This Will Be Your Favorite Integral

The golden ratio is a pretty famous math inequality. BUT have you heard of (what I call) the Golden Integral?

This Golden Ratio Integral integrates to a pretty remarkable result!

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

#math #brithemathguy #goldenratio

[BriTheMathGuy]

🎓Become a Math Master With My Intro To Proofs Course!

[soulsilencer]

"To solve this we have to know a few things about the golden ratio." Yeah the lack of information about phi isn't the problem here :D

[vitorcurtarelli]

I don't think you need to use the factorial definition of the Gamma, since the factorial per sè isn't defined for non-natural numbers. You could just use the property that x•Gamma(x) = Gamma(x+1) and that (1/phi) = (phi-1) at 3:42 for a more "formal" solution.

Great video tho! Loved the reupload fixing that mistake at the start, shows a great care and love for the craft.

[penta]

When in doubt, the answer is probably 1 or 0. This helped me more times in my math undergrad wayyy more than it really should’ve lmao

[GaryFerrao]

wow. you re-uploaded just to fix the a/b = (a + b)/b. wow.

[francescomantuano]

you really have one of the most entertaining and underrated math-related channel, keep up the work cause we all love it!!

[rodriguezzamarripadiego]

The fact that the integral is equal to 1 make my head explode, nice video

[violintegral]

Dr Peyam also did a video on this integral, taking a completely different route. He actually guessed an antiderivative of the function instead. It wasn't an intuitive solution, but I still found it pretty surprising that a complicated function with irrational powers like this one has an elementary antiderivative. It shows just how special phi is!

[robertmeadows]

Another great video from Bri. This is a great solution and a really nice use of the Beta and Gamma functions, however, I evaluated the integral without using any special functions:

Let u=1+x^phi, then, x=(u-1)^(1/phi), so dx=(1/phi)*(u-1)^(1/phi -1).
Since, phi^2 = phi + 1, dividing by phi and taking 2 from both sides gives, 1/phi - 1 = phi - 2.
Using this to simplify the expression for dx gives, dx = (1/phi)*(u-1)^(phi-2) du.
At our bounds we have, x=0 -> u=1, x=inf -> u=inf.
Re-writing the integral in terms of u gives:

I = int( du ) from 1 to infinity.

By pulling the constant out the front and multiplying the integrand by (u^-2)/(u^-2) gives,

I = (1/phi)*int( du ) from 1 to infinity

Rearranging the integrand, by noticing that the powers of u are now equal, and simplifying gives:

I = (1/phi)*int( (u^-2)*(1 - 1/u)^(phi-2) du ) from 1 to infinity,

Now let v = 1 - 1/u, so dv = u^-2 du. When u=1 -> v=0, u=inf -> v=1. This then gives:

I = (1/phi)*int( v^(phi-2) dv) from 0 to 1.

Integrating using the power rule and plugging in the bounds gives:

I = (1/phi)*(1/(phi-1)) = 1/(phi^2 - phi)

Since phi^2 = phi + 1, then subtracting phi from both sides gives, phi^2 - phi = 1, which reduced our integral to:

I = 1

Hope that's clear enough for people to follow. I honestly didn't think I was gonna be able to solve it but then I noticed the beautiful substitution of v = 1 - 1/u.

[judedavis]

Anyone else get the shivers whenever he says: ‘pheee’ instead of phi

[thatkindcoder]

This seems like a pretty brutal integral, don't know why I'd;
*Evaluates to one, and uses the Beta and Gamma function*
I could base a religion out of this

[danipent]

01:20 oh wow thanks nobody has ever told me that I tend towards infinity 😊

[masterclash]

It took a while, but I figured out how to do this with only u-substitution!

Starting at the beginning, I set u=1+x^phi, and after following similar steps from the video, I rearranged the function to get:

1/u^phi * 1/(u-1) * x du all over phi.

Solving for x in terms of u gets (u-1)^(1/phi), and plugging back in and dealing with exponents gets us:

1/u^phi * 1/[(u-1)^(1-1/phi)] du all over phi.

Plugging in 1/phi = phi - 1 for the second term gets us:

1/u^phi * 1/[(u-1)^(2-phi)] du all over phi.

Bringing the second term to the numerator and splitting up the exponent, we get:

1/u^phi * (u-1)^phi * 1/(u-1)^2 du all over phi.

Because the first two terms have the same exponents, we can combine the terms inside of them:

(1 - 1/u)^phi * 1/(u-1)^2 du all over phi.

We want the second term in terms of 1 - 1/u, so taking a 1/u^2 out of it and flipping the signs (no change because -1 is squared), and combining with the first term, we get:

(1 - 1/u)^(phi-2) * u^-2 du all over phi.

Lastly, doing w-substitution where w = 1 - 1/u and dw = u^-2 will yield an integrand of:

w^(phi-2) dw all over phi.

And an anti derivative of:

[w^(phi-1)]/(phi-1) all over phi, integrated from 0 to 1 - 1/(inf^phi + 1) if we did our u-subs correctly.

Doing our strategy of taking the limit as a variable (b) goes to infinity for indefinite integrals and plugging in, we get:

[(1-1/[b^phi + 1])^(phi-1)] / (phi-1) all over phi where b goes to infinity.

Because b goes to infinity, the numerator just turns to 1, leaving us with:

1/(phi-1) * 1/phi

And finally, plugging in phi - 1 = phi^-1 for the first term nets us our magical answer…

1

Very cool!

[saketram]

Yayyy thank you so much Bri !
I wanted this!

[alexjaeger]

Dejà-vu, I've just been in this place before...

[jameslalonde]

Bro it always amazes me how you solve these abstract type integrals. Whenever I run into integrals like these I just jave no idea what to do but you make it look so simple. Good job

[pwsk]

OMG! I'm a lover of integrals and your videos make me live them... You are amazing! 😃

[willyh.r.]

Aesthetically elegant. Thanks 4 sharing.

[Monkieteam]

I did by substituing for u = 1/(1+x^phi), after that there are some cancelations and the integral becomes very simple, without The need to know special functions. I would say however it is certainly more stylish to recognise the beta/gamma functions in there

[kjl]

This looked so scary but if you think about the definition of the golden ratio in terms of the sides of a rectangle it’s no wonder why integrating it gives such a nice number