a general integral family
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- Опубликовано: 11 фев 2025
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Gamma Functions, gnarly π’s, all that was missing was a backflip.
Thank you, professor.
This is integral is the way to prove the gamma function reflection formula. I did it using the power of contour integration!
Hi there, could you please take the trouble to outline the process? can’t figure it out myself
yo this integral pops off son
Easier method is to sub u=x^b and use the beta function
Well, not all arguements hold unless you assume a,b>0 in the first place. I thought we're gonna find all real a and b where it converges. First time you do it on 1:27 where you assume b>0. And then when you use limit comparison you also assume b>0, otherwise you couldn't use it (separate case would be needed).
He said it was for the viewers to verify
A video suggestion: details on the labert W function. Can it be differentiated, can it be expressed explicitly somehow, what properties does it have?
Yes
The usage of a-1 instead of a kind of gives away that this is specifically structured for something magical to happen
Interesante todo. En muchos libros te ponen a calcular integrales de ese tipo, sin dar el fundamento de las constantes involucradas. No es la primera vez que él convierte la integral de una variable a una de dos variables como auxiliar, llegando al resultado elegantemente. ❤
Hmmm... the integral from 0 to infinity is larger than the one from 0 to 1, if the integrand doesn't end up hitting 0 and be negative from there, right?
This is a very useful formula...thanks Prof. 🌩⚡
Amazing result seriously
@1:20 that assumes b is non-negative. If b were negative this wouldn’t necessarily be true.
but b>a>0 so both a and b are non-negative, or am I missing something?
@@PhilBoswell why is b>a>0?
13:24
So fast
Awesome integral and even more awesome solution but this integral can be more conveniently solved by the method of complex analysis (with the aid of the pizza contour )
I think this series also converges as 0>a>b.
Was there supposed to be an assumption that b is positive? At 1:20 for instance it's assumed xᵇ < 1 if x
I have same question
I think that assumption was made for the sake of proof, and later it was shown to be implied by the restriction on a...possibly?
@@Jack_Callcott_AU The thing is in the beginning he implicitly assumed b>0 to prove a>0, then afterward used the fact that a>0 to prove b>0. So I think he accidentally used some circular reasoning there.
@@Bodyknock Thanks for the reply. I'll have to watch the video again and think about it a little bit more. I must say I really did enjoy the video, and it is an important formula that he derived. 👍
@@Bodyknock yes I’m confused by that circular reasoning as well. He never really showed why b can’t be negative, he said it clearly won’t converge then but not sure why
Mellin transform! Mellin transform!
Missing a gamma flip!
A fun but not too difficult variant of this is to determine convergence or divergence of the series
\Sum_{n=1}^{\infty}\int_1^{\infty}\frac{x^{a-k}}{1+x^b}dx
Wouldn't a simpler start have been
INT x^(a-1) / (1+x^b) d x =
1/a INT 1 / (1+x^b) d x^a =
1/a INT 1 / (1+u^(b/a)) d u
which then only requires the function 1/(1+u^k) to be integrated over u, i.e. with a single variable.
Admittedly, I am not sure how to do that, but looks like it might be some standard integral (wolfram alpha does it for specific b/a > 1).
We can use B(x,y)=INT t^(x-1)/(1+t)^(x+y) dt=Gamma(x)Gamma(y)/Gamma(x+y)
@@Otanes_Yu How is the integrand 1/(1+t^k) related to 1/(1+t)^k then?
@@landsgevaer set t^k=u to get 1/(1+u) then it is related to B(x,(1-x))
But doesn’t the sign in the comparison test(2:10) have to be less than for the test being applicable? I mean the integral is convergent if the bigger one is convergent and not smaller.
At the very beginning integral from 0 to 1 would easily converge with a = 0 and b = -1
This reminds me of Baker-Nachbaur type integrals of the second kind.
Solve without gamma function
In fact, you have proved the main formula for Beta function.
When backflip?
Very nice.
Wouldn't this be easier with the beta function? Just substitute X^a.