Feynman's technique is INSANELY overpowered!!!

[@maths_505]

Note: from the 18:30 mark onwards the right hand side includes arctan(t/2) and not arctan(1/2) untill we actually plugged in t=1. Apologies for the error, I subconsciously drifted to the target case of I(1) while carrying out all the math correctly 😂

[@meteor3033]

"We are gathered here today..." LMAO THE CULT OF FEYNMANN INTEGRATION I LOVE IT

[@dutchie265]

Towards the end, it is written inverse tan of 1/2. However, this should be inverse tan of t/2. Final result stays the same, as t=1.

[@emanuellandeholm5657]

Thanks!

[@frankargenti]

ja that 's not a arctg he forgot the 1/t cause he cant do the math

[@frankargenti]

SO many smart people and no one pointed out the mistakes he made ...... wow

[@元兒醬]

2:30 put t into sin^2tx and you'll get a quicker answer

[@erwanquintin3057]

Excellent video, thanks. I would mention that those I’’’ integrals are Laplace transforms. Never a bad time to point out the power of those things.

[@fengshengqin6993]

Wow ,really crazy integral !

[@gibbogle]

Excellent. A very surprising answer.

[@manskiptruck]

Also works really nicely if you define the parameter inside the sine and just immediately differentiate twice with respect to the parameter. You end up integrating 2/x^(2)cos(2a ln(x)) which can be done by parts, where [a] was my parameter.

[@davidarencibia7628]

I was scared to integrate cos(2nlnx) and then integrate that more times

[@wagsman9999]

Nice. I fear I may never be skilled enough to know when to apply Feynman's technique. But it is really cool.

[@debblez]

whenever you have a monomial in the denominator

[@two697]

Parameterising it as e^(-u)sin^2(tu)/u^2 actually makes the ensuing working out much nicer. You get that I''(t)=2/(1+4t^2) with I(0)=I'(0)=0 and directly end up with the nicer looking solution of arctan(2)-1/4ln5

[@joshuaiosevich3727]

lmao, I did the same thing

[@shayanfazeli8211]

love the channel, keep on keeping on!

[@Sugarman96]

Here's how I'd do it with the Fourier transform. After the first u sub, you have an exponent times sin(u)/u squared, so the integral I is half the integral of e^-|u| times sin(u)/u squared, so you use Plancherel theorem to turn it in the integral of a triangle windows function times 1/(1+x^2), which is already easier

[@yaroslavdon]

From 2:40 , you could probably solve it using the Laplace transform and its properties.

[@Charliethephysicist]

But is it not just the same procedure to evaluate the Laplace transform?

[@nicogehren6566]

brilliant solution. thanks man !

[@sadraderhami2628]

It's funny that I got the exact same answer, differently. I put the alpha in the sin function and got the second derivative of it instead of changing variables first. ( which actually involved some really cool tricks that I learned from this channel as well). So anyway thanks a lot for your great content and I'd love to see more(especially if you would upload some tutorials about differential equations)

[@MathOrient]

Year 2050: "Feynman's technique is still working 😆"

[@vladimir10]

Awesome vids, dude! What I especially like is you approach, balanced between regorousity and fun. Don't stop! Can you crack professor Penn's 'members only' recent sum?

[@maths_505]

I'll have a look at it

[@maths_505]

Could you send me a link? I can't seem to find the members content part of his page

[@vladimir10]

@@maths_505 Neither can I find it. I saw it yesterday in my KZbin feed and I don't see it anymore. It involved summation of Reinmann zeta of n , all over pi^n, and the RHS was something ridiculously simple like 1\(exp(2)-1). I'm not sure about the details though.

[@ksmith-s9f]

I think you have an error at about the 15:25 point...You should have 0=1/4(1)+c, which implies that c=-1/4. Although, I could be wrong.

[@GreenMeansGOF]

12:44 Is Dominated Convergence justified?

[@holyshit922]

y = ln(x) Integration by parts with dv=1/y^2dy and u=e^{-u}sin^2(u) Express sin^2(y) as cos(2y) Calculate Laplace transform and evaluate it at s=1

[@zunaidparker]

In your final step you could have written -¼ ln5 which is slightly more elegant. Another great video! Just keep control of your script and editing...you got a bit nervous in the middle there and started rambling 😜

[@maths_505]

I don't like using scripts....it makes things too organized and is not the optimal way to carry out the reverse cowgirl formulation of integral calculus😂😂😂

[@absurdengineering]

@@maths_505 I support this motion :) I love the quirky format. If I wanted a dry integral calculus recitation, there’s plenty of those around. Dry works for some people, quirky works for others. Me quirky. Me likes.

[@Sty5A467]

Where and how could I learn those ways to solve integrals?? I mean how to learn gamma function and beta function and Feynman technique and laplace transformation...etc??

[@maths_505]

"You have come to the right place my child"

[@absurdengineering]

Just stay tuned to this channel I guess? 😂

[@rayquazabtc9131]

Sir I didn't understand the part when you found the value of c please explain

[@MrWael1970]

At the minute 18:07, the factor of 8 outside the integral should be divided by 4 not by 2. So, What about using Lobachevsky integral instead of Feynman technique from the beginning. Anyway, this is very interesting integral and you have a talent to solve such complex integrals.

[@maths_505]

The table of anti derivatives says that we need to divide by a=2 and not a²=4

[@MrWael1970]

@@maths_505 the form of 1/(1+t^2) is the derivative of arctan. Thus, if we took 4 as a common factor from the denominator, then we divide 8 by 4. Thanks

[@maths_505]

@@MrWael1970 factor out 4 from the denominator and you get 1/(1+t²/4). Write t²/4 as (t/2)². Now you need (1/2)dt as the differential element so insert the ½ in the numerator and to balance that out we need to multiply outside the integral operator by 2. That gives you 2/4 outside the integral operator which is ½. Multiply that by the factor of 8 we already have then we get the factor of 4. You're right about factoring the 4 from the denominator but we also need to address the form of the differential element.

[@MrWael1970]

@@maths_505 thanks, I got confused, but now it is clear. Regards.

[@maths_505]

@@MrWael1970 it's cool....the signs of great mathematicians include forgetting constant factors cuz they're completely hypnotized by the beautiful structure 😂

[@joshuaiosevich3727]

theres an easier way to go about this integral IMO, first we start by setting ln(x) to x getting us the integral from 0 to infty of sin^2(x)e^-x/x^2, then we actually define our integral function by putting the parameter t inside of the sin term instead of the exponential term, note for later that I(0)=0, for I'(t) we end up getting sin(2tx) inside so note again that I'(0)=0, we will have no worries over constants in this scheme. Then we differentiate one more time and we get the integral from 0 to infty of 2cos(2tx)e^-x which is equal to 2/(1+4t^2), then we integrate twice and plug in one. edit: my solution is no better, it's just what I came up with before I watched your solution. You literally taught me every advanced integration technqiue I know, you're the GOAT.

[@ayhanozkan5071]

I am a 56 years old electrical engineer and this channel gives me more pleasure than having sex...duh! His choice of integrals and diff equations are top notch. Keep up the good work.

[@ericthegreat7805]

Can we do IBP with e-u and sin2u/u2 then the term containing sin is a Dirichlet integral.

[@Anonymous-Indian..2003]

It can easily be done by using Laplace transform.....

[@nura8578]

ok, this is cool

[@jorditoral8478]

In which cases wouldn’t it be possible to switch the differential operator with the integral one?

[@maths_505]

If there were problems with boundedness or convergence. Dirichlet's convergence theorem for integrals is a pretty good criteria to test whether or not the integral converges or not.

[@fasihullisan3066]

I failed to solve by L Hospital rule that limit t approaches to zero tlnt^2/(t^2+4) is equal to zero.

[@tueur2squall973]

It's just easy to see , you compare the growing rates of your functions! Any power of t will dictate its behavior over the behavior of an ln function. So the product t*ln(t²/t²+4) , inside the ln , when t approches 0 , the whole thing converges to 0 , then we have ln(0) , so it goes to -infinity but the whole thing is multiplied by a power of t that actually wins there cuz "it goes to 0 way faster than the ln function". And you should be able to verify this tesult using l'Hospital rule

[@calcul8er205]

You can split it up as 2tln(t)-tln(t^2+4). Using the fact that the limit as t approaches 0^+, tlnt approaches 0 gets you to the answer. That fact can be proven using L'Hopital's Rule

[@giuseppemalaguti435]

Uso la funzione integranda I(a) =sin(alnx) ^2.... Derivo 2 volte... I=arctg2-ln5/4 se non ho fatto errori di calcolo...

[@Mr_Mundee]

mistake: the answer is actually arctan(2)-ln(5)/4 not arctan(2)+ln(5)/4

[@adriansison1503]

Nice

[@petterituovinem8412]

a great channel to satisfy my integration fetish

[@honestadministrator]

Writing z for ln(x) one gets Integration (0, INF) ( sin^2(z)/( z exp(2z)) dz

[@padraehsani3908]

Woooooow

[@david-melekh-ysroel]

You're going so fast, I can't well understand what you're doing ☹️☹️

[@BruceWayne-mk9km]

Plz tell me, whether Feynman's technique is essentially solving an integration using Leibniz's rule?

[@thomasblackwell9507]

It is the same. Just Feynman made it famous.