Integrating Lambert W Function
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- Опубліковано 9 лют 2025
- In this video, I showed how to integrate Lambert W function using integration by parts and U-substitution. The process is quite similar to the one employed in integrating ln(x)
Derivative of Lambert W Function
• Derivative of Lambert ...
D - I method of integration by parts
• D I Method Integratio...
That is the cleanest in-use chalkboard I have ever seen.
And maybe he uses Hagoromo chalk.
Damn. The way you present is so smooth. I love it
You are a most gifted teacher
I just discovered this channel today, your explanations are very clear and is very obvious you have a real passion for math. Your content is amazing, please keep bringing these amazing videos
Welcome aboard!
This was very useful. Please do a follow up video on the GLOG function. GLOG seems to be related to LambertW but is more suited to a different class of problems. In particular, the Colebrook friction factor equation can be solved directly with the GLOG function.
I just read this comment today. I never heard of the GLOG function. I am currently reading up on it. I will make a video when I fully understand it. Thank you.
11:49 "this is you, remember", Yes! It's me! I love how you are talking to me in this video 😅
Lol
Man, I love your videos. You got a talent for teaching
I appreciate that!
This was quite wholesome to watch! Keep it up man!
Mon cher ! Vous êtes vraiment très fort ! Et vos explications sont très claires ! Merci bcp !
Thank you!
We can also use the fact that the integral of an inverse function f^-1 (x) =
xf^-1(x) -F(f^-1(x)) + C
in this case f(x) is xe^x and F(x) is xe^x-e^x
that term at the end x/W(x) is the same as e^W(x)
Danke!
Thank you!
I LOVE YOUR VIDEOS MAN, THEY MAKE ME LOVE CALCULUS
Bro got predicted, good video as always
You have impeccable handwriting.
Is this an ASMR math or am I missing something?
I was amazed at myself when I saw that I was able to reach it before seeing the video, but I did not use changing the variable. I used the logic that I put that W(x) is the function that connects xeⁿ to x and not vice versa, and I put an integral for x, but with dxeⁿ
your math is very organized and concise! loved the integration by parts table. thanks!
Just discovered this channel and I have to say I loved the way you explained this!
I think a lot of students wouldn’t be as afraid of math if they had a professor like you, this is a marvelous integral 🙏🏻
Welcome!
i fell in love with your channel!
Great video, thanks! Methinks I should've started with the dW(x)/dx video, though.
Good call!
Me, at first: "Oh, no. This is going to be intimidating." Me, at the end: "Hey, that wasn't so bad, after all."
That is good teaching.
amazing explanation
I loved the video, awesome!!!
I have a PhD in math and I've never seen a DI table. It isn't very different from what I learned, I just didn't learn to write it that way. I think if I ever taught again I would start by teaching the way I learned it, since it uses first principles, but then teach the DI table after the first test.
I agree.
The mad lad did it.
The integral of any inverse function:
x*f-¹(x)-F(f-¹(x))+c ( f-¹(x) is the inverse function, F(x) is the antiderivative of f(x))
The integral of xe^(x)=e^(x)(x-1)
So the integral of w(x) is
x*w(x)-e^(w(x))(w(x)-1)+c
=x*w(x)-e^(w(x))*w(x)+e^(w(x))+c
=x*w(x)-x+e^(w(x))+c
Beautiful work 👍👍👍
Thank you! Cheers!
Next video: Use the Lambert W function to show for which cases do we have x^y=y^x, when x isn't equal to y, for instance 2⁴=4² and √3^√27=√27^√3.
Especially when fixing a value for one variable, like y=2, when the solutions are x=2, x=4, and x~=-23/30.
Great video veny clear and the enthusiasm is contagious. loved the music at the beginning (4:49): Is that African percussion?
Great video
Underrated 🔥 🔥 ❤
I really like your videos,are you using Hagoromo chalk?The writing looks very smooth
I do sometimes.
nice! Gotta sub
Thanks for the sub!
The u-substituiton is a good idea! Another way would be to integrate by parts ∫1*W[x] dx = x*W[x] - ∫x*W'[x] dx and then rewrite W' .
But your way is better.
Why does W(x)e^W(x) return x instead of W(x)? Thank you.
Select y such that x=y*e^y. Then W(x) = W(y*e^y) = y by definition of the Lambert W function. So W(x) *e^W(x) =y*e^y. And that is equal to x by the selection.
u perfect as always !!
Great video!
How can you integrate something that is not even a function?! What does it mean
W is a function
Is there a formula for anti-derivative like the one for derivative (first principle)?
This function is not an elementary function, so I have hege doubts. Never tried it yet.
@@PrimeNewtons so there's one for elementary functions?
Riemann sums
@@fusuyreds1236 pretty sure it's for definite integral
@@clemberube6681 right
I'd take that first solution and rationalize the denominator. So you'd get
x*W(x)^2 - x*W(x) + x + C
Beautiful!
You could factor the x to get x(W(x)^2 - W(x) + 1) + C, but I like the top one better
What I found strange is that the RambertW function is also called a Productlog function. If call it Productlog, it might think it's a function created by multiplying log, so I personally think it's more appropriate to call it Rambert than Productlog.
Lambert, BTW
The productlog comes from the fact that W is the inverse of xe^x, thus the product part of the name. Like a log, but not quite, and this specifies (very imperfectly) how
That is really cool
Nice video!
Great video, and I want that cap.
Thank you! 😊❤
laplace transform
fourier series
fourier ..
Красунчик! Респект!
Wasen't that integration by parts leaves the last part as integral? I think the formula should ends with (...) +2e^u-integral e^u du (since i didn't do integrals for a time please forgive me if i am wrong)
Man could you integrate 3rd root tanxdx? I want to see how to do it in a simple way cuz you explain things nicely
There's no nice way for that. It's messy all the way.
Nice q👌🏼
Please give an example for this integral... thanks
Never stop learning🤓
Don't see any links in the description!
🤥
Fixed. Thanks
I used integration by parts first
Int(LambertW(x),x) = xLambertW(x) - Int(x*LambertW(x)/((1+LambertW(x))*x),x)
Int(LambertW(x),x) = xLambertW(x) - Int(LambertW(x)/(1+LambertW(x)),x)
Int(LambertW(x),x) = xLambertW(x) - Int(((1+LambertW(x))-1)/(1+LambertW(x)),x)
Int(LambertW(x),x) = xLambertW(x) - Int(1,x) + Int(1/(1+LambertW(x)),x)
Int(LambertW(x),x) = x(LambertW(x) - 1) + Int(1/(1+LambertW(x)),x)
Int(1/(1+LambertW(x)),x)
u = LambertW(x)
x=u*exp(u)
dx = (u+1)exp(u)du
Int(1/(1+LambertW(x)),x) = Int(exp(u),u)
Int(LambertW(x),x) = x*(LambertW(x) - 1) + exp(LambertW(x))+C
It's not Lambert W function ❌
It's bprp fish 🐟 function ✅
The first answer means you have to worry about case where W(x) = 0.
Now padé aproximation for lambert function
Integration by parts is merely flipping axes. Simple. The "DI" method is an artefact.
I understand that W(x*e^x)=x. But why is W(x)*e^W(x)=x?
Because:
LHS: W[W(x)*e^W(x)] = W(x)
RHS: just W(x)
@@allozovsky ah thanks I see
How could you write: W(x)e^W(x)=x ?
Instead it should be: W(xe^x)=x.
Isn't it?
Inetgrate W(x)^f(x) dx ; where f(x) is the gamma function❗
I can't believe this bruh 💀
*blocked