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raabe 4K
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- Опубліковано 18 сер 2023
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minor mistake at the end, it's alna-a+0.5ln(2π)
True!
VERY TRUE!!
I graphed it, it was off to the right by one
All you needed to do was add one to every a individually
Another way of approaching this integral is to think of the integral as a function of a and take the derivative. From the fundamental theorem of calculus, f'(a) = ln(G(a+1)) - ln(G(a)) = ln(a), where G(a) denotes the gamma function at a. Integrating, we see that f(a) = a ln(a)-a+C. To find the constant, note that f(0) = C = int [0,1] ln(G(x)) dx. By making the substitution x -> 1-x, we see that 2C = int [0,1] ln(G(x)) + ln(G(1-x)) dx. Since ln(G(x))+ln(G(1-x)) = ln(G(x)G(1-x)), we can make use of the reflection formula, making the integral become int [0,1] ln(pi)-ln(sin(pi x)) dx, which is evaluated in the video.
This proof is much better:)
Brilliant
raabe 4K indeed
just kidding, thank you for the video
4:45 this argument is fundamentally wrong. For example x^2+y^2 also has 'nice symmetry built into it', but its partial derivatives are 2x and 2y respectively, which makes them not equal. What's crucial here is the linearity of the term x+y
I was confused by this statement too
A clearer argument is that ∂x and ∂y are equal for any function of the form f(x+y) because they're both equal to f'(x+y).
He was obviously referring to the symmetry upon differentiation. Don't berate the guy because you don't comply with his (valid) notion of symmetry.
@@davidutoob thank you this helps explain this a lot!
I didn't understand. I think he just change y to x and change x to y
Isn't that ln[sqrt(2Pi) (a/e)^a)] ? Using log rules, this is = ln a^a - ln e^a + ln sqrt(2Pi) = a ln a - a + 0.5 ln(2Pi) (the corrected result from above)
1:31 to 6:24 - a much simpler way of doing all of that would be the integral of something from a to a+1 = (integral of same function from 1 to a+1) - (integral of same function from 1 to a), and then substitute the terms, then use the fact the gamma function has the same value at 0 and at 1 to line up everything.
substitute what terms?
If someone can prove that a certain integral value where the borders are computable numbers (including +- infinity) is non-computable, that would be great.
I really like the idea of the zeroth integral. It’s a nice hallmark.
Interesting how the square root of 2pi shows up in the answer. I wonder if this has got something to do with Stirling’s approximation…?
As you can switch the two integrals around, I think you can also switch limits and integrals, meaning you could probably replace Gamma(x) by Stirlings approximation, and pull the limit outside (ln is continuous).
I think it's just because you get that a lot from the Gamma function (see Wallis product). So it's more so a different symptom of the same thing.
this reminds me of the Stirling approximation
12:26
Gnarly and fun. Just what I wanted for my Saturday morning.
Thank you, professor.
that's actually a really neat integral problem. i enjoyed that the most in a while
Or more directly, substitute x by x+a, use Feynman trick to derive the integral in respect to a, then integrate within the integral in respect to x to get log(gamma(a+1)/gamma(a))=log(a).
Integrate to get back thrle original integral to get alog(a) - a + c.
Constant c is defined when a=0 which is a known integral that can be solved using inversion formula of the gamma function.
for a = 2 the value is 1.61, close to the value of i in the integral of (1 / ((1 + x power i) power i)) dx on 0 to infinity (which you did earlier) in power 2/3, as if it were the Kepler's law for the lunar ellipse axis' libration. So this Raabe integral is a measure of temporal distance between the two subsequent harmonics of the lunar librations showing up over time (but quite a long time, several centuries).
Doesn't this only show its true for a>=1, and still need to be done for the case: 0
Michael, just wanted to ask when the Math Major videos will continue?
It should be a*ln(a)-a = Integral(ln(y)dy) from 0 to a. The limit of y*ln(y)-y -> 0 when y->0.
Raabe 4K
Isn't that method also called Feynman's trick ? Or am I getting confused ?
Yes, euivalent more or less.
رائع جدا كالعادة
Don't you love symmetric functions?
On the thumbnail, it says “Rabbe’s Formula”, not “Raabe’s Formula”. it’s probably a mistake
Is that how you pronounce Raabe? I've been saying Ra-bay for a while 😅
Raabe was Swiss so it's probably more like "Rah-buh"
V good
You are difficult teacher you jumping the steps of the solution not algorithmize the steps
It is better for you to watch how math 505 develops his solutions
As a huge fan of both Penn and maths505, I personally don't see either of them "jumping the steps" harder than the other, it's just that they use different methods (I can't recall maths505 ever using pictures for switching up double integrals, or using many double integrals at all) and sometimes assume different prerequesites, which is fine.