How Richard Feynman would evaluate this monster log integral

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!

[@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

[@antoniomora4537]

​@@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

[@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?

[@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

[@AnsisPlepis]

this was incredibly satisfying to watch. awesome video!

[@riadsouissi]

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

[@ublade82]

Feynman's Technique: Knowing the answer to everything

[@nicogehren6566]

beautiful solution. keep rocking the integrals.

[@matthew.y]

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

[@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.

[@DaveJ6515]

Cool! Enjoyed it from start to finish

[@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 ? 💀

[@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

[@edcoad4930]

Gloriously pleasing. Chapeau!

[@amrendrasingh7140]

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

[@ihatethesensors]

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!

[@maths_505]

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

[@alanrodriguez9365]

Wow, thank you for the fun ride!

[@daddy_myers]

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. 😂😂

[@manstuckinabox3679]

It's those sneaky egyptians! we knew they went irrelevent after the dawn of the 1st century...

[@osamaattallah6956]

This reads like a bot wrote this, surpirsed about complex numbers and pi lol

[@urosmarjanovic663]

@@osamaattallah6956 What is next, that pesky "e" number? Incredible!!!

[@bubbazanetti4577]

@@urosmarjanovic663 That damned Oily Macaroni Constant that likes to jump scare at random times. 0.577215664901532860606512090... WTF is that all about???

[@SiphonSoulsX]

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).

[@georgesheffield1580]

Thanks for showing this .

[@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.

[@user-ne1en4mf6i]

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

[@DominicProMax]

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

[@geraltofrivia9424]

Beautiful

[@AmanBansal-xb8uk]

Just subbed, great channel!!

[@Ezy.Kemistry]

Fantastic class

[@drstrangecoin6050]

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

[@maths_505]

Already done

[@seegeeaye]

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

[@uhbayhue]

That indeed was awesome 👌

[@bjrnleonsrenriedel8585]

The solution is so beatiful😮

[@unidentifieduser5346]

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

[@hadikareem2335]

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

[@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

[@himanka1roy237]

thank you sir❤

[@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)

[@ajskilton]

Wow 👌 👏, thank you 👍

[@aerialwinston9932]

Cheery cheery cheery color, and voice is a service

[@nablahnjr.6728]

nothing is more overpowered than guessing the solution

[@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.

[@meeharbin4205]

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?

[@maths_505]

It's mostly hit and trial but it works pretty well with logarithmic bois especially on this interval.

[@spaceface2918]

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.

[@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

[@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

[@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.

[@thefeynmantechnique]

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.

[@maths_505]

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!

[@maths_505]

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

[@francischang]

Great video! What drawing app are you using?

[@NaumRusomarov]

super neat.

[@mattiagiardini7245]

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?

[@maths_505]

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.

[@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

[@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.

[@newplayer3259]

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

[@robertorossano6442]

that's Feynman's technique™ !

[@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

[@wilurbean]

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

[@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

[@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.)

[@schizoframia4874]

Why is there a parental advisory sticker😂

[@maths_505]

Why not😂😂😂

[@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.

[@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"

[@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

[@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).

[@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.

[@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 😂

[@JYT256]

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

[@twistedcubic]

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

[@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.

[@wynautvideos4263]

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

[@verma.shaurya]

What tablet / software do you use for your videos?

[@Ryezn5057]

Random, which app do you use to record the lecture ?

[@darksoul.0x7]

I remember learning this in my applied mathematics 1 class

[@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 😂

[@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.

[@zahari20]

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

[@ashrafulhaque6476]

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

[@ahmadnoorbig5191]

What app are you using?

[@joshelguapo5563]

Man this video is nostalgic

[@maths_505]

How so?

[@nitroxide17]

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

[@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.

[@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 😂

[@ghk27]

take Residue theorem into consideration and expand the integration core?

[@xxthelinkxx3296]

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

[@maths_505]

😂😂😂

[@paulelliott9487]

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?

[@Fictionarious]

To think that I once thought long division was complicated

[@vozestoica8436]

What app is he using?

[@user-hh1lk8ks4n]

What does it looks like ?

[@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

[@ulisesbussi]

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

[@yusuke4964]

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

[@sakinano99]

using I for both the final solution and the integral function is potentially very confusing

[@steffen8544]

Nice video! Where was feynmans trick?

[@maths_505]

Taking the derivative of the integral function and integrating that

[@Aerxis]

@@maths_505 so, Leibniz trick?

[@maths_505]

@@Aerxis no way mate That's the Leibniz rule...its only called Feynman's trick cuz he popularised it

[@cavesalamander6308]

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

[@alexdefoc6919]

Hey what do you use to draw?

[@maths_505]

It's an s pen

[@emilfrei6303]

What happens if the log is not defined? Don't you have to be more precise bevor coming along with log?

[@maths_505]

There's nothing wrong with the behaviour of the log in this case

[@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)

[@AndDiracisHisProphet]

Nice

[@JamesJoyce12]

Some of us would argue it is the Risch algorithm.

[@oussamawahbi4976]

PI just has to show it self everywhere

[@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

[@alexdefoc6919]

12:41 i just realised i went mad for math. I laughed so hard when he forgot he putted a there

[@soundcloudslave2790]

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

[@silverfox1754]

Man i still have problems visualising the countours before integrating