A golden ratio integral

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Full solution development for this ridiculously awesome integral making use of the golden ratio and leading to a beautiful result.

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Very satisfying integral. The fact that η(2) popped in both sub integrals is nice and makes you think about whether or not it could have been arrived at from that integral before splitting it.

Sugarman
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Hi,

The final result can be simplified into : sqrt(5) * pi^2/12

"ok, cool" : 2:31, 2:58, 11:26,

"terribly sorry about that" : 3:55, 4:23, 6:05, 6:08, 10:10, 10:13, 13:11, 14:18 .

CM_France
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This integral possesses beautiful properties indeed. When replacing the golden ratio by a generic power z in C, we obtain the closed form: I(z) = (z^2 +1)/ (12*z) * Pi^2. And this yields the elegant reflexion formula I(z) = I(1/z). Its only zero seems to occur when z = +/- i 🙂

DD-cend
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This is a salivating beauty, and I cannot think of another way of describing it.

slavinojunepri
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This was gorgeous 😍. Thanks for the amazing result.

leroyzack
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You could simplify even more at the end since 1/(phi -1) is just phi. It turns into 3*phi-phi^2 = 3*phi - (phi + 1) = 2*phi - 1 = sqrt(5)

MichaelDruggan
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15:30 Here you could save a few steps by substituting phi - 1 = 1/phi.
You end up with: (1/phi^2 + 1) / (1/phi) = (phi^2 + 1) / phi = 1/phi + phi = phi - 1 + phi = 2*phi - 1 = 1.

hanbali-khadim
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Where do you find such integrals? They're all really cool. Do you have any textbooks you can recommend which have integrals like this?

jiiktaar
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**slaps roof of video** this bad boy can fit so much nice cancellation taking place

somerandomletters
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Im so early there isn't even audio

maxmoedough
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13:47 gamma(z) should be gamma(s), no? Also at 11:10 shouldn't it be t^(phi-2)?

esphix
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How many years work in integral department (Years of experience)

MRGamesStreamer
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Take it one step further by relating 'phi to kewness of fruit trees, thereby expanding the integral repetoir of Golden Ratios.

M_i_i_
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11:13 shouldn't it be "phi -2" instead of "phi -1"? Cool result nevertheless.

Chester
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Mashallah! I've said it once, whoever has given you the name Kamal (perfection) has depicted you exactly! Please thank him/her for me.😊

trelosyiaellinika
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Si arriva facilmente a I=Σ((-1)^k/(k+1))π/sinπ(k+1)Φ..poi, boh..tu hai usato un metodo diverso .io, invece, ho usato..la serie logaritmica, la funzione beta, e poi la gamma reflection.. poi mi sono bloccato..ah ah...forse ho trovato l'errore:non si può sviluppare in serie logaritmica perché ln(1+x)...x, tra 0 e 1, è maggiore di 1..

giuseppemalaguti
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Just out of curiosity, Where do you get these integrals? Like what book/s?

guy_with_infinite_power
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It's cheating to put phi in the intergrand I feel. Not surprising that phi pops out in the result.

zunaidparker
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