the integral that Feynman('s trick) couldn't solve

[@goodplacetostop2973]

17:37 "your" 💀💀💀

[@ericknutson8310]

I just wrote a paper on this trick for my senior undergradd project. Everywhere I look now I see Feynman because my browser history thinks its my favorite thing ever.

[@CamiKite]

Very interesting ! We can see numerically that the limit of the integral of ln(x)*cos(ln(x))/x^t converges to 1 if we let t->1, but the integral diverges for t=1, that's crazy !

[@Ensivion]

amazing video, this is a great pitfall that can happen when trying these kinds of problems, misusing theorems that have caveats that you didn't account for.

[@PawelS_77]

13:35 Shouldn't it be +cos u?

[@itzakehrenberg3449]

Yep!

[@PRIYANSH_SUTHAR]

Yes sir.

[@samueldeandrade8535]

Done on purpose to make people comment "Shouldn't it be +cos u"?

[@alexicon2006]

​@@samueldeandrade8535Proof for that claim? Any history of clickbait with the channel? Any leads following one to believe in this youtuber resorting to such trickery?

[@giorgioripani8469]

Yes but the final result won't change

[@gp-ht7ug]

Hi Michael you should explain better the Feynman technique. It’s very interesting but it’s not well explained in internet

[@gamer_paul2503]

He already did that, this is actually based on a theorem of another mathematician, which he made a video about a long time ago.

[@Alan-zf2tt]

As gamer_paul said before me plus! There are no limitations upon watcher from making further personal research no?

[@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ]

Look up Leibniz's integral rule. It's the general form of the theorem exploited by what's called "Feynman's trick".

[@DanielWalvin]

Hi Michael, love your videos! 😁 The thumbnail says "Surely your joking" instead of "Surely you're joking" - I'm not joking 😛

[@TomFarrell-p9z]

Good thing this is a math channel; not an English channel! 🙂

[@CM63_France]

Yes, kinna joke 🤣

[@alexkaralekas4060]

Damn i will have trust issues with Wolfram alpha now

[@byronwatkins2565]

At 6:50, that denominator should be t^2+2t+2.

[@finite1731]

Reminded me of a competition question where the gave an example of "proving" 1=2 using DUTIS but the reason why it didn't work was again because it didn't satisfy the requirements for DUTIS, it was quite cool because doing integration bee's you kinda come to expect the question to just allow for DUTIS.

[@Alan-zf2tt]

Ah well! If it is competition hints and reminders type video a refresher on qualifying conditions is very, very important

[@SuperSilver316]

The associated Sine Integral also seems to give Wolfram trouble, as it returns 0 for this value, but if you make your substitution you will also get a divergent integral.

[@txikitofandango]

I've seen a few videos that use Feynman's trick but this is the first one I can recall where the goal was to find I'(t). Other times, the strategy was to get a nice formula for I'(t), integrate it, and then evaluate it.

[@Alan-zf2tt]

comment made at 0:20 in video: you know its gonna get complicate when the integral has a big 'I' in front of it. comment made at 17:37 wicked! 🙂

[@SuperSilver316]

I wonder if Wolfram is performing an Abel Sum when it’s numerically integrating? The machinery at work here is very interesting 🤔

[@seherkasimoglu4596]

Feynman..❤

[@CatholicSatan]

In the late 19th century, mathematicians were awash with so-called "pathological" functions breaking the rules of integration. Mostly, this consisted of functions discontinuous somewhere (eg: at all rational points but not irrational points) to test such as the rules of Riemann integration. In this case of differentiation under the integral, it comes down to how misbehaved can a derivative be? Volterra, for example, in 1881 constructed a function that had a bounded derivative at all points (it was differentiable everywhere) but whose derivative was so discontinuous that the (Riemann) integral of the derivative did not exist. Later on, however, Lebesgue, with his bounded/dominated convergence theorem, allowed this particular pathology to disappear. I'm no mathematician but I guess there's a test that can be applied to see if the Feynman trick would work. Or, if not, then Wolfram will continue to give the odd wrong result!

[@DOTvCROSS]

Integrate from 1 to e^1, = cos(1)+sin(1)-1=2(tan(1/2)-tan(1/2)^2)

[@gibbogle]

"your joking"??

[@jazzjohn2]

In addition to advancing mathematics, let's not leave behind the English language.

[@MrApril-07]

* you're joking

[@shohamsen8986]

But ln x is continuous on (0,1) and cos is a continuous fn thus cos(ln x) is also continuous on (0,1). The same goes fro 1/x. Shouldnt that suffice for integrals as you can remove the point 0 from your region of integration and still get the desired value of the integral.

[@TheEternalVortex42]

That's true in general for integration, but the theorem for swapping the order of differentiation and integration requires continuity at the endpoints.

[@przemysawkwiatkowski2674]

It seems the function infinitely many times changes sign near zero. :-(

[@shohamsen8986]

@@TheEternalVortex42 that m8ght be one version of the theorem. I feel like there maybe a version where continuity at the end points may be sacrificrd . If such a version cant exist, then there should be some reason. Maybe the version has to do with approximating the region of integration with subsets of thw original set.

[@shohamsen8986]

@@przemysawkwiatkowski2674 not sure why infinite differentiability would be required. If we need to differentiatebunder the integeal sign once, then only C1 should be required

[@blabberblabbing8935]

​@@shohamsen8986 @przemysawkwiatkowski2674 didn't mention differentiability but the original function which changes sign around 0 infinitely many times as looking at the plot can show

[@OssamaAzuzaui]

Integration by substitution : Am I a joke to you? u=lnx => du=dx/x

[@gp-ht7ug]

For a moment I thought it would be usin(u)-integral of(-1*-cos(u))…

[@richardchapman1592]

Love your skills at extra high maths. That is why i tell you of an angle of inves😅t

[@richardchapman1592]

investigation that Google doesn't want any old free loader to benefit from. It concerns the possibility of rewriting the methods of making derivatives by using saw tooth functions instead of circular ones.

[@rocky171986]

So Wolfram is wrong?

[@KogularajK.]

Yes..

[@bobh6728]

On Wolframalpha instead of 0 for the lower limit, try 0.1, 0.01, 0.001 etc. The value of the integral is all over the place, both positive and negative.

[@bobh6728]

Graphing the function on Desmos looks like it shoots to infinity and then drops 0 at x=.2, reaches a minimum of -1.5 at about 0.33, then reaches 0 at x=1, so it seems like it may converge. But when you zoom in on x =0, you see that it oscillates over and over.

[@MatondoMaduhu-s9d]

Integration is my favorite 🎉🎉🎉🎉🎉🎉🎉. This is more funny

[@giorgioripani8469]

But why do you think Wolfram got it wrong?😮

[@KogularajK.]

Continuity

[@FleuveAlphee]

Evidence that algorithms are not replacing mathematicians, yet.

[@giorgioripani8469]

​@@FleuveAlpheewhat do you mean? Wolfram uses deterministic algorithms. They are the definition of Mahematics. Is like saying calculators will never replace them 😅

[@APaleDot]

@@giorgioripani8469 No, saying algorithms are the definition of mathematics is like saying calculators are the definition of arithmetic.

[@jay_13875]

GPT-o1 argues that the Riemann integral diverges, but that it can be regularized using Mellin transform theory to yield a value of 1. I have no idea if that makes any sense, but it might explain why we got a value of 1 in the first attempt.

[@nirajmehta6424]

GPT is useless in these situations. it's often right, but occasionally confidently incorrect. if you can't discern when it is either, then it's answer is actually less useful than simply not knowing

[@jay_13875]

@nirajmehta6424 do you know what GPT-o1 is and have you ever used it? It's much better at math problems than the old GPT-4o. The main issue I have with it is that it often arrives at the right answer but doesn't explain its reasoning steps well.

[@SuperSilver316]

A Laplace Transform consideration can also be made to his substituted integral near the end. Make the appropriate substitution to get I = -int(cos(t)tdt, 0

[@giorgioripani8469]

@jay_13875 Could you provide the prompt you have used and the amswer you got?

[@edmundwoolliams1240]

It isn't "Feynman's trick". He didn't invent it, he learned the technique of integration by differentiation under the integral sign from another calculus book. Did you even read Surely you're joking Mr Feynman?

[@SuperSilver316]

Yeah he followed the way of Leibniz, these tools have always been there, he just popularized them.

[@TheEternalVortex42]

That's still what it's called. Most things in math aren't actually named for their original discoverer anyway

[@antormosabbir4750]

6 colours is bad, 7 colours is good. It symbolizes rainbow, not filthy stuffs

[@pietervandenakker419]

Everybody has the right to have stupid opinions. Everybody should have the right to express those stupid opinions. But if you are smart, you keep those stupid opinions for yourself, and don't make everybody aware of your stupid opinions.