let's solve an impossible integral

[@mcalkis5771]

Given how often you use them, a mini course on special functions would be most fitting and welcome.

[@CM63_France]

Hi, You can simplify a bit, knowing that g(z+1)=z g(z) , you can write : g(1 -1/n) = -g(-1/n) / n so : I = -g(-1/n) / (n^(1/n)) "terribly sorry about that" : 1:15 , 4:32 , 4:38 , 13:36 , "ok, cool" : 2:05 , 3:35 , 4:30 , 4:38 , 6:12 , 7:36 , 9:45 , 12:09 , 12:56 ,.

[@yoav613]

No,you will solve this impossible integral,i'l just watch and enjoy😂😊

[@random22453]

why introduce the lambert W function in the first place? we know that ye^y= 1/x^n through this alone we can find dx in terms of dy and then boom you get the same result

[@Anonymous-Indian..2003]

Bhai maine bhi aise hi Kiya, Vichaar milte hamaare 😂 Btw, you JEE aspirant ?

[@liamturman]

Because it’s cool :(, don’t be mean to our cool niche functions, they need to see the light of day sometimes too.

[@liamturman]

You should try to generalize the indefinite integral of any inverse function!

[@ranaranino5959]

For N=2 we obtain a nice closed form sqrt(2 π)

[@MrWael1970]

Very nice result. Thanks for solving this monster integral.

[@Mathematician6124]

Hey friend 😊. It was easy for me as I solved many such vedios of yours.

[@maths_505]

Always love to hear from you my friend

[@lakshay3745]

12:45 If I'm not wrong you forgot to write a factor of N outside the factorisation so technically it should be N ^2-1/N right?

[@xanterrx9741]

Absolutely beautiful integral

[@thundercraft0496]

It's interesting that this is the first integral in your channel where i can actually do it LoL

[@أَنْفَالٌ-غ2ح]

1:49 i didnt understand the inverse function , is there a video that explain it ?

[@spaghetti1383]

When I did the problem, I simplified further with Euler's reflection formula. I don't know if it was worth it.

[@zabsup]

babe wake up new maths 505 video just dropped

[@shawan7089]

Plz make a problem sheet on integrals

[@BigTonyPersonal]

Is there any easier integrals that involve this type of recursive integral? This is really cool but so much algebraic manipulation and W function

[@maths_505]

Well we don't do easy integrals here quite often. We do really tough ones that teach people new concepts. I encourage you to do further research about the Lambert function. Trust me you'll enjoy it insha'Allah

[@BigTonyPersonal]

@@maths_505 Oh I would gladly do an integral like this, it's just I like challenging my less experienced friends with interesting integrals. Was wondering if you or anyone else had any pretty ones like this that could be doable for someone with around A-Level experience ☺️

[@gesucristo0]

What’s the limit as N -> \infty?

[@seanshameless0]

Infinity, it’s just simplifies to N as N approaches infinity

[@davode76166]

It would be beautiful for n=2

[@evankalis]

Could you do a video on techniques for nonlinear or just tricky differential equations? I got spoiled by laplace 😭

[@petterituovinem8412]

you get a nice result when N = 2

[@SuperSilver316]

She W on my Lambert til I Function

[@stefanalecu9532]

Lambert W rizz

[@Kritic_lyr]

ze best pre-exam procrastination regime

[@pluieuwu]

whoa that is SERIOUSLY cool ❤

[@threepointone415]

Ohhh-kei cool!

[@Calcprof]

Matheamtica didn't get this one. (But you did)

[@maths_505]

As usual😎😂

[@giuseppemalaguti435]

=I(W(x^(-n))dx)=???

[@albert3344]

Okkkkkk cooooool

[@vladthemagnificent9052]

terribly sorry about that

[@maxvangulik1988]

y=ln(1/(x^N•y))=-ln(x^N•y) e^-y=y•x^N ye^y=x^-N y=W(x^-N) I=int[0,♾️](W(x^-N))dx ye^y=x^-N (y+1)e^y•dy=-N•x^-(N+1)•dx W(x^-N)=y x^-(N+1)=(ye^y)/x x=(ye^y)^-(1/N) I=-sgn(N)•1/N•int[0,♾️](y^-(1/N)•(y+1)e^-(y/N))dy t=y/N Ndt=dy I=-sgn(N)•N^-(1/N)•int[0,♾️](t^-(1/N)•(Nt+1)e^-t)dt I=-sgn(N)•N^(1-1/N)•int[0,♾️](t^(1-1/N)•e^-t)dt-N^-(1/N)•int[0,♾️](t^(-1/N)•e^-t)dt I=-sgn(N)•N^(1-1/N)•gamma(2-1/N)-N^-(1/N)•gamma(1-1/N) I=sgn(N)•((1/N-1)N^(1-1/N)-N^-(1/N))gamma(1-1/N) I=sgn(N)•N^-(1/N)•(1-N-1)•gamma(1-1/N) I=-sgn(N)•N^(1-1/N)•gamma(1-1/N) yay i got one edit: if N is positive, the bounds are [♾️,0]. If N is negative, the bounds are [0,♾️]. If N=0, the bounds are [w,w], where w=W(1) is transcendental, and I becomes 0 because the bounds are the same. Thus, there's a factor of sgn(N).

[@maxvangulik1988]

It wasn't until the end of the video that he stated that N is a positive integer. Still, I think the sgn(N) is p cool