Math for fun: integral of ln(x) from 1 to ? is equal to 2
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- Опубліковано 9 лют 2025
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I want the area under the curve y=ln(x) from 1 to some number t to be 2, but how can we achieve this? Not only do we have to use calculus integration by parts, but we also need to use the Lambert W function to solve the resulting equation for us.
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Bonjour prof vous êtes un excellent talant vivant en mathématiques.. j'aimerais être comme vous mais pourriez vous me dire comment être aussi brillant que vous....je souhaite être meilleur c'est pourquoi je me confie à vous...
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In Finding Nemo they could have just used the Lambert-W function and then they would easily find Nemo
They need to heve Nemo in the exponent first and I doubt they had that
Yay the fish is back
Fun fact, the general form with an area of N under the curve is t = (N-1)/W((N-1)/e)
that's what I thought right away when I saw the question :)
What an interesting problem! I could see things like this being used for many applications, like how long you need to drive to drive a certain displacement.
the fish function is kinda bad imo, you give away a medium fish and a small fish for a big fish but I personally find that since you have to eat the fish in one go or it'll go bad the m fish and s fish serve a better purpose for consumption and if you are collecting fish, I doubt you want to decrease the number of fish you have unless your tank is running out of space, but bigger fish generally need more space to be healthy so I wouldn't use this as a solution even then
Ahh isnt that fishy
@@mitochondria1674 LoL
so true goat!!
Wow first real practical application of the Lambert W function I've ever seen that could come up in everyday calculus! Thanks so much!!
Hey bprp, I just wanna say a massive thank you for your videos, not only are they super entertaining for me but when I started watching them a couple years ago I had no idea what was going on, not about calculus, complex numbers or anything. But slowly I started picking things up and I started to basically get a online class from you, whilst being entertained with great content. Now that my curriculum is more lined up with your content and european style (IB by the way), I actually use the stuff you use a lot and it helps me get really good grades, so yeah sorry for the paragraph but thanks :)
For anyone wondering you can simplify the result like this:
= e^(W(1/e) + 1)
= e * e^W(1/e)
= e * W(1/e) e^W(1/e) / W(1/e)
= 1 / W(1/e)
I didn't understand why e*W(1/e)e^W(1/e)=1
I got the last result first ;)
@@riccardomarino9786 because, by definition, W(x)*e^W(x)=x. It is the true property of Lambert’s W function.
Even I got the last result first.
@@riccardomarino9786 by the definition of inverse functions f(f^-1(x)) = x
if f(x) = x*e^x and f^-1(x) = W(x), then:
x = f(f^-1(x)) = f^-1(x)*e^f^-1(x) = W(x)*e^W(x) -> W(x)*e^W(x) = x
in this case we have e * W(1/e)*e^W(1/e) = e * 1/e = 1
Thanks to your videos I was actually able to solve this one on my own, never thought I would like math this much. Thank you for reigniting my interest in math :)
Integrals like this are so satisfying to solve, especially when the result is as elegant as 2! This video does an amazing job breaking down the steps clearly. I’ve been practicing similar integral problems, and SolutionInn has been super helpful for finding more examples to work through.
I am amazaed by this guy just evaluating e rised to an exponent accuretaly in his head, truly amazing
It's crazy how after watching so many of your videos, I can solve problems like these without any problem at all in a matter of seconds. Lambert W fuction really is so good afterall.
That’s awesome, cheers!
hey bro... Where are you from?? I'm from Bangladesh... Your videos are so helpful for me... Some days later my year final examination... I'm derepressed 😭😭.... Take my love ❤
He's from Taiwan
Such a common problem man, thank you for the solution.
Nice question and good answer, love it.❤
You can also use parametrical equations to integrate ln x
y=t
x=exp(t)
$ydx
dx=exp(t)*dt
$t*exp(t)dt
You don’t get out of integrating by parts but it’s a little more intuitive.
Fun fact, I actually did this question and t can also be write as the 1/LamberW(1/e), i feel like the method you used was harder than mine though but its fun to see that e^( LampbertW(1/e) + 1) is the same as 1/LamberW(1/e) seems like it wouldn't be true😀
perhaps it's just not nice to have W in the denominator :)
It is actually a simple consequence of the product log function being an inverse of
f(x) = x * exp(x)
In this channel, the most common usage is
W( f(x) ) = x
but you should not forget the other usage which is
f( W(x) ) = x
W(x) * exp( W(x) ) = x
then you simply divide by W(x) to get
exp( W(x) ) = x / W(x)
and that identity can be applied to this problem to show the equality that you mention:
exp( W(1/e) + 1) = e * exp( W(1/e) )
exp( W(1/e) + 1) = e * (1/e) / W(1/e)
exp( W(1/e) + 1) = 1 / W(1/e)
@@XJWill1 I used a more elementary friendly way to get to my result here it is : We have gotten to the point x ln(x) - x = 1, we use logarithm rules and expand the x to ln(e^x) now we get ln(x^x) - ln(e^x) = 1 lograithm rules: ln(x^x/e^x) = 1 take out the ln, we get x^x / e^x = e, now take denominator to the other side and multiply: x^x=e^x * e now take the root of x on both sides: x= e^(x/x) * e^(1/x) we get x= e^1 * e^(1/x) take the e^1 to the other side: x/e = e^(1/x) , now take the x to the other side: 1/e = 1/x * e^(1/x) as you might have noticed you can take the lambert W on both sides,
W(1/e) = W(1/x * e^(1/x)) simplify W(1/e) = 1/x hence x= 1/W(1/e), yours is obviously much more simpler but as a highschooler myself, I find that this way is much simpler, though I see how your idea provides for a much better and simpler, and thanks for the proof btw.
@@romanbykov5922 Hmm not sure, 1/x is usually a pretty friendly function
Instead of using the lambert function and then solving numerically, you can just use recursion to solve numerically. Since t.lnt-t=1 then t=(1+t)/ln(t). Let ans equal some starting number e.g. 10 and then ans=(1+ans)/ln(ans) and repeat until it converges e.g. 4.777239, 3.694211, 3.592233,] 3.591122, 3.591121. It converges pretty quickly and easy to do on a calculator.
Paused the video, tried it first, comes out approx 3.591. 1 / (Fish 1/e)
Edit : Fixed typo
yay a heart!
The W function is my favorite function I never learned about in a proper math class!
Evaluate the surface integral ∬SF⋅dS
for the vector field F(x,y,z)=x^2i+y^3j−xzk, where S
is the part of the paraboloid 0.6+1.1x^2+y^2−z=0
that lies above the disk x^2+y^2≤1
and has upward orientation
Can you do a clear video of the Lambert function for high school students? Thank you.
Gosto muito da condução dos seus vídeos, da forma como faz matematica.
please try this equation: e^i^x=x
You mean e^(ix) = x
or e^(i^x) = x
@@hellohabibi1 ... As the OP wrote it it means the second.
@@hellohabibi1, the second one, the tower of powers
t can be further simplified to just 1/W(1/e). I still have not taken full calculus classes but I do love the lamber W function
I knew the answer was going to use Lambert W function the second I saw the problem)
How? I’m new to lambert W, how do you recognize it so early?
@@onetwo7191 It's a practice thing, kinda like how you can expect to solve a quadratic whenever powers of 2 are involved. The Lambert W function appears naturally in natural log and e problems, and since the integral of ln(x) is probably some weird product of natural logs and such, it's not a bad guess that the Lambert W function would come into play
I did the integral then used Newton-Raphson iteration to approximate t as 3.59, which is correct to 3 significant figures. I think that's more satisfying than using the W.
@@sweettoy3824W is an analitic function, sometimes is better to have one
Integration of lnx, by parts:
u = lnx, u' = 1/x
v' = 1, v = x
2 = [uv - integration of u'v]
2 = [xlnx - integration of 1]
2 = [xlnx - x]
Upper bound is t, lower bound is 1
2 = tlnt - t - ln1 + 1
2 = tlnt - t + 1
1 = tlnt - t
1 = t(lnt - 1)
Iiii have no idea how to simplify and/or solve this
Dam I saw it from the thumbnail and thought we got to choose the lower bound too and had an answer since we could freely choose one
Not my first Lamberto tho. I see ln, I know.
Get integral and factor out a b to be nicer.
b(lnb-1) = 1
Now set up lambo by creating an e^ term. Then have it match what’s next to it by multiplying e^-1.
e^(lnb) (lnb-1) = 1
e^(lnb-1) (lnb-1) = e^-1
Lambo:
lnb-1 = W(e^-1)
Finish:
b = e^[W(e^-1)+1]
The result can be simplified to 1/W(1/e)
the lambert w function is bprp's favorite math concept
I guarantee the answer uses Lamdba W function!!
you mena lambert?
you mean mean?@@DotDotEight
@@sowndolphin5386 yes my bad, i didnt notice that typo
Whenever I see problems that use the Lambert's W Function, I just say, nah too lazy, and end up using the Newton-Raphson Method 😂😂
My solution was pretty similar to that in the video, but I put the answer in a neater form (as some others have in their comments)
∫₁ˣln t dt=[t(ln t-1)]₁ˣ (using integration by parts)
=x(ln x-1)-1(0-1)
=x(ln x-1)+1
So we need
x(ln x-1)+1=2
x(ln x-1)=1
x(ln x-ln e)=1
x ln(x/e)=1
(x/e)ln(x/e)=1/e
ln(x/e) e^[ln(x/e)]=1/e
So (as 1/e>0) there is a unique real solution, which is
ln(x/e)=W₀(1/e)
x/e=e^W₀(1/e) [⇒x=e^[W₀(1/e)+1] ]
But W₀(a)e^W₀(a)=a, so e^W₀(a)=a/W₀(a)
So x/e=(1/e)/W₀(1/e)
x=1/W₀(1/e).
plz make a video and explain a simple solution for Morley's theorem.🙏
Если использовать формулы для функции Ламберта, то ответ можно упростить:
t=e^(1+W(1/e))=e*e^W(1/e)=e*(1/e)*1/W(1/e)=1/W(1/e)
Получаем: t=1/W(1/e)
😜
Здесь использовали формулу: e^W(x)=x/W(x)
Could you use series expansion of ln(x) instead of W?
i dont know about W function cuz i am in high school . i tried and got t should be between 3 and 4
Integral entre 0and1(-ln(x))^1/2
this answer actually nicely simplifies to 1/W(1/e)! (not factorial, just excited)
I got the solution, but I had trouble checking it.
Lambert W function ❎
bprp's fish function ✅
😊
You can put the answer in an alternate form as follows: t=1/W(1/e), which looks better in my opinion.
X = e^(W(3/e) + 1)
no
@mhm6421 i saw end of the video, i got it wrong sorry
hey bprp, could you try making a vid about tetralog someday? I wld appreciate if you do.
Nice!
Is there any way to expand ( a+b+c+d)^7
Troll question!
Try this insanely hard integral :
Integral between 0 and pi/2 of tan(x)^(1/(x^x))
Bro plz try this d/dx(log base x of n)
Can you solve for y?
x = log(base b) of [ (W(y' /lnb)) /lna ]
normal person : W(xe^x) = x
bprp : W(🐟e^🐟) = 🐟
Just checking in to see how you pronounce ln x.
😆
W still feels like cheating. Or "here magic happens".
I've even asked WolframAlpha this question, but it seems to bug out and ask me to reask. When I formulate it another way, it just shows me a numeral approximation but nothing else. Could you try and solve it, please? Thank you!
x^x^x = 2
Hey Steve! Show us a HARD analysis problem - at the 899 (end of PhD or even postdoc) level. 😈
Though I know how to solve this I came to see because to see bprp's Lambert fish function. 😅
Try this very difficult integral : ∫_0 ^(pi/2) tan(x)^(1/x) dx.
So you want a VERY difficult question ? Solve the differential equation : ∫_0 ^pi ( ∫_y' ^y'' ) dx dt is y'''(e).
Lambert what?
Your help please Integral of ((x^2 + sin x)/(x^2 +1))dx
Well did it after expanding logt upto t-1 or even upto of power 3 from there we can get an approx value 1+ (2)^1/2 instead using lambert transformation
I have to say, I don't remember what the antiderivative of ln(x) was...I suppose if it was that straightforward, lol, there wouldn't be a video about this...maybe it doesn't have a straightforward derivative?...I don't know, lol...
I don't think I've actually understood what is the W function
It's called Lambert W function, also known as the Product Log function. I'm not an expert on the topic myself, but I think of it as an inverse function of
f(x) = xe^x
Its just a function we define as the inverse function of y=xe^x; simalar to how if we have y=e^x, its functional inverse is y=ln(x)
This function allows us to solve equations that can't be solved using normal algebra techniques. Such as equations where we have a variable outside and inside an exponent or logarithm.
bprp has a couple of videos about the Lambert W function on his channel.
@@Bayerwaldler... He is incomplete. It is wrong to never evaluate W(x)s by calculator or computational for higher level Numerical Analysis Senior level University computer math courses. How did he find W(1/e) to get his approximate value of 3.59112? There are no lookup tables that will equal a log(x), ln(x), or trigonometry type equivalent table for x values.
From Wikipedia the Lambert Function is a summation formula. W(x) = summation of n=1 to infinity [((-n)^(n-1))/(n!)](x)^n for |x| < 1/e. Wikipedia states outside this range the solution for 1/e
What if I want that area to be i?
can you please come to Iran someday? pleaaaaaaaaase
i tried it; my brain was hurting i was so angry but i got 3.5911223...
I don't know the relationship between this number and e or ln
Please make a detailed video on fields medal
I am the first one commenting on your video
Math for fun fact: integral of ln(x) from 0 to 1 equals -1
Can you prove that why
(-)(+)=-
(-)(-)=+
Please Make a video with a youngest an and famous mathematician and physician subbrno isaac bari
Troll, go home
The first one to comment on! 🕶
Stop making easy exercices, I feel that I am good at math
B = (84 - A)/(1 + A) (A and B both are positive integers. A, B > 1 ) then find the value of AB.
I dont understand the W function, can someone explain? It doesnt seem to give any information at all. Why not just invent some function K so that t = K(2)?
I think i'm first ?
I am the 300th viewer
first
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jesus touched me when i was 6
Odin said I can go to Valhalla
you lot have been saying the world will end any day now for almost 2000 years
@@jackwoods7070 So what? Quoting from a book that is more than 2000 years old and was written by ignorant people proves nothing.
@@bjornfeuerbacher5514 Why do you say they were ignorant? Have you achieved omniscience, so no one will ever be able to say you were ignorant?
Нашёл t≈3.5911215, решал через функцию Ламберта W(x).