Integrating Lambert W Function
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- čas přidán 21. 12. 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...
That is the cleanest in-use chalkboard I have ever seen.
And maybe he uses Hagoromo chalk.
You are a most gifted teacher
Damn. The way you present is so smooth. I love it
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!
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)
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
Man, I love your videos. You got a talent for teaching
I appreciate that!
This was quite wholesome to watch! Keep it up man!
It was great keep up with the good work. Am watching from Cameroon
I LOVE YOUR VIDEOS MAN, THEY MAKE ME LOVE CALCULUS
11:49 "this is you, remember", Yes! It's me! I love how you are talking to me in this video 😅
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.
You have impeccable handwriting.
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
Bro got predicted, good video as always
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)!
The mad lad did it.
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.
amazing explanation
Great video
i fell in love with your channel!
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!
Great video, thanks! Methinks I should've started with the dW(x)/dx video, though.
Good call!
I loved the video, awesome!!!
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.
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
Great video!
Nice video!
u perfect as always !!
Beautiful work 👍👍👍
Thank you! Cheers!
Great video veny clear and the enthusiasm is contagious. loved the music at the beginning (4:49): Is that African percussion?
That is really cool
I really like your videos,are you using Hagoromo chalk?The writing looks very smooth
I do sometimes.
Is this an ASMR math or am I missing something?
Great video, and I want that cap.
nice! Gotta sub
Thanks for the sub!
Danke!
Thank you!
Красунчик! Респект!
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👌🏼
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)
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
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.
laplace transform
fourier series
fourier ..
How can you integrate something that is not even a function?! What does it mean
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
Please give an example for this integral... thanks
Why does W(x)e^W(x) return x instead of W(x)? Thank you.
Now padé aproximation for lambert function
Don't see any links in the description!
🤥
Fixed. Thanks
Inetgrate W(x)^f(x) dx ; where f(x) is the gamma 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.
It's not Lambert W function ❌
It's bprp fish 🐟 function ✅
I can't believe this bruh 💀
*blocked