Integral of ln(x) with Feynman's trick!
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- Опубликовано: 29 сен 2024
- 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
@@Invincible2203 HI
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
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
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.
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.
Wow, the method looks fire, and the way you're teaching it is even more fire!
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: ruclips.net/video/Y6ZQMgk3A8s/видео.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
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
I actually think that the formula of udv is more suitable for this integral , but the idea of the video is nice too
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
Thanks for this. I now know how to change the difficulty settings for calculus 2. Got so bored at easy mode.
Differentiating Integration of Integral
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
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
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!
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
Differentiating under the integral sign (in a rigorous way) was known for hundreds of years before Feynman.
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.
Tizio caio, hai il mio rispetto🇮🇹🇮🇹🇮🇹🇮🇹
This is brilliant, thank you
Excellent explanation! Thanks.
I'm sorry, but you should use $\ln x$ instead of $ln x$ on your preview
sound like analytic continuation
I would solve definite integrals by typing it in a calculator
Too much like being back at Uni for comfort!
Its newton leibnitz 2nd form taught in B,tech 1st year
Thank You
don't we have a primitive for ln(x) ? xln(x)-x ? So we just have to works with limits
Everything in the world is exactly the same so this is also the same as itself
How do you know math being a left-handed ?
I also don't have a plus one.
Why give Feynman credit for Leibnitz' rule?
By Integration by parts it will be easyer
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?
Very nice!
Just use primitiv of ln(x) = x ln(x) - x and your done ;)
No? Put 0 in ln(x) and then tell me how finite is your undefined value...
Thomas Timothy Martinez Eric Williams Patricia
Integral of ln is x.lnx-x
that was awesome
Sign is function
Please don't use your finger to wipe the board, oil from the skin ruins the pen and the board
I will rather use graph rotation to evaluate this integral.
large
Nice!
Leibniz rule*
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
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...
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. 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.
thanks for a well-thought-out and well executed video. i' m a mathematician myself and i really enjoyed it.
"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"^^)
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
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?
I love the fact that it's Feynman's "trick", and not his theorem or other such posh word. He would have loved that!
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.
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
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
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
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)
i love your t-shirt "Calculus Finisher" which i feel like to win calculus marathon ^^
Nice alternative, nonetheless looks slower with the respect of integration by parts
We cas add 1_1 it will be ln x + 1 -1 = lnx + x/x -1 = (xlnx)' - x'
Hmmm just a linear transformation would suffice
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.
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.
If it's Leibnitz rule, surely it's Leibnitz trick, not Feynman's.
Correct, but it was popularized in modern times by Feynman, and it isn't taught often in standard calc classes, but is brought up in physics.
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.
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 ❤️
Integration by parts meanwhile making this video 1 minute long only
But this technique is wayyy cooler
Am I the only one who is feeling uncomfortable by the way he holds the pen
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.
You can make substitute x=u+1 and then define I(t)=ln(ut+1) and then I(0) = 0
This is Leibniz's technique NOT Feynman's. Feynman took this from a book on Advanced Calculus. The real name of the technique is "Leibniz Integral Rule". To attribute this to Feynman is complete BS. He didn't come up with it.
The book he got this method from as well a several other gems is: Advanced Calculus by Woods, 1926.
Very beautiful demonstration! But you have to be careful with the bound x=0! In almost every formula or transformation of your theory you have to exclude x=0! And the definition of intrgral_0^1 lnx is
Lim_a->0^+ Integral_a^1 lnx dx. But if you calculate that definition with the antiderivative x lnx - x you find with the rule if De L'Hópital lim_a->0^+ (a lna) =0 and the limit value really is -1 . So a strict mathematical analysis of this indefinite integral Integral_0^1 lnx dx really equals -1. I am delighted! 😀
1:00, 2:20 And more foundational: Of course, one hast to justify WHY you can interchange the integration with the differentiation for t in the first place... but physicists and so on ;).
HOW VERY INTERESTING
MY eyes hurt so much watching integration videos nonstop for 12 hours+ but just one more and I shall
Finally sleep
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...
(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!
My favorite way to evaluate this integral is Symbolab
I simply "knew the answer was x ln(x) - x"
At 0 it approaches 0
At 1 it's -1
So it must be -1
(Watched video to check I was right)
You can just do it properly mathematical with limits. Not difficult but way more stylistic. Tricks like that are funky and technically need to be proofcheck every fucking time. Because they are hard to proof in general because of convergence. Sure you can make a statement about it but good luck memorizing it.
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.
x=e^(-u)
x in (0,1) implies u in (infinity,0)
dx=-e^(-u)du
ln(x)=-u
integral is just -Gamma(2)=-1
I haven't been able to keep up since 4:25. Why can we say that I(t) = Integral x^t? dx? I've been able to do it myself until the point where I(t) = ln(t) + c
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).
Hmm, I see what you guys are all suggesting, but my alternative approach would be to sit down and have a beer! This is WAY over my head!!
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!
Wow noice love mathz from ❤️ 💙 💜 💖 💗 💘 ❤️ 💙 💜 💖 💗 💘 ❤️ 💙 💜 💖 💗 💘 ❤️ 💙 💜 💖 💗 💘 ❤️ 💙 💜 💖 💗 💘 ❤️ 💙 💜
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
6:37 “FEYnd” lol 😂😂😂
Man this is underrated.😂😂
I think, you need the proof, that your original integral converges, you need convergent majorant I'(t) independent on t
Why everyone on youtube forgets to proof the unifotm convergance of those integrals? It the main and hardest part of feynmam technique
if you aren't allowing integration by parts i could always make an "educated guess" ;)
you keep putting the marker in the cap every 5 seconds lol just leave it out man let it be free
d/dx (x ln(x) ) = 1 + ln(x)
d/dx ( x ln (x) - x) = ln (x)
Integration (0, 1) ( ln ( x) )
= (value at x = 1 ) ( x ln ( x) - x)
- (value at x = 0) ( x ln ( x) - x)
= ln( 1 ) - 1 = -1
Herein we used the following
(as x ---> 0 ) ( x ln (x) )
= (as z ---> 0 ) ( exp ( z) z )
= 0
Feynman method is very helpful to evaluate some integrals but here you are using the wrong example
No integration by parts? How about integration by knowing the answer by heart bc its the most basic integral?
Oh yeah, instead of IPP, you derive under integral sign without any security