

[@francischang]

Great video! What drawing app are you using?

[@242jemmy]

why do we replace -1 with i when i = -1^1/2 ???

[@speeshers]

That indeed was awesome 👌

[@kokojamba232]

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?

[@maths_505]

For definite integrals it doesn't matter

[@kokojamba232]

@maths_505 thank you, but why?