This Will Be Your Favorite Integral

[@BriTheMathGuy]

🎓Become a Math Master With My Intro To Proofs Course! www.udemy.com/course/prove-it-like-a-mathematician/?referralCode=D4A14680C629BCC9D84C

[@vitorcurtarelli254]

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.

[@BriTheMathGuy]

I try :) Thanks so much for watching!

[@YossiSirote]

I did not see that one coming. 😀

[@uggupuggu]

factorial is defined for non-natural numbers, you are wrong

[@vitorcurtarelli254]

@@uggupuggu objectively, no. n! is defined as n*(n-1)*...*2*1. Using the property that n! = n*(n-1)!, we can use the Gamma function to non-natural numbers, but by definition the factorial is defined only for positive integers. Saying that (1/2)! = √π/2 is an abuse of notation.

[@soulsilencer1864]

"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

[@BriTheMathGuy]

😂

[@Anokosciant]

i loves when he still explain basic algebra and uses beta and gamma function

[@Fire_Axus]

YouFeeAreIrr

[@penta4568]

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

[@BriTheMathGuy]

Facts.

[@tsan_jey]

Who can't relate?

[@muskyoxes]

Math is too hard then, i guess, because physics is just the study of things that are zero

[@quentind1924]

What about e and pi ?

[@angelmendez-rivera351]

@@muskyoxes what?

[@francescomantuano1505]

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

[@BriTheMathGuy]

Wow, thank you!

[@GaryFerrao]

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

[@rodriguezzamarripadiego9625]

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

[@BriTheMathGuy]

Thanks a ton!

[@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!

[@BriTheMathGuy]

For sure!

[@muskyoxes]

I don't think the antiderivative itself is elementary, just that it has a nice value for these particular endpoints.

[@violintegral]

@@muskyoxes no, it's elementary lol. Go watch his video if you don't believe me!

[@xinpingdonohoe3978]

​@@muskyoxes x(1+x^φ)^(-1/φ)+c

[@saketram9354]

Yayyy thank you so much Bri ! I wanted this!

[@BriTheMathGuy]

You're so welcome!

[@hsjkdsgd]

Absolutely incredible

[@BriTheMathGuy]

Glad you thought so!

[@robertmeadows4852]

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( (1/phi)*(1/u^phi)*((u-1)^(phi-2)) 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( (u^-2)*(1/u^(phi-2))*(u-1)^(phi-2) 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.

[@mchmch6185]

Hi Robert. I basically found the same thing but after following Bri's first stage of u=x^phi rather than your u=1+x^phi. You then get the integral of 1/E where E is u^(2-phi)*(1+u)^phi = u^2*(1+(1/u))^phi and then a v=1/u substitution seems maybe a bit more obvious, followed by v -> v-1. OK, I'm doing v=(1/u)+1 rather than your final v=1-(1/u) [and getting a final integral from 1 to infty rather than 0 to 1], but maybe it is easier to spot with the positives.

[@judedavis92]

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

[@MichaelRothwell1]

I still do, even though I am used to the fact that in Britain its pronounced phy and in the US its pronounced phee.

[@zeozen]

"fee" is closer to greek pronounciation:D

[@nordicexile7378]

Yeah, just sounds wrong to me. I mean, we say "pie" instead of "pee" for "π", don't we?

[@thatkindcoder7510]

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

[@willyh.r.1216]

Aesthetically elegant. Thanks 4 sharing.

[@BriTheMathGuy]

Glad you like it!

[@danipent3550]

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

[@pengutiny6464]

Oh-

[@manucitomx]

Wow! Loved this.

[@BriTheMathGuy]

So glad!

[@pwsk]

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

[@BriTheMathGuy]

Glad you like them!

[@masterclash9959]

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!

[@ass123qdwqdw]

why cant to tkae x = 1??

[@sahiljain1504]

Put 1+ 1/x^phi = u. Solves in second

[@alexjaeger]

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

[@jameslalonde4420]

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

[@MemyBurosi]

This vid needs to be watched over and over

[@BriTheMathGuy]

Glad you think so!

[@law26504]

An amazing video man. Keep it up!

[@BriTheMathGuy]

Thanks a ton!

[@amirb715]

no need for gamma or beta function at all....it's much more simpler. @2:30 just let z=u/(1+u) then u=z/(1-z) and it becomes a simple algebraic integral.

[@simon39wang43]

This is absolutely amazing

[@BriTheMathGuy]

glad you thought so!

[@simon39wang43]

@@BriTheMathGuy thanks, I subscribed

[@gabitheancient7664]

indeed, that is now my favourite integral, will save in my favorites omg

[@edoardomartignoni3222]

Frickin awesome indeed… 👍👍

[@BriTheMathGuy]

Glad you thought so! Thanks for watching :)

[@kjl3080]

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

[@jonathandambrosio4628]

Great videos! So well made and fun to watch.

[@BriTheMathGuy]

Glad you like them!

[@RedTitan5]

Wow... New info... Thank you for sharing...sir

[@BriTheMathGuy]

So nice of you!

[@jrcarlyon680]

Please please keep your hands still. Thanks so much🙏

[@giuseppemalaguti435]

grande spiegazione..bravo

[@YaBoyUneven]

I swear every time a problem turns out like that I get a mathgasm, there, I said it

[@khaledchatah3425]

WHAT THE ACTUAL FUCK. THAT WAS SO FKIN SATISFYING

[@yurfwendforju]

though I wont be able to remember in 2 minutes, it is indeed my favorite integral at the moment.

[@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

[@bhz8947]

My math is very rough (what I remember from my C.S. degree), but integration has always seemed so ad hoc to me, like a collection of disjointed heuristics. Is it basically a tradeoff between the complexity of how we represent formulas vs. the complexity of the algorithm we use to integrate them?

[@Павал-л8ч]

Like in proving theorems, there is probably no general algorithm in solving integrals analytically

[@BernardBolanowski]

at 1:52 x is just u^1/phi or a radical with index of phi

[@miguelcerna7406]

Fascinating. I'm fascinated.

[@BriTheMathGuy]

Glad to hear it :)

[@buzyparticals3753]

When he said "pheee", I felt that

[@SuperYoonHo]

sum intergral

[@ishanagarwal766]

Mind= so blown even the quarks got split

[@BriTheMathGuy]

🤯

[@lukandrate9866]

Instead of using the beta function it is possible to just transform u^(φ-2)du into u^φ*d(-1/u), substitute t = -1/u and then it simplifies and we get what we wanted

[@chimetimepaprika]

Dude, this year has been like math explosion festinghouse. Nice.

[@aidenwinter1117]

1:39 with respect to me? Never thought I’d get any respect at all 🥺

[@Ben-rd3mg]

What a brilliant integral

[@GlorifiedTruth]

Mind blown, but more in a makes me want to cry way. I need to brush up on my integrals!

[@netanelkomm5636]

4:00 but how do you know that properties of factorials apply to non integer inputs?

[@zahari20]

Please do not confuse people! The beta function is written by B(x,y). Capital Greek beta.

[@Nebulisuzer]

beta is ß not B

[@holyshit922]

My way for calculating this integral Indefinite integral Int(1/(1+x^φ)^φ,x) can be quite easily integrated by parts Int(1/(1+x^φ)^φ,x)=Int((1+x^φ)/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=Int(1/(1+x^φ)^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x) Now we can integrate Int(1/(1+x^φ)^(φ-1),x) by parts with u = 1/(1+x^φ)^(φ-1) and dv = dx After integration by parts we can add integrals we will get Int(1/(1+x^φ)^φ,x)=Int(1/(1+x^φ)^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)-Int(x*(-(φ-1)(1+x^φ)^(-(φ-1)-1))*φ*x^(φ-1),x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(φ(φ-1)x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(φ*1/φ*x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+Int(x^φ/(1+x^φ)^φ,x)-Int(x^φ/(1+x^φ)^φ,x) Int(1/(1+x^φ)^φ,x)=x/(1+x^φ)^(φ-1)+C If we want to calculate this integral using antiderivative we have to calculate the limit Now we need to calculate limit limit(x/(1+x^φ)^(φ-1),x=infinity) To calculate this limit all we need to do is some algebraic manipulations limit(x/(1+x^φ)^(φ-1),x=infinity)=limit(1/((1+x^φ)^(φ-1)/x),x=infinity) =1/limit(((1+x^φ)^(φ-1)/(x^(1/(φ-1))^(φ-1))),x=infinity) =1/limit((1+x^φ)^(φ-1)/x^(φ)^(φ-1),x=infinity) =1/limit(((1+x^φ)/x^φ)^(φ-1),x=infinity) =1/limit((1+1/x^φ)^(φ-1),x=infinity)

[@Samir-zb3xk]

An interesting fact that I realized from playing around with this integral is that Γ(φ+1)=Γ(1/φ) even though φ+1≠1/φ

[@noyadishon6649]

THIS IS SO COOL

[@BriTheMathGuy]

😎😎😎😎

[@Shreyas_Jaiswal]

Yes, now I can integrate any expression, just by introducing a new function. 😀😀

[@geraltofrivia9424]

This is beautiful

[@threepointone415]

This is a very cool integral, but my favorite is probably the integral from -infinity to +infinity e^-x^2 dx, which comes out as √pi

[@tusharchilling6886]

I am amazed to see that mathematicians have come up with shortcuts for integration as well to save our time. Although I did not understand the technical stuff, I loved how this abstraction and clever substitution of beta function made this problem so beautiful. I remember studying gamma functions in my college. These are some college level concepts

[@nandanmadhuj7172]

I substituted x^(phi)=tan^2(theta). Nice problem.

[@BriTheMathGuy]

Cool!

[@TheTorturer666]

Favoritegral.

[@BriTheMathGuy]

😎

[@storeksfeed]

I literally just randomly googled this video lol. But I like it :)

[@BriTheMathGuy]

Great to hear!

[@ethanstrumwasser8798]

awesome video as always! only thing wrong with them is that they end :P

[@BriTheMathGuy]

Thanks! 😄

[@dipayandasgupta7506]

This was on cantor dust level 1…

[@monx94]

Nice!

[@BriTheMathGuy]

Thank you! Cheers!

Make some videos about zeta function , please

[@mathematicsmi]

kzbin.info/www/bejne/qZrdn5Vvqpqrnqs

[@coast-guard-1cargo-spectio552]

(You should pay a fine for calling Phi, Fee.)

[@BriTheMathGuy]

👮‍♂️

[@chanderule605]

I mean, that's how every language but english calls it

[@chjxb]

this is not maths but literature

[@BriTheMathGuy]

🧐

[@shanmugasundaram9688]

I wonder the golden ratio has so many identities.

[@PianoBounty]

I don’t think you can actually use the factorial for non-natural numbers. Why not use the properties of the gamma function instead?

[@tusharchilling6886]

What if b is larger? It just makes b=0. And if b is 0, a should be a negative number? And also, a+b/b will tend to infinity which means a/b=infinity which just means a=0 x infinity But a is negative, right? The golden ratio formula has a contradiction

[@tsan_jey]

I am watching this and pretend I understand everything after the word "integral."

[@jatloe]

That was very cool :)

[@BriTheMathGuy]

Glad you thought so :)

[@elwind762]

I clickbaited because the differential variable wasn’t specified in the thumbnail :(

[@BriTheMathGuy]

🥲

[@sharpnova2]

is it ok to undo the analytic continuation of factorials (namely Γ(x)) when your argument isn't a natural number? it'll usually work, but you'll need to stipulate caveats regarding the evaluation of factorials, enforcing their arrangement as quotients of successive or integer-modulo arguments to the factorial function. Γ(x) also has the property that Γ(x + n) = x^n * Γ(x). so you could have just evaluated that directly. very pretty problem. i hardly ever see people talk about β(x, y)

[@aadilhasan8319]

feynman’s technique for a parameter k in place of phi, so differentiating under the integral?

[@Swaaaat1]

At the beginning i thought this was an ex joke.

[@catsarecool9773]

I hit the like button as soon as I heard you were pronouncing phi correctly LOL

[@cheesecak11857]

it's now my favourite integral lol

[@alibekturashev6251]

some people pronounce φ as phai and others as pheeh. which one should i use? i’m not an english speaker and i’m confused

[@GuillermoMartinez-eq7kt]

Great video! But a Gamma function is equivalent to its argument factorial only when the argument is an integer, which is not the case for Phi. Is this correct?

[@kennethgee2004]

ah what so the answer of one seem pretty reasonable because the integral of 1/x is ln x and the limit of (1+1/x)^x as x goes to infinity is e. The answer to ln e is 1. As an aside, 0! is not equal to 1 despite popular opinion. The gamma is not an exact answer for generalized factorials. The definition of factorial requires that the domain is the set of natural numbers. The gamma function is generalized in that it is an analytic continuation, but it loses meaning when things are more generalized.

[@lukandrate9866]

Γ(1+n) = n! for all natural n Γ(1+0) = integral from 0 to inf of e^(-t) dt = 1 implies 0! = 1

[@Fire_Axus]

where is the video right here?

[@simarjeetsingh2589]

Yo the algorithm finally recommending some good channels!

[@user-wx8bm1pg1d]

Damn it. I was hoping this didn't require a function I don't know of

[@BriTheMathGuy]

There are other ways to do it!

[@navsha2]

I guess that the golden ratio has infinite possibilities ( If you can change the value of the variables such as x)

[@decreasing_entropy3003]

Isn't Γ(x)=x! iff x belongs to the set of natural numbers? Here, φ is not a natural number, so shouldn't we not be using notion of factorial?

Where's the plot? I need a plot of the distro! Where is my plot!!

[@alexandrepatrot]

Nice video, but notice you should have employed the properties of Gamma rather than the factorial at the end of the calculations. Gamma(phi-1)= (phi-1)Gamma(phi-2) and the result follows. Observe that factorial emerges only for natural numbers which is not the case for phi.

[@Dream-op5th]

How do you get that black background?...do you literally record it in a black planted rookm with a light on you?

[@BriTheMathGuy]

Green screen :)

[@joyli9893]

Golden integral

[@fasihullisan3066]

very nice

[@BriTheMathGuy]

Glad you thought so!

[@igvc1876]

you might want to reconsider showing yourself in the picture.

[@BriTheMathGuy]

What makes you say that? (genuinely asking)

[@alperenerol1852]

There is a famous saying, if there is a nasty integral the answer is a gamma or beta function.

[@sethdon1100]

If you love the Gamma Function, prove that gamma(z)*gamma(1-z)=pi/sin(pi*z) for all non integer z

[@MaxBerson]

Shout out for pronouncing the Greek correctly!

[@lapizuko]

Instead of using some properties of the beta function which feels quite.. obscure I guess, you could've really proven that result by deriving the prof of that property in that specific case or something.

[@NeagVasileNV-P]

Hello! :) Could you tell me, please, the name of the software that you're using in the presentation? :) TY!

[@thetruealex7478]

Does this integral have a practical use?

[@tamazimuqeria6496]

New integral idea Int from -infinity to infinity (e^((-(x-m)^2)/(2(n)^2))dx

[@rubenvela44]

Golden triangle

[@ZeroJingz]

am i the only one who didn't understand a thing but still watched till the end! 🤣

[@Zi7ar21]

Nah Rendering Equation is my favorite

[@otesunki]

....i swear ive seen this....