We use the Gamma Function, the Leibnitz Rule and Euler's reflection formula like they're the easiest things on earth and still need to debate whether Pi is now equal to 0 or not. Also at 18:03 the german boi came through once more: What is a "diskriptschen"? :D
Michael Empeigne This is not arctan(x), this is arctan(x!). This is definitely not elementary. To integrate this, one would need to integrate by parts. For example, if the bounds of integration are x = 0 and x = t, then one could antidifferentiate 1 to x and differentiate arctan(x!) to x!ψ*(x)/[1 + (x!)^2]. From this, one can conclude the integral of arctan(x!) from x = 0 to x = t is equal to the difference of x!·ψ*(x)/[1 + (x!)^2] evaluated at the points x = 0 and x = t, all minus the integral of x·x!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t. The difference of x!·ψ*(x)/[1 + (x!)^2] evaluated at the points x = 0 and x = t is equal to t!·ψ*(t)/[1 + (t!)^2] + γ/2, and the integral of x·x!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t is equal to the integral of (x + 1)·x!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t minus the integral of x!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t. The latter can be evaluated with a simple substitution. Let y = x!, so dy = x!·ψ*(x)·dx. This leaves the integrand 1/(1 + y^2), with bounds of integration y = 1 and y = t!. This integral is equal to arctan(t!) - π/4 by the fundamental theorem of calculus. The former is equal to the integral of (x + 1)!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t. Therefore, the integral of arctan(x!) from x = 0 to x = t is equal to t!·ψ*(t)/[1 + (t!)^2] + arctan(t!) + γ/2 - π/2 minus the integral of (x + 1)!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t. This is the furthest one can go in evaluating this integral. The latter involves the shifted factorial, which implies that one could write a linear recurrence relation to encode the value of the integral, but it is trivial to note that such recurrence relation has non-elementary solutions.
Damn I just derived this yesterday and thought it was the coolest thing ever. Are you also going to use this to get the generalization for hyperfcatorials?
Alekk Getting the generalization is rather trivial from this, though. For example, if you sum the equation with t running from 0 to s - 1, you get that the integral of ln[Γ(x)] from x = 0 to x = s is ln[sqrt(2π)]·s - s(s - 1)/2 plus the sum of t·ln(t) from 0 to s - 1. t·ln(t) = ln(t^t), and the sum of natural logarithms is the natural logarithm of the product. Therefore, the sum of ln(t^t) from 0 to s - 1 is equal to the natural logarithm of the product of t^t from 0 to s - 1. Given the convention 0^0 = 1, this product is simply equal to the hyperfactorial of s - 1 for integer s. Hence, this gives us an immediate generalization defined from the Gamma function. Call this product K(s), and let K(s) be the analytic continuation of the shifted hyperfactorial. Then the hyperfactorial is equal to sqrt(2π)^s·e^[s(s - 1)/2]e^I, where I is the integral of ln[Γ(t)] from 0 to s. Alternatively, e^I is equal to the standard geometric (product) integral of 0 to s of Γ(s). This geometric integral is also equal to the continuous analogue of the product of factorials, which is equal to the superfactorial. Thus, this equation relates the generalizations of the factorial, superfactorial, and hyperfactorial.
Aphrontic Alchemist No, not at all. The integral that he derived in the video has x running from t + 1 to t, not from 0 to t, so this is not *the* integral of ln(Γ(x)). However, it is correct that forward difference of the antiderivative of ln(Γ(x)) is just the antiderivative of ln(x) plus a constant.
I think this formula is super interesting, because if you sum both parts of the formula, with t running from 0 to s - 1, for example, you get the integral of ln(Γ(x)) from 0 to s, which is the antiderivative of ln(Γ(x)), being equal to ln(sqrt(2π))s - s(s - 1)/2 plus the summaton of t·ln(t) with t running from 0 to s - 1. This last summation is simply equal to ln(K(s)), where K(s) is the K-function, the analytic continuation of the hyperfactorial. This is simple to derive, since t·ln(t) = ln(t^t), and the sum of logarithms is the logarithm of a product, and the product of consecutive t^t's is the K-function. Therefore, what Raabe's formula implies is that the antiderivative of ln(Γ(x)) is C + ln(sqrt(2πe))·x - x^2/2 + ln(K(x)). This is crazy cool! It is a very unorthodox method for finding an antiderivative, and nonetheless to do it for a function that one would think you could not express its antiderivative in terms of other "elementary" functions. It is also cool because it indirectly relates factorials to hyperfactorials. To create a direct relationship between the factorials and hyperfactorials, simply exponentiate this equation. By doing so, you change the logarithmic integral into a product integral. The product integral from 0 to t of Γ(x) is equal to K(x)·e^(-[x^2 - ln(2πe)·x]/2) = K(x)·e^[-x(x - 1)/2]·sqrt(2π)^x, meaning that the product integral of factorials is the hyperfactorial times some exponential function. And the most beautiful part about this is that this is totally consistent with Stirling's approximation. In fact, if you take the product on both sides of Stirling's approximation, you get an expression that gives the asymptotic approximation to this exact equation, which is probably since products asymptotically approximate product integrals. I think this truly is S-tier.
Papa Flammy, great vid as usual.. I'd like to see where these integrals occur in science/physics, gives a little more context about why they needed solving :)
meanwhile i just see that as \int (x ln x-x) dx, if i feel i got the time i will add to it +1/2ln (2\pi x) but usually that's just too insignificant anyway
An Unknown does the first tower even converge? I doubt it does tbh, looks more like a 4 cycle on repeat. And the second one is a really weird function like below or at 0 its not defined, then it’s 1 up until 1 and after that it’s plus infinity. Reminds of a bit of the limit of n to infinity of x^n except that that one is defined at 0, but oh well. (I assumed you were talking about the infinite power tower, if not then sorry for this comment)
An Unknown I am really confident the first doesnt have a single finite value, and how would you integrate an infinite function like that? Like for every finite tower it’s doable and I’m sure it has a pattern but you probably can’t do that for infinitely many Xs
@@tipeg8841 y=x^x^x^....... Actually means y=x^y, 😃 i^i^i^i^..... is must having a finite value though I don't know the answer. If it is equal to y, Then y= i^y
An Unknown I already know that but plug in any number and see it doesn’t work if it’s not inside the radius of convergence of the series. Just being able to say x = y^(1/y) doesn’t mean it actually IS that value, take x=10 and it’s obvious it that doesn’t converge. So you first need to find it’s radius of convergence before getting your hopes up. I’m not an expert on the answer but you can google it too if you’re really curious 🤔
I did this before watching, but i can't figure out how to get rid of the constant after integration. My result is I(t) = t * (ln(t) - 1) + c Does anyone have a particular pair of values (x,y) such that I(x)=y? That would be enough for me to get a full solution. Edit 1: Ok i now at least have I(0) = c, but still can't find the value for I at x=0 Edit 2: Got it now, c=0 Edit 3: Wait wtf? Why is my constant wrong? Oh maybe because the argument for setting c=0 used the assumption that the factorial is a continuous function, which on the interval [-1,1] it is absolutely not. Whoops! Edit 4: Wait but it does look continuous there. Can someone tell me what i did wrong? For context, my approach was to use the recursive definition of the factorial ( so x! = (x-1)! * x ), logarithm properties ( ln(a * b) = ln(a) + ln(b) ) as well as the linearity of integrals, which yielded I(t) - I(t-1) = Integral from t-1 to 1 of ln(x)dx I then used the fundamental theorem of calculus to tell me that ln(x) = d(I(x))/dx, did the improper integral and got I(t) = t * (ln(t) - 1) + c
Вообще мне не нравится ведущий и издёвки что он говорит и показывает. Я не в коем случае не хотел бы чтобы в жизни ко мне относились как-то так. Материал как по мне здесь бывает редкий и я в какой-то мере хочу его знать. Но это пожалуй единтсвинное за что можно любить данный канал.
![Biblical Greek Grammar [Demonstrative Adjective] This is the work of the God (15) (Gen1:26, 6th day)](http://i.ytimg.com/vi/UXm6Qdxgs3Y/mqdefault.jpg)
![A Powerful Elementary Integral [ (Almost) Impossible Integrals #1 ]](/img/n.gif)
I watched all 20 minutes and I couldn't stop smiling at your excitement
So... any chance of an Integral tier List video?
We use the Gamma Function, the Leibnitz Rule and Euler's reflection formula like they're the easiest things on earth and still need to debate whether Pi is now equal to 0 or not.
Also at 18:03 the german boi came through once more: What is a "diskriptschen"? :D
19:43 as well
Last time I was so early, calculus did not even exist
Joaquin Sotomayor Escudero ....was it before time existed?
Last time I was this early, Newton and Leibniz still liked each other.
Beautiful indeed. Papa Raabe would be proud.
Your way of explanation of complex mathematics is pretty awesome hats off
Integrate arctan(x!)
*insert the excuse me wtf meme*
Not elementary.
integrating arctan x is easy.
Michael Empeigne This is not arctan(x), this is arctan(x!). This is definitely not elementary. To integrate this, one would need to integrate by parts. For example, if the bounds of integration are x = 0 and x = t, then one could antidifferentiate 1 to x and differentiate arctan(x!) to x!ψ*(x)/[1 + (x!)^2]. From this, one can conclude the integral of arctan(x!) from x = 0 to x = t is equal to the difference of x!·ψ*(x)/[1 + (x!)^2] evaluated at the points x = 0 and x = t, all minus the integral of x·x!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t. The difference of x!·ψ*(x)/[1 + (x!)^2] evaluated at the points x = 0 and x = t is equal to t!·ψ*(t)/[1 + (t!)^2] + γ/2, and the integral of x·x!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t is equal to the integral of (x + 1)·x!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t minus the integral of x!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t. The latter can be evaluated with a simple substitution. Let y = x!, so dy = x!·ψ*(x)·dx. This leaves the integrand 1/(1 + y^2), with bounds of integration y = 1 and y = t!. This integral is equal to arctan(t!) - π/4 by the fundamental theorem of calculus. The former is equal to the integral of (x + 1)!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t. Therefore, the integral of arctan(x!) from x = 0 to x = t is equal to t!·ψ*(t)/[1 + (t!)^2] + arctan(t!) + γ/2 - π/2 minus the integral of (x + 1)!·ψ*(x)/[1 + (x!)^2] from x = 0 to x = t. This is the furthest one can go in evaluating this integral. The latter involves the shifted factorial, which implies that one could write a linear recurrence relation to encode the value of the integral, but it is trivial to note that such recurrence relation has non-elementary solutions.
14:34 When the effects of math kick in
Mathgasm ma boi 😅
Like your enthusiasm. I also loved neat (algebraic) methods of solving integrals when I was an undergrad.
Damn I just derived this yesterday and thought it was the coolest thing ever. Are you also going to use this to get the generalization for hyperfcatorials?
Flammable Maths damn
Alekk Getting the generalization is rather trivial from this, though. For example, if you sum the equation with t running from 0 to s - 1, you get that the integral of ln[Γ(x)] from x = 0 to x = s is ln[sqrt(2π)]·s - s(s - 1)/2 plus the sum of t·ln(t) from 0 to s - 1. t·ln(t) = ln(t^t), and the sum of natural logarithms is the natural logarithm of the product. Therefore, the sum of ln(t^t) from 0 to s - 1 is equal to the natural logarithm of the product of t^t from 0 to s - 1. Given the convention 0^0 = 1, this product is simply equal to the hyperfactorial of s - 1 for integer s. Hence, this gives us an immediate generalization defined from the Gamma function. Call this product K(s), and let K(s) be the analytic continuation of the shifted hyperfactorial. Then the hyperfactorial is equal to sqrt(2π)^s·e^[s(s - 1)/2]e^I, where I is the integral of ln[Γ(t)] from 0 to s. Alternatively, e^I is equal to the standard geometric (product) integral of 0 to s of Γ(s). This geometric integral is also equal to the continuous analogue of the product of factorials, which is equal to the superfactorial. Thus, this equation relates the generalizations of the factorial, superfactorial, and hyperfactorial.
That's reason, why Math is queen of science!!! Beautiful INTEGARAL 😍
Flammable mathematic's mindblowers
A very spicy integral papa...I absolutely love your work. Thank you for all you do. Seriously.
Really epic integral!
So in other words, the integral of ln(Γ(x)) and of ln(x) are off by a constant, ln(2π)/2
Aphrontic Alchemist No, not at all. The integral that he derived in the video has x running from t + 1 to t, not from 0 to t, so this is not *the* integral of ln(Γ(x)). However, it is correct that forward difference of the antiderivative of ln(Γ(x)) is just the antiderivative of ln(x) plus a constant.
19:43 Dressed by Hugo Boss
Amazing, so beautiful!
That is the most beautiful integral☝️☝️☝️🙌
Wow .... It was great. I just enjoyed.
Thank you so much *🔥Flammable Maths🔥* .
I think this formula is super interesting, because if you sum both parts of the formula, with t running from 0 to s - 1, for example, you get the integral of ln(Γ(x)) from 0 to s, which is the antiderivative of ln(Γ(x)), being equal to ln(sqrt(2π))s - s(s - 1)/2 plus the summaton of t·ln(t) with t running from 0 to s - 1. This last summation is simply equal to ln(K(s)), where K(s) is the K-function, the analytic continuation of the hyperfactorial. This is simple to derive, since t·ln(t) = ln(t^t), and the sum of logarithms is the logarithm of a product, and the product of consecutive t^t's is the K-function. Therefore, what Raabe's formula implies is that the antiderivative of ln(Γ(x)) is C + ln(sqrt(2πe))·x - x^2/2 + ln(K(x)). This is crazy cool! It is a very unorthodox method for finding an antiderivative, and nonetheless to do it for a function that one would think you could not express its antiderivative in terms of other "elementary" functions. It is also cool because it indirectly relates factorials to hyperfactorials. To create a direct relationship between the factorials and hyperfactorials, simply exponentiate this equation. By doing so, you change the logarithmic integral into a product integral. The product integral from 0 to t of Γ(x) is equal to K(x)·e^(-[x^2 - ln(2πe)·x]/2) = K(x)·e^[-x(x - 1)/2]·sqrt(2π)^x, meaning that the product integral of factorials is the hyperfactorial times some exponential function. And the most beautiful part about this is that this is totally consistent with Stirling's approximation. In fact, if you take the product on both sides of Stirling's approximation, you get an expression that gives the asymptotic approximation to this exact equation, which is probably since products asymptotically approximate product integrals. I think this truly is S-tier.
Damn! Mate that is an absolute beaty. I'm so happy to watch this. Tank you
4:27 couldn’t one just use substitution to solve that integral since the integrand is on the form f’(x)/f(x) which would yield ln[ f(x) ]
What a mad boy ! Thank you very much.
Lovely person, please make some videos using The epsilon delta for limits (including the derivative), more series and integrals.
One of ur best video papa.....;) loved it
Papa Flammy, great vid as usual.. I'd like to see where these integrals occur in science/physics, gives a little more context about why they needed solving :)
in 7:30 you indicate that the in the numerator Gamma(t+1) = t.Gamma(t), but Gamma(t+1) = (t+1).Gamma(t).
@Hamish Blair yes
Γ(t+1) = t! = t . (t-1)! = t . Γ(t)
when your natural logs looks like a limit
Papa flammy what is the intro song at 0:22?
He uses a blackboard and chalk, from there onwards I know the guy has class, respectable and very clever work!
I love your writing board so can you tells details about your board.
soon! :)
@@PapaFlammy69 thanks
What intro song does flammy use?
B O I 👏
This looks a lot like Stirling’s approximation. Is that derived from this?
Trey Forest You can derive the Stirling's approximation from this, but it is not how it was originally derived when first discovered.
A coil of inductance L Henry and a capacitor of C Farad are connected in series if I=I0, Q=Q0 when t=0 Find: Q and I when t
Can someone explain the meme at the start? that one sure went over my head
Wot in tarnation¿¿
where is papa?? behind ∫ln(Γ(x)dx ?
Hey boy nice solution I solved this beautiful by using the Legendre formula 😎
meanwhile i just see that as \int (x ln x-x) dx, if i feel i got the time i will add to it +1/2ln (2\pi x) but usually that's just too insignificant anyway
the integral of ln(t!) from x-1 to x asymptotilally equal to ln(x)! itself
Can you prove apollonius' Theorem?
Have you looked at how similar this result is to the Stirling’s approximation? Put everything under the log and have a look
Can you talk about the Finite value of power tower of i (imaginary),
i^i^i^i^i^i........ =?
And,
Integral of
F(x)=x^x^x^x^x^x^x^.........
Please...
An Unknown does the first tower even converge? I doubt it does tbh, looks more like a 4 cycle on repeat. And the second one is a really weird function like below or at 0 its not defined, then it’s 1 up until 1 and after that it’s plus infinity. Reminds of a bit of the limit of n to infinity of x^n except that that one is defined at 0, but oh well. (I assumed you were talking about the infinite power tower, if not then sorry for this comment)
@@tipeg8841 Hey bro,
Both of them are with infinite power. I desired to have a Finite value of it.
An Unknown I am really confident the first doesnt have a single finite value, and how would you integrate an infinite function like that? Like for every finite tower it’s doable and I’m sure it has a pattern but you probably can’t do that for infinitely many Xs
@@tipeg8841 y=x^x^x^.......
Actually means y=x^y, 😃
i^i^i^i^..... is must having a finite value though I don't know the answer.
If it is equal to y,
Then y= i^y
An Unknown I already know that but plug in any number and see it doesn’t work if it’s not inside the radius of convergence of the series. Just being able to say x = y^(1/y) doesn’t mean it actually IS that value, take x=10 and it’s obvious it that doesn’t converge. So you first need to find it’s radius of convergence before getting your hopes up. I’m not an expert on the answer but you can google it too if you’re really curious 🤔
Beautiful😄
papa gone mad at 6:14 and became δδ
PAPA FLAMMY : you remind me about me when I was your age. I
Like you a lot!
> Stirling's Approximation yeet ~ Engineers
Haha "oh what the fuck that was not a nice gamma function"
Hah! Lucky refresh!
Delicious Broccoli Raabe.
P.S. Someone explain the intro meme to me pls
I think theres a mistake when you say that (x+1)! = x!x it's equal to x!(x+1)
Aidan Sgarlato gama of x+1 is equal to x!
I'd love some pie
10:05 we're going to put it - down DOWN -
cit.
This is nicer than 69.
One could also view this as the double integral of the digamma :)
This was a bit too trivial. But thanks anyway, it was a fun solve.
Epic
nice
You are so fucking funny xD xD
I did this before watching, but i can't figure out how to get rid of the constant after integration.
My result is I(t) = t * (ln(t) - 1) + c
Does anyone have a particular pair of values (x,y) such that I(x)=y?
That would be enough for me to get a full solution.
Edit 1: Ok i now at least have I(0) = c, but still can't find the value for I at x=0
Edit 2: Got it now, c=0
Edit 3: Wait wtf? Why is my constant wrong? Oh maybe because the argument for setting c=0 used the assumption that the factorial is a continuous function, which on the interval [-1,1] it is absolutely not. Whoops!
Edit 4: Wait but it does look continuous there. Can someone tell me what i did wrong?
For context, my approach was to use the recursive definition of the factorial ( so x! = (x-1)! * x ),
logarithm properties ( ln(a * b) = ln(a) + ln(b) ) as well as the linearity of integrals,
which yielded I(t) - I(t-1) = Integral from t-1 to 1 of ln(x)dx
I then used the fundamental theorem of calculus to tell me that ln(x) = d(I(x))/dx, did the improper integral and got I(t) = t * (ln(t) - 1) + c
Im fuckin mind blown...
no srsly.... wtf
integaral
whats the meme all about? :O
Wowe!
Номер 3869 из Демидовича, если что
I came
M’lady
My favourite SS boy
I am the 666th viewer
meh
I dont understand this shit
Вообще мне не нравится ведущий и издёвки что он говорит и показывает. Я не в коем случае не хотел бы чтобы в жизни ко мне относились как-то так. Материал как по мне здесь бывает редкий и я в какой-то мере хочу его знать. Но это пожалуй единтсвинное за что можно любить данный канал.