Integral of ln(x) with Feynman's trick!
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- Опубліковано 8 вер 2019
- 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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I feel like destroying an ant with a cannonball
May be, destroying the ant with cannon ball makes it easier to target the elephant 🐘
😂😂
@@PrasannaKumar-zx7gr oh nice
@@priyanshutyagi3688 HI
My favorite way to evaluate this integral is with my right hand.
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
My favorite way to evaluate this integral is to recognize that this is just negative the integral of e^x from negative infinity to zero.
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
Wow. Splendid. I think this “trick outside the box” is in fact a much more intuitive approach at understanding Feynman integration in general. Really nicely done. Thank you!
Steve K.
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.
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...
thanks for a well-thought-out and well executed video. i' m a mathematician myself and i really enjoyed it.
Well, my trick for this would be to swap coordinates.
y = lnx ; x = eʸ
When lnx → 0⁺, y → -∞; when lnx = 1, y = 0.
And lnx is negative in the (open) interval of integration, while eʸ is positive in its interval. So there will be a sign change, and:
∫₀¹ lnx dx = - ∫₋₀₀⁰ eʸ dy = -1
Done. But it could be argued that this is equivalent to integration by parts.
EDIT: And I see mine is not the only comment using this trick.
Fred
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.
Wow, the method looks fire, and the way you're teaching it is even more fire!
X^t was a brilliant idea in the solution.
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
Pretty slick trick.
Knowing how the integral from -inf to 0 of e^x is 1, it makes sense intuitively that this integral evaluates to -1.
x^t was really clever
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
"remember a partial derivative means everything not t is a constant" I love scaring my students about multivariable calc and when we start with partial derivatives they all go like "wait: it´s that easy?" They were expecting some ultimate hell. And then i always go like "i never said the maths would be hard. but getting that curved d right will be a nightmare for anybody not used to cursive" They usually want to kill me^^ and yeah that was a really clever idea to just reverse Feynman´s technique. (and please call it a technique trick sounds cheap. Usub for me is a "trick" elevated to the status of a technique by usefulness. Feynman´s technique is the same incredibly useful and also a very elegant use of the leibnizrule. (strictly speaking they are not the same as you know better than me. It is strictly speaking allowing us to switch Integral- and differential operators.Feynman is still rthe guy who thought "well I can crack a couple of tough nuts with that"^^)
Thank you Mu Prime! That x^t is awesome!
This was really neat!
I used your method to evaluate the indefinite integral of ln|x|^k. I used the Leibniz rule for integration and the general Leibniz rule. I let I(t) = integral of x^t such that the nth derivative of I(t) = integral of the nth partial derivative of x^t = integral of ln|x|^n * x^t dx. I then used the power rule to evaluate I(t), and the general leibniz rule to evaluate the nth derivative of that. I simplified where I could, replaced every n with a k and t with a zero, then ended up with a finite sum (assuming k is finite). Very nice I thought.
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 :)
Wonderfully useful way feynman technique. Thank you for the video sir!
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.
Honestly, I didn't expect this video could be *Super Cool* ... But it was!
*Great* ... *Sure Cool* ...
(Actually you teach another way of thinking about Math and it's great too.)
I love Feynman (he's my favorite scientist) and he brings me here and now I'm so happy.
Great example and definitely the different example of Feynman's Method.
Thank you so much ❤️
While trying this myself I just realised that you can also find it by realising that since exp(x) is the inverse of ln(x), which means its the graph reflected in y=x. If you interpret the required integral as the area, it will be te same as the integral of exp(x) from -inf to 0. this has a nice convergence to 1. Since ln(x)< 0 on [0,1] this means the area is -1. and hence the integral is -1. Might sound a shabby but ey.
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.
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
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.
My favourite way to integrate ln X is to call it 1 times ln X . Then you can call the integral of 1xln(X) xln(x) - integral of x(1/x) which turns out to be xln(x)-x+c
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
Real clever mate. Well explained.
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.
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
Are there any integral questions we can't solve by laplace but we can solve by feynman's technique?
Here's an example of a difficult integral with Feynman's technique: ua-cam.com/video/Y6ZQMgk3A8s/v-deo.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
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
Hi, could you provide a link to learn this technique?
i love your t-shirt "Calculus Finisher" which i feel like to win calculus marathon ^^
This is brilliant, thank you
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.
"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
Integration by parts works and saves time
Thank you, its hard to find a full fledged explanation of this technique online.
This is a pretty scatterbrained explanation tbf
This integral equations of ln(x) and others are well explained and comprehended from the professor's explanation. Truly yours Sir!
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
i like this
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.
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
Here we can also think the integral given is the area traced the the curve lnx and -ve y and +ve x axis so its inverse fuction e^x curve will have the same area between -infinite to 0 so the integral becomes Integration of e^x from 0 to -infinite (as y becomes +ve in this)= -1
Brilliantly done thanks for it sir
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
how do we know we can perform the partial derivative inside the integral?
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!
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.
You have yet to learn your real name!
Calc. 3 is way easier than Calc. 2 so you'll be fine!
@@TrinidaddyGdom where lmao
love it brother
that was awesome
Explicas de maravilla, deberías de buscar alguien que ponga subtítulos en español, por que a veces los latinos no tenemos buen inglesb
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?
Isn't ln(x) one of the integrals we should have memorized? It's -x+x•ln(x). Evaluated at 0 is 0. Evaluated at 1 is -1+1•ln(1). ln(1) is zero, so the integral is equal to -1+0-0=-1.
Really interesting!
Very well explained. Good job.
Круто! Не знаю, что ты рассказывал, но в целом твои действия понятны и без комментариев)
Why can you derivate under integral sign? And why the function I(t) is continuous in t=1?
6:37 “FEYnd” lol 😂😂😂
Man this is underrated.😂😂
How do you know you can put the derivative inside the integral?
I was watching this blackpenredpen video about how he integrates integral from 0 to 1 of (x-1)/lnx dx where the question was from a JEE test and its hard to wrap my head around the fact that you can integrate an integral at the bounds where the function is not defined itself. I tried it myself using taylor series expansion for e^u after making a substitution of x=e^u and then kind off finding limit putting A and B in as the bounds where A tends to 1 and B tends to zero. But that did not work. He used the Feynman method but its hard to wrap my head around the fact the integral had a finite value which otherwise seemed nonexistent.
That is because integrals only care about intervals, not points.
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?
The constant would certainly disappear after the ensuing differentiation, but would this be correct nonetheless?
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?
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!
@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
Thanks for this. I now know how to change the difficulty settings for calculus 2. Got so bored at easy mode.
Why just dont use: lnx=lnx*1, and use udv=uv-vdu
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!
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.
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.
You can make substitute x=u+1 and then define I(t)=ln(ut+1) and then I(0) = 0
This is surprisingly cool.
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.
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!
Thanks kid....well done!
Integral is a kind of an area. How it can be negative if the function isn’t set at the zero?
When the function is negative (the area is below the x-axis), we define the integral to be negative!
Дифференцировать под знаком интеграла можно только если он сходится равномерно. Differentiating under a sign an integral is possible only if he meets evenly.
Не используй машинный переводчик.
I could watch this video til the end thanks to the cute instructor. Wish I'd had a cute math tutor like him
Excellent explanation! Thanks.
Nice one sir..
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
You might be right! I've never looked into the origin of the technique.
Let 0
I love the fact that it's Feynman's "trick", and not his theorem or other such posh word. He would have loved that!
Is not like “theorem” is posh, is just a common and more precise way to characterize some statements.
That's brilliant 👏... thanks for this beautiful tip (outside the box 🤪). 😉
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.
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.
Nice work✌️✌️
Pretty cool! 😎
It's just eazier to integrate the function and evaluate
But this trick is very useful for function whose integral isn't elementary(like e^x²)
Impressive
you're a great teacher
On peut remplacer 0 par une variable "t" et en calcule la limite lorsque t tend vers 0+ c'est plus simple
Isnt the integral of lnx like xlnx-x at least thats what i learned at school ?
And as long as you know this you can solve any integral with any sort of lnx with integral rules.
Good work!
My favorite way to evaluate this integral is Symbolab
Forget the mathematics. You are a fine THINKER. You could do anything in life.
Since ln(x) has a vertical asymptote at x=0 how can we find the area from 0 to 1?
It turns out that, because of the way that ln(x) approaches negative infinity, the area is still finite! It's similar to the idea of taking the integral of e^(x) from negative infinity to zero, which is also finite.
@@MuPrimeMath Thanks. I guess it's no stranger than the Gabriel's Horn solid of revolution.
ok, but how do you know that the initial integral converges? If it diverges, you can get some bad results, I think.
Maaaan help, how come the area under the function is negative?
The function ln(x) is negative between 0 and 1, so the integral is negative as well!
Now the really important question is where do I get that wallpaper? 😆
Could you not just take the integral of e^x from -inf to 0?
You absolutely could do that! I thought it would be fun to challenge ourselves by trying to solve the integral with only Feynman's trick, since we have to be a little creative.
I figured it out, but I kinda cheated by trying to do it by parts to try and see where I am going. Anyway, I managed, but it took longer than I want to admit, considering it was an approach so simple.
Edit: and the reason was : I didn't bother evaluating the polynomial term (x^t) to get -1. Perhaps looking at the basic things might make you see the path to solving something seemingly unbreakable. Lesson learned again, I thank you.
Good stuff!
I actually think that the formula of udv is more suitable for this integral , but the idea of the video is nice too
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.
for finding a point of I(t), can't we just plug in I(1/x) so that the integral is zero?
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!
Nice approach.... Love from Bangladesh 🇧🇩
clearly explained
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.
Differentiating under the integral sign (in a rigorous way) was known for hundreds of years before Feynman.
Bravo!
Can you give the name of your board?
writeyboards.com/
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.
Differentiating Integration of Integral