Integral of ln(x) with Feynman's trick! 

May be, destroying the ant with cannon ball makes it easier to target the elephant 🐘

😂😂

@@PrasannaKumar-zx7gr oh nice

@@priyanshutyagi3688 HI

Mine I with my left(yes I'm replying after 2 years)

Selim Akar🔥🔥🔥 görse duygulandırdı

I can’t do math rn because my right hand is busy 😢

@@herobrine1847uhh 🤨

@@lakshshastry8278 your hand is next

Rotating the graph is the graph of the inverse. ln(x) and e^x are inverses.

Actually it is the reflection.

@@SMEEST55 Yes, a reflection in the line y = x. Fred

It would be nice to explain intuitively( without formulas) why e^-x and xe^-x have the same area from 0 to infinity..(as this integral suggests)

Simone Dartizio Draw it on a piece of paper and see for yourself. Its pretty much just reflected on both the x and y axis

Thank you. That exact thing had crossed my mind.

@@mrnogot4251 By the way, recently I've watched video where it was told that in 99% cases you can just try to use the technique and, if you got the finite answer, then everything is OK and differentiation under the integral sign is allowed.

@@regulus2033 physicists usually never bother checking that things converge lol

I suggest you to search about the dominated convergence theorem and a corollary that (roughly speaking) states that if you have a function f(t,x) such that its partial derivative with respect to t is, in absolute value, uniformly bounded by an integrable function g(x), then the Feynman technique holds

@@regulus2033 Maths is not about being correct 99% of the time...

You missed a y there (that was supposed to be in front of e^y)

@@That_One_Guy... I missed more than that. My 3rd line doesn't follow from my 2nd line. Those 2 lines should have been: y = lnx ; x = eʸ When x → 0⁺, y → -∞; when x = 1, y = 0. The rest is OK; it works as written. All I'm really doing, is flipping the graph about the y=x axis, then integrating the area w.r.t. y. Fred

@@That_One_Guy... you are right, although not matters on result

How the hell did you write those symbols

@@andreaq6529 Some of them are available in macOS using option & command keys, or Keyboard Viewer; some of them I've copy-pasted from other people's posts. In Windows, there's a way to generate them as Unicode characters, but I don't know the details of that; I think you can discover them by poking around the web a bit.

It's a more general way of producing ln()s inside integrals. So if you want (ln(x))^2 in an integral, you just introduce x^t, evaluate the integral and then differentiate it twice

It's a more general way of producing ln()s inside integrals. So if you want (ln(x))^2 in an integral, you just introduce x^t, evaluate the integral and then differentiate it twice

that's called integration by parts and this video isn't about it

@@BLVGamingY you missed his joke

@shragdharkunal1258 if yes, i apologize profusely. may i get a short explanation

I did the same thing

hi this is very old, but i hope you could possibly explain in more detail what you did to solve it this way please :)

​​​​@@joo_21applying king's property/change of coordinates gives the integrand ln(1-x). Using the Taylor series of ln(1-x) gives -sum((x^n)/n) from n = 1 to inf. Consider the sequence of functions {f_n} where f_i (x) = (x^i)/i. We see that this is a sequence of polynomials which are continuous over Real numbers, and the limits of integration are finite, hence the functions in this sequence are integrable wrt x and their integrals do not diverge. Hence by fubini's theorem, we can interchange the integration and summation and we get (with the integral now): -sum(integral((x^n)/n)) = -sum((1^{n+1} - 0) / ((n+1)n)) = -sum(1/((n+1)n)), from n = 1 to inf. This is a standard telescoping sum. Consider the partial summation -sum(1/((n+1)n)) from n = 1 to k. We can rewrite this as -(sum(1/n) - sum(1/(n+1))) from n = 1 to k, which is equal to -(1 - 1/(k+1)). Taking the limit of this partial sum as k goes to infinity yields the summation: -sum(1/((n+1)n)) from n = 1 to inf which is -1

"x.ln(x) - x" is a primitive of ln(x)

Noureddine I know.

The same way Feynman did!:D

Wait how are you learning calc if you haven't learned precalc? Or did you mean your school hasn't covered precalc yet?

@@pbj4184 The second one. One year and smt ago we were at precalc at school, now we are covering derivatives

i like this

what a negative area means?

@@reycali6124 it means that the area is under the x axis.

it's just a fancy notation to make you understand where it's located on the plan.

This is a pretty scatterbrained explanation tbf

Is not like “theorem” is posh, is just a common and more precise way to characterize some statements.

Here's an example of a difficult integral with Feynman's technique: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Y6ZQMgk3A8s.html

sure, Fresnel Integral, Gaussian Integrals, Ahmed's integral, integral of Sinx/x from 0 to \infty to name a few exceptional ones.

@@farhannoor3935 The last one's called the Dirichlet integral

You have yet to learn your real name!

Calc. 3 is way easier than Calc. 2 so you'll be fine!

@@TrinidaddyGdom where lmao

To prove that we can switch the partial derivative and the integral, we would have to establish that x^t and x^t lnx (the derivative) are both continuous on the region of integration. For rigor, we would have the integral go from some lower bound "a" to 1, then take the limit as a approaches 0 so that lnx is continuous on the whole range of integration!

Man this is underrated.😂😂

ou encore calculer directement la primitive de lnx en posant u'=1 et v=lnx on a diretement xlnx-x et par le calcul evident de la limite aux bord de l'integrale on obtient -1

@@telemans107 Bien vu ! Pour la limite pathologique xlog(x) au voisinage de 0, faire le changement de variable x=1/y, ce qui ramène à la limite de -log(y) /y au voisinage de l'infini, laquelle est égale à 0, résultat bien connu.

Не используй машинный переводчик.

No pienso que nadie tenga el conocimiento y el tiempo para traducirlo, pero ¡me encanta que usted esté mirando mis videos!

@@MuPrimeMath guess I've got both of them

yo hablo español y un poco de ingles y creeme que se entendió todo de verdad explicas genial amigo sigue así

@@MuPrimeMath Why not just do t tomes ln of x at 5:31 as opposed to x^t..it's simpler and I think what most ppl would've thought of?

Once you have the base knowledge to solve a problem, the solution comes when you look at it from the right perspective. In the case of the integral in this video, the perspective is to look at the integral of ln(x) as the ending point instead of the starting point; that gets us to the solution. You have to try a method for a while, but once you end up just banging your head against the wall, think about the ways that you could flip the problem on its head; that's usually the best way to figure it out!

Mu Prime Math thank you for your ideas. Wish you the best

When the function is negative (the area is below the x-axis), we define the integral to be negative!