Integrating Lambert W Function

[@xizar0rg]

That is the cleanest in-use chalkboard I have ever seen.

[@Arriyad1]

And maybe he uses Hagoromo chalk.

[@haniyasu8236]

Damn. The way you present is so smooth. I love it

[@bazboy24]

You are a most gifted teacher

[@pierotezen4272]

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

[@PrimeNewtons]

Welcome aboard!

[@pablomandelo4311]

Man, I love your videos. You got a talent for teaching

[@PrimeNewtons]

I appreciate that!

[@Misteribel]

11:49 "this is you, remember", Yes! It's me! I love how you are talking to me in this video 😅

[@TelepathShield]

Lol

[@uggupuggu]

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)

[@atmonotes]

your math is very organized and concise! loved the integration by parts table. thanks!

[@DiabaLompoSouleymane]

Mon cher ! Vous êtes vraiment très fort ! Et vos explications sont très claires ! Merci bcp !

[@bpr214]

You have impeccable handwriting.

[@strawberrycake8253]

This was quite wholesome to watch! Keep it up man!

[@mathiasarrua1207]

I LOVE YOUR VIDEOS MAN, THEY MAKE ME LOVE CALCULUS

[@makramaarid6598]

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

[@pekorasfuturehusband]

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 🙏🏻

[@PrimeNewtons]

Welcome!

[@paulmichaud7565]

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.

[@Amoeby]

Is this an ASMR math or am I missing something?

[@codigodesenior3995]

i fell in love with your channel!

[@faustobarbuto]

Great video, thanks! Methinks I should've started with the dW(x)/dx video, though.

[@PrimeNewtons]

Good call!

[@SirBeYou]

Bro got predicted, good video as always

[@niccolopaganinifranzliszt3556]

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

[@jaimeduncan6167]

Great video veny clear and the enthusiasm is contagious. loved the music at the beginning (4:49): Is that African percussion?

[@stefanriegel2963]

Danke!

[@PrimeNewtons]

Thank you!

[@niloneto1608]

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.

[@renesperb]

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.

[@usernameisamyth]

amazing explanation

[@RogerLmao]

I loved the video, awesome!!!

[@DaMonster]

Beautiful work 👍👍👍

[@PrimeNewtons]

Thank you! Cheers!

[@flowingafterglow629]

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

[@vanecrnacki3875]

I really like your videos,are you using Hagoromo chalk?The writing looks very smooth

[@PrimeNewtons]

I do sometimes.

[@erichlf]

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.

[@anglaismoyen]

The mad lad did it.

[@김상문-o9m]

u perfect as always !!

[@anotherelvis]

Great video

[@clemberube6681]

Is there a formula for anti-derivative like the one for derivative (first principle)?

[@PrimeNewtons]

This function is not an elementary function, so I have hege doubts. Never tried it yet.

[@clemberube6681]

@@PrimeNewtons so there's one for elementary functions?

[@fusuyreds1236]

Riemann sums

[@clemberube6681]

@@fusuyreds1236 pretty sure it's for definite integral

[@fusuyreds1236]

@@clemberube6681 right

[@AbsolutelyNoOne251]

How can you integrate something that is not even a function?! What does it mean

[@guh967]

Great video!

[@거미남자_spidy]

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.

[@livikolumina5220]

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

[@wagsman9999]

Great video, and I want that cap.

[@ryanchiang9587]

laplace transform fourier series fourier ..

[@spicca4601]

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)

[@fortpile]

nice! Gotta sub

[@PrimeNewtons]

Thanks for the sub!

[@KarlFredrik]

Nice video!

[@Rom-1k]

Thank you! 😊❤

[@HashemAljifri515]

Man could you integrate 3rd root tanxdx? I want to see how to do it in a simple way cuz you explain things nicely

[@PrimeNewtons]

There's no nice way for that. It's messy all the way.

[@codex8797]

That is really cool

[@baselinesweb]

Why does W(x)e^W(x) return x instead of W(x)? Thank you.

[@martenjanderuiter5474]

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.

[@MASHabibi-d2d]

Please give an example for this integral... thanks

[@voice4voicelessKrzysiek]

Don't see any links in the description!

[@PrimeNewtons]

🤥

[@PrimeNewtons]

Fixed. Thanks

[@AndriiBilous]

Красунчик! Респект!

[@holyshit922]

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

[@alexuserectus1607]

Nice q👌🏼

[@juanosorio8148]

Now padé aproximation for lambert function

[@tirtharajbanerjee]

How could you write: W(x)e^W(x)=x ? Instead it should be: W(xe^x)=x. Isn't it?

[@TanmaY_TalK]

It's not Lambert W function ❌ It's bprp fish 🐟 function ✅

[@wayneosaur]

The first answer means you have to worry about case where W(x) = 0.

[@chandranisahanone]

Inetgrate W(x)^f(x) dx ; where f(x) is the gamma function❗

[@serae4060]

I understand that W(x*e^x)=x. But why is W(x)*e^W(x)=x?

[@allozovsky]

Because: LHS: W[W(x)*e^W(x)] = W(x) RHS: just W(x)

[@serae4060]

@@allozovsky ah thanks I see

[@Necrozene]

Integration by parts is merely flipping axes. Simple. The "DI" method is an artefact.

[@weo9473]

I can't believe this bruh 💀

[@Nzargnalphabet]

*blocked