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

Another integral with Feynman's trick: • It took me 3 hours to ...
We can integrate ln(x) with integration by parts, but are there other sneaky ways to do it? Thanks to Tizio Caio for requesting this challenge!
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[@IustinThe_Human]

I feel like destroying an ant with a cannonball

[@PrasannaKumar-zx7gr]

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

[@aboutneetphysics1334]

😂😂

[@Invincible2203]

@@PrasannaKumar-zx7gr oh nice

[@aashsyed1277]

@@Invincible2203 HI

[@regulus2033]

I think, just in case someone doesn’t know one “but” of Feynman’s technique, you should remind that it is necessary to check the uniform convergence of considered integral in order to be able to differentiate it under the integral sign.

[@mrnogot4251]

Thank you. That exact thing had crossed my mind.

[@regulus2033]

@@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.

[@AshFeatherRocks]

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

[@luisescarcega]

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

[@frenchimp]

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

[@178fahimahmed7]

X^t was a brilliant idea in the solution.

[@pbj4184]

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

[@decib3ls_dB]

When I saw the thumbnail I tried to do it without parts and here is my approach. 1: Use kings law for definite integrals to convert it to the integral of ln(1-x). Expand that(Taylor series) integrate the polynomials and then plug values. You will get a infinite series that is fairly easy to evaluate and hence get the answer -1.

[@animeshmajumdar66]

I did the same thing

[@joo_21]

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

[@idrisShiningTimes]

​​​​@@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

[@tiziocaio101]

Thank you. This helped a lot. I’m trying to learn Feynman’s technique, but in my class at school we didn’t even study logarithms and exponential so I’m learning calculus by myself.

[@nournote]

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

[@tiziocaio101]

Noureddine I know.

[@TheEcwin]

The same way Feynman did!:D

[@pbj4184]

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

[@tiziocaio101]

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

[@최문규-o4d]

We can solve this by inverse function relationship between exp(x) and lnx Intrgral 0 to 1 lnxdx= -[integral 0 to infinite exp(-x) dx = exp(-x)]0 to infinite = -1

[@juicetime910]

i like this

[@starpawsy]

This feels like something I should have learnt (or at least memorised) in 1971, when Vince Pauley introduced me to the wonder of integral calculus. Alas, that was too long ago.

[@egeyaman4074]

Are there any integral questions we can't solve by laplace but we can solve by feynman's technique?

[@MuPrimeMath]

Here's an example of a difficult integral with Feynman's technique: kzbin.info/www/bejne/j2e9goCdoJh0btU

[@farhannoor3935]

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

[@pbj4184]

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

[@anthonyjulianelle6695]

My students would be asking why I was making it so hard when Integration by Parts does it much easier. I usually introduce differentiating inside the integral with an example where the integrand does not have an integrand which is an elementary function.

[@WizardCell]

Wow, the method looks fire, and the way you're teaching it is even more fire!

[@merveille271]

This integral equations of ln(x) and others are well explained and comprehended from the professor's explanation. Truly yours Sir!

[@JohnBoen]

I just started to reacquaint myself with calculus, and new videos are beginning to roll through. I've watched quite a few hours of mathy stuff this month, and you do great work. You provide excellent descriptions of the thought process. That was a great explanation of the Fineman approach. I can tell - I understood your explanation while you gave it, and I could almost replicate it on a piece of paper without looking back to the video. Where were you 35 years ago - when I could have used you in DifEq?!? Great work :)

[@derblaue]

My intuitive awnser to ln(x)dx from 0 to 1 was to integrate e^x from ln(0) = -inf to ln(1) = 0. e^(-inf)-e^0 = -1

[@ianrobinson8518]

Feynmans method can be expressed conceptually as differentiation in t followed by integration in x followed by integration in t to back out the original differentiation. Your method merely reverses the order of operations, viz integration in t followed by integration in x followed by differentiation in t to back out the original integration. One might say it’s still Feynmans method but in reverse. However note there is one other difference. There is no requirement to evaluate C since it essentially vanishes with the differentiation at the end. Which approach will prove effective depends on whether the derivative or antiderivative of the function is easier to integrate. Yours is a clever insight of what’s going on under the hood with Feynmans method. This is the second time I’ve seen a video on the unorthodox usage but I’ve seen a few other more complicated written examples where it comes into play.

[@evat267]

Cant wait to use this in my multivariable calc class for absolutely no reason since we havent done this yet and Im not sure we will

[@tgx3529]

I think, this solution is very complicated. If you need the result without integration by parts and without substitution, you can use , exp x is inverse function to ln x. Then - integral(exp x) from -infinity till 0 = integral from lnx for x = 0 till 1

[@Superman37891]

6:37 “FEYnd” lol 😂😂😂

[@saicharanritwikchinni9608]

Man this is underrated.😂😂

[@abdelwahabazeddine7035]

Sans diminuer de l'importance de la méthode de Feynman qui évite le recours à l'analyse complexe, mais dans le cas présent c'est comme tirer une mouche avec un canon. La méthode classique consiste à faire le changement de variable log(x) =u, x=exp(u), dx=exp(u)du Soit int(uexp(u)du) entre les bornes -inf et 0. Une intégration par parties donne directement le résultat recherché -1.

[@telemans107]

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

[@abdelwahabazeddine7035]

@@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.

[@edsonstarteri6313]

Let 0

[@domc3743]

You can make substitute x=u+1 and then define I(t)=ln(ut+1) and then I(0) = 0

[@russellthescout9639]

Thanks for this. I now know how to change the difficulty settings for calculus 2. Got so bored at easy mode.

[@RigoVids]

I have yet to take calc 3, I’m wearing with baited breath for spring semester to come, and I hope this kind of content will become available to me soon. I’m seeing partial derivatives, the part of calculus I didn’t get to in high school, and I am finally gonna start learning new stuff. It’s been ages since I’ve felt motivated like this.

[@MichaelKingsfordGray]

You have yet to learn your real name!

[@TrinidaddyGdom]

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

[@prodbytukoo]

@@TrinidaddyGdom where lmao

[@Комрон-л8ф]

I actually think that the formula of udv is more suitable for this integral , but the idea of the video is nice too

[@sihamamir548]

it's very easy to evaluate this intégral by parts you complicated it man

[@gloystar]

Well I guess it's much easier if you just use ( t ) and limit as t approaches 0, similar to the way we deal with improper integrals. Good video though!

[@shafin3365]

Nice approach.... Love from Bangladesh 🇧🇩

[@aleksandervadla4840]

Why just dont use: lnx=lnx*1, and use udv=uv-vdu

[@MuPrimeMath]

This video was meant as a fun challenge to see if we could solve the integral with Feynman's trick, without using integration by parts!

[@tiziocaio101]

Aleksander Vadla I was chilling and I was thinking a non standard way to integrate this integral. I tried Feynman’s technique but I failed, so I asked this guy.

[@eugenioguarino2651]

Interesting and smart. But there is a much easier way to calculate that integral: knowing that logarithm is the inverse function of the exp, that integral equals minus integral from minus infinity to zero of exp(x) dx, which is easily seen to equal minus 1.

[@spudhead169]

I'm sure this is considered thinking outside the box to you, but to me it's climbing out of the box, getting on a rocket ship and landing on Pluto. Feynman's technique scares me as it is without futzing around with it like that. I just hope I can eventually reach this kind of level.

[@mathematicslover1936]

Using integration by parts U=ln x first function V=1 second function Apply u. V formula of integration with get X(ln x - 1) ৷¹ 0 =-1

[@BS-bd4xo]

I was quite surprised that the integral was not ∞. But it does make sense, cuz the integral from -∞ to 0 of e^x is 1. Really good video.

[@daeyounglim1310]

I could watch this video til the end thanks to the cute instructor. Wish I'd had a cute math tutor like him

[@ciiil8802]

Differentiating Integration of Integral

[@harish6787]

Brilliantly done thanks for it sir

[@omchavan5664]

Is it only me with this unique methode Take xlnx, differentiate it you get lnx + 1 Now integrate again so Int(lnx) + int(1) = xlnx Int(lnx) = xlnx - x = x(lnx-1) Now put limits zero to 1 x(lnx - 1) tends to zero when x tends to zero And at 1 the value is -1 So answer -1

[@euva209]

Here's an inside the box way of getting the result without using by parts. Since the integral of e^x from infinity to 0 is symmetric to that of ln x from 0 to 1, we know the area of the latter will be the negative of that of the former.

[@jonetyson]

Differentiating under the integral sign (in a rigorous way) was known for hundreds of years before Feynman.

[@thelosts9940]

Isnt the integral of lnx like xlnx-x at least thats what i learned at school ?

[@thelosts9940]

And as long as you know this you can solve any integral with any sort of lnx with integral rules.

[@riveenwickenham8392]

As a left-handed person, this made me want to be right-handed...

[@nablahnjr.6728]

if you aren't allowing integration by parts i could always make an "educated guess" ;)

[@Aramizyera23]

I'm sorry, but you should use $\ln x$ instead of $ln x$ on your preview

[@tatawhillman3783]

This is brilliant, thank you

[@peterchindove7146]

Very well explained. Good job.

[@labiquette4821]

You can't derivate the function inside the integral without any justification, you have to use the theorem of derivation of an integral with parameter, there are hypothesis that you must verivy before using it

[@ЛеонидРубинштейн]

Дифференцировать под знаком интеграла можно только если он сходится равномерно. Differentiating under a sign an integral is possible only if he meets evenly.

[@IoT_]

Не используй машинный переводчик.

[@jojo1124]

Hi, could you provide a link to learn this technique?

[@pedromonteiro1556]

Excellent explanation! Thanks.

[@victormagaud2967]

don't we have a primitive for ln(x) ? xln(x)-x ? So we just have to works with limits

[@pawebielinski4903]

This is surprisingly cool.

[@Gianni_X]

Tizio caio, hai il mio rispetto🇮🇹🇮🇹🇮🇹🇮🇹

[@TschumiQu]

for finding a point of I(t), can't we just plug in I(1/x) so that the integral is zero?

[@MuPrimeMath]

No because t is a constant inside the integral, so it can't depend on x. It needs to stay the same as x goes from 0 to 1!

[@mathphschjhb7749]

sound like analytic continuation

[@renesperb]

I can't see the the advantage of forcing a certain method , in particular in such a simple case. Write the integrand as 1* lnx and integrate by parts and you are finished immediately.

[@bikashbayan1945]

For the last time, this is Leibnitz rule of differentiation inside an integral sign. Don't know why people call it otherwise.

[@bernhardriemann3821]

I did this in my head with a slight error, I considered exp(x) as the inverse of ln(x) and knew that it is the same as finding area under curve of exp(x) from -infinit to 0, but the error I made was the graph of ln(x) was below x axis so it should be -1 instead of +1.

[@faresberarma3349]

Feynman method is very helpful to evaluate some integrals but here you are using the wrong example

[@angeloritofasanaro9850]

Distributive large

[@leeraphael5081]

need to check why we could interchange the derivative, limit and integration.

[@swedishpsychopath8795]

I think the point where he messed up is that he mixed up integrate with derivate. You can't just write I(t) = xxx, and then write I'(t) = something. The point is to find the original expression that led to xxx, not to derivate the expression. Instead he should write: i(t) = xxx, and then find I(t).

[@janekgroe4304]

No integration by parts? How about integration by knowing the answer by heart bc its the most basic integral?

[@tcoren1]

Isn't this actually the regular version of feynman's trick, with the way you originally tried to solve it being the alternative? In physics I originally learned the trick as a way to quickly calculate the integral x^2*e^(-x^2) by taking the derivative of e^(-ax^2) with respect to a. Since feynman was a physicist it seems likely this is how the trick came to be

[@MuPrimeMath]

You might be right! I've never looked into the origin of the technique.

[@Pernambrock]

How do you know math being a left-handed ?

[@archangecamilien1879]

Perhaps use the symmetry of ln(x) and e^x, lol, or something...it should be the same as the integral from -1 to 0 of e^x, no?...Lol...perhaps I'm making some mistake here...

[@jamessquare]

(xlnx)'=1+lnx. So lnx=1+lnx-1=(xlnx)'-1. Then one antiderivative of lnx is xlnx -x, which you now evaluate at 1 and 0 to get -1. Look, Ma. No integration by parts!

[@tanmaymishra9576]

Its newton leibnitz 2nd form taught in B,tech 1st year

[@micha6589]

ok, but how do you know that the initial integral converges? If it diverges, you can get some bad results, I think.

[@MichaelKingsfordGray]

Too much like being back at Uni for comfort!

[@Karkov695]

Why can you derivate under integral sign? And why the function I(t) is continuous in t=1?

[@pk1pro]

Integration by parts meanwhile making this video 1 minute long only But this technique is wayyy cooler

[@kkahdnshsndjd4574]

Isnt it same as minus infinity to 0 e^x

[@odionmario5490]

Integration by parts works and saves time

[@debjitsarkar7128]

Awsome..to see a common thing in a different way is science

[@TriuneWorshipper]

I’ve never used integration by parts for this you should just know integral(ln(x))dx is always just 1/x (evaluated from a to b)

[@MuPrimeMath]

That's the derivative, not the integral!

[@TriuneWorshipper]

@@MuPrimeMath hehe 😉

[@folsomboy86tbone15]

I didn't read through every comment, so apologies if this is a repeat. When you integrated wrt t within the x integral, why didn't this produce "+ C" in the antiderivative, which would in turn integrate to "+ Cx" and then + C(1-0) when evaluated?

[@folsomboy86tbone15]

The constant would certainly disappear after the ensuing differentiation, but would this be correct nonetheless?

[@quentinremy2091]

I also don't have a plus one.

[@Triadii]

I would solve definite integrals by typing it in a calculator

[@pandabearguy1]

Everything in the world is exactly the same so this is also the same as itself

[@davidwebster9788]

Why give Feynman credit for Leibnitz' rule?

[@eytansuchard8640]

Good work!

[@__-xh3uw]

Minor nitpick: @ 1:55 it should be integral(d(I(t)))

[@obasimatictutorial]

Nice one sir..

[@gfrvgxhj4931]

On peut remplacer 0 par une variable "t" et en calcule la limite lorsque t tend vers 0+ c'est plus simple

[@absolutelymath3399]

I like this video. :) But I have a question. How do you deal with problems that you are stuck on and that bug you all the time? Just though you may have ideas.

[@MuPrimeMath]

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!

[@absolutelymath3399]

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

[@ayyu12]

U can also do it without byparts by realising that integration of (xf’ + f)dx = xf + C. So now u need to create something like that with Lnx. U can do that by adding and subtracting 1, then writing 1 as x.1/x -> integration(lnx + x(1/x) - 1). So now the first two terms fit the form I’ve mentioned exactly, and u can integrate -1 with ease too.

[@robertveith6383]

Spell out the word "you."

[@angeloritofasanaro9850]

We cannot Imagine this yech

[@sanketgore2515]

Nice work✌️✌️

[@jacklabestia]

Really interesting!

[@aupadhyay]

Why we should bother so much?

[@UnevenerGgc2]

Couldn’t you use the fact that the integral from a to b of f(x) = integral from a to b of f(a+b-x) and say that I(t) = integral 0 to 1 of ln(1-tx), I(0) = 0 and continue with feynmans?

[@MuPrimeMath]

Yes you could do it that way. You actually get an interesting infinite series at the end!

[@robertl.crawford4369]

Thanks kid....well done!

[@listentome5583]

I just finished calc 1, so this might seem like a dumb question, but are we supposed to know that the partial derivative x^t lnx = lnx, in order to solve this integral?

[@MuPrimeMath]

We know that the partial derivative of x^t with respect to t is x^t lnx because it's the same as the derivative of a^t, which is a^t lna. After that, we plug in t=0 to get lnx by itself!

[@kiryls1207]

@S Raaj K a^t = (e^lna)^t = e^(lna*t), so the derivative of a^t (in t) is the derivative of e^(lna*t) => e^(lna*t) * lna = a^t * lna

[@gauravaswal25]

Graph ?

[@gabrielalmeida5047]

Thank You

[@vashushukla1727]

love it brother

[@CringeBasedDuality]

Terrence Tao did this on his diapers

[@ignacio4067]

Actually I(0) wouldn't be helpful, for Ln(abs(0)) doesn't exist, would it?

[@MuPrimeMath]

Even though ln(0) doesn't exist, we can still take the integral of ln(x) from 0 to 1! To do this rigorously, we would first take the integral of ln(x) from a to 1, where a is some positive real number. After that, we take the limit as a→0+. I skipped that step since we get the same answer either way.

[@ignacio4067]

@@MuPrimeMath Thanks, though I meant the integral of Ln(tx+1) (3:23 aprox.) , yet I think then I(t) might not still be Ln(abs(t)) + C as with I(t) = integral from 0 to 1 of Ln (tx). There you're not taking the integral but evaluating the primitive Ln(abs(t)) + C at t=0 Could you still resort to taking your limit? Sorry if I'm talking nonsense Which time do you refer to?

[@leif1075]

Isnt anyone else wondering why not just use t times ln x at 5:19 instead of x^t times ln of x?

[@MuPrimeMath]

Because knowing the integral of t*ln(x) dx requires knowing the integral of ln(x), whereas we can do the integral of x^t dx without knowing the integral of ln(x).

[@stepone3040]

Would be better if you introduce the technic before you use it cause not everyone knows the proposition

[@franconuzzi4000]

In this case the parts method is more fast.

[@chainsawmanfan3002]

If I could be smart by being left handed