Integrating Lambert W Function
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- Опубликовано: 21 дек 2023
- 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...
![Lambert W Taylor Series Expansion [ Lagrange Inversion Theorem ]](/img/1.gif)
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
your math is very organized and concise! loved the integration by parts table. thanks!
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!
Is this an ASMR math or am I missing something?
Man, I love your videos. You got a talent for teaching
I appreciate that!
Mon cher ! Vous êtes vraiment très fort ! Et vos explications sont très claires ! Merci bcp !
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)
I recommend you to do the integral of the integral of the integral of the Lambert W function of a quadratic. Those who want to see the video 👇
Haha. Nope!
@@PrimeNewtons at least the integral of the lambert of the quadratic?
@@Shadowslayer6000 Dafuq even means Lambert W function of a quadratic? Give an example
W(2x^2 + 3x -5)
@@niloneto1608 W(n, x^2+x+1) where n is an integer, I would like n to be 0 and hey you even have a solution for it !!!!!!! (just not real but COMPLEX)!
This was quite wholesome to watch! Keep it up man!
It was great keep up with the good work. Am watching from Cameroon
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
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.
I LOVE YOUR VIDEOS MAN, THEY MAKE ME LOVE CALCULUS
i fell in love with your channel!
11:49 "this is you, remember", Yes! It's me! I love how you are talking to me in this video 😅
I loved the video, awesome!!!
Bro got predicted, good video as always
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.
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
You have impeccable handwriting.
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 actually prefer the first solution. It shows that W's integral can be expressed in a balance of arithmetic operations and their inverses. Like a who's who of elementary math showing up in the integral of a somewhat niche operator.
Great video!
u perfect as always !!
amazing explanation
Great video, thanks! Methinks I should've started with the dW(x)/dx video, though.
Good call!
Great video
The mad lad did it.
That is really cool
Nice video!
Beautiful work 👍👍👍
Thank you! Cheers!
Great video, and I want that cap.
nice! Gotta sub
Thanks for the sub!
Great video veny clear and the enthusiasm is contagious. loved the music at the beginning (4:49): Is that African percussion?
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.
Danke!
Thank you!
Красунчик! Респект!
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
Nice q👌🏼
I really like your videos,are you using Hagoromo chalk?The writing looks very smooth
I do sometimes.
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.
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
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)
Please give an example for this integral... thanks
How can you integrate something that is not even a function?! What does it mean
laplace transform
fourier series
fourier ..
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
Don't see any links in the description!
🤥
Fixed. Thanks
Why does W(x)e^W(x) return x instead of W(x)? Thank you.
Now padé aproximation for lambert function
Inetgrate W(x)^f(x) dx ; where f(x) is the gamma function❗
How could you write: W(x)e^W(x)=x ?
Instead it should be: W(xe^x)=x.
Isn't it?
It's not Lambert W function ❌
It's bprp fish 🐟 function ✅
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
Integration by parts is merely flipping axes. Simple. The "DI" method is an artefact.
I can't believe this bruh 💀
*blocked