raabe 4K

[@arahatchikkatur1906]

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.

[@БорисНазаров-х7к]

This proof is much better:)

[@shadow-ht5gk]

Brilliant

[@bot24032]

minor mistake at the end, it's alna-a+0.5ln(2π)

[@Jack_Callcott_AU]

True!

[@tolberthobson2610]

VERY TRUE!!

[@Nzargnalphabet]

I graphed it, it was off to the right by one

[@Nzargnalphabet]

All you needed to do was add one to every a individually

[@zalut_sky]

raabe 4K indeed just kidding, thank you for the video

[@s4623]

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.

[@spiderjerusalem4009]

substitute what terms?

[@coc235]

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

[@Happy_Abe]

I was confused by this statement too

[@davidutoob]

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).

[@Jacob.Peyser]

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.

[@Happy_Abe]

@@davidutoob thank you this helps explain this a lot!

[@khoozu7802]

I didn't understand. I think he just change y to x and change x to y

[@neomooooo]

I really like the idea of the zeroth integral. It’s a nice hallmark.

[@wolframhuttermann7519]

If someone can prove that a certain integral value where the borders are computable numbers (including +- infinity) is non-computable, that would be great.

[@rainerzufall42]

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)

[@manucitomx]

Gnarly and fun. Just what I wanted for my Saturday morning. Thank you, professor.

[@anonym95642]

this reminds me of the Stirling approximation

[@DmitriStarostin]

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).

[@ManyWaysMA]

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…?

[@skylardeslypere9909]

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).

[@Noam_.Menashe]

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.

[@riadsouissi]

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.

[@demenion3521]

that's actually a really neat integral problem. i enjoyed that the most in a while

[@goodplacetostop2973]

12:26

[@vinesthemonkey]

raabe 4k

[@enire8477]

Doesn't this only show its true for a>=1, and still need to be done for the case: 0

[@Happy_Abe]

Michael, just wanted to ask when the Math Major videos will continue?

[@eytansuchard8640]

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.

[@Redfox0928]

Isn't that method also called Feynman's trick ? Or am I getting confused ?

[@riadsouissi]

Yes, euivalent more or less.

[@Noam_.Menashe]

Don't you love symmetric functions?

[@minwithoutintroduction]

رائع جدا كالعادة

[@VaradMahashabde]

Raabe 4K

[@BurningShipFractal]

On the thumbnail, it says “Rabbe’s Formula”, not “Raabe’s Formula”. it’s probably a mistake

[@LeonhardEulerShades]

Is that how you pronounce Raabe? I've been saying Ra-bay for a while 😅

[@MDMajor]

Raabe was Swiss so it's probably more like "Rah-buh"

[@nasseerkassim1771]

V good

[@abdoshaat3304]

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

[@patricius6378]

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.