This integral breaks math

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[@evankassof9773]

"I would say that's real progress, but now to make some complex progress..." is a genuinely great joke. Your videos are awesome!

[@waddy1057]

You can just say it is 2pi straight away by the complex mean value theorem (the contour integral one) Edit: 2*pi*f(a) = int f(a + e^ix) where the integral runs from 0 to 2*pi. Which can be proved from Cauchys theorem

[@nicolastorres147]

The fact that this crazy complex integral evaluated to tau 🤯

[@alexdefoc6919]

What is tau? (ik what is tau)

[@amritlohia8240]

@@alexdefoc6919 tau = 2*pi...

[@ahsgdf1]

Great problem, great solution. But mine is simpler in that it does not require the reflexion formula. Let e^(i x) -> z, dx-> -i dz/z, then the integral becomes I = - i Int{|z|=1}(1/z 1/Gamma(1-z)dz). Since 1/Gamma(1-z) has no singularity for |z|0) = 2 pi*1 = 2 pi qed. Summary: the Gamma function effectively drops out of the integrand.

[@theelk801]

did you even need to do the manipulations at the start before doing the sub? you could just start with the sub and have the contour integral of 1/(iz Γ(1-z)) and then from there the residue is super easy to compute since the gamma function has no zeros

[@maths_505]

Well now that you mention it .....I guess not. But do you really expect me to miss out on a chance to apply the reflection formula?😂

[@theelk801]

oh I get it, but I still have to clock in for my job as the residue guy

[@maths_505]

@@theelk801 and you do an excellent job indeed.

[@HaliPuppeh]

OMG dude, I LOST it at "The first thing we should notice about this integral is.. WHAT THE HELL!?!?!?". I'm laughing so hard I can't see.

[@kucckumelon2837]

My eyes are opened master. Thank you for this integral

[@Trimza42]

Hey @Maths 505, awesome integral! I was wondering what software you use as a blackboard? I am hoping to get into tutoring soon but the software I'm currently using is trash.

[@maths_505]

Samsung notes.

[@AzharLatif-d4z]

By letting Gamma(1-e^ix) =u and integrating the given integral we still arrive at 2 pi in one go?

[@MyphysicsMaster0337]

I'm very thankful for making these videos by spending your time and energy Sir

[@GreenMeansGOF]

You can replace the upper limit of integration by the whole integral lol

[@maths_505]

😂😂😂

[@TMH2007]

The best video you uploaded till date!(I guess)

[@manstuckinabox3679]

Hey! it's a problem that's Not from gamelin's book! That was something new, nice!

[@roeelazar]

Wow! One of the finest integrals in this channel

[@prosimion]

can you show a simple base change differential for this statement/integral. I'm not a mathematician so it's a bit of a task to work it out. thank you none the less

[@shelumieldilen]

Wow. Amazing integral.

[@MrWael1970]

Amazing result. What about when we integrate over real domain? Thanks.

[@maths_505]

I haven't found a solution using purely real techniques. Hopefully I find one sometime.

[@AliAkl-un2ys]

Nice ,poles of gamma function

[@herbertdiazmoraga7258]

without a doubt, this is one of the best videos of the channel!

[@maths_505]

Hell yeah🔥 Thanks bro

[@illumexhisoka6181]

I realized something (Maybe a shortcut) if we continue with a substitution and put the limits of integration it will be the integral from 1 to 1 so for a second it will look like the value of the integral is zero (and I am not sure why it's not) but since e^(ix) Is 2π periodic 2π also do the same effect as 0 Actually any number in the form 2kπ will do (When k is an integer) so the value of the integral "must"(I feel like) Be in the form of 2kπ is there a way to proof that k have to be 1 to give the value of the integral (I am illumi x hisoka account but I changed my picture and name please tell me if you recognize my account)

[@maths_505]

No bro that's not how it works. The problem is essentially 2 dimensional: we're integrating along a curve in the complex plane so just substituting the endpoints of the interval doesn't make sense.

[@illumexhisoka6181]

@@maths_505 I whenever I see a complex numbers in a integral problem I never think that it's special I will just treat it like any other number (Of course using it's propties and Euler formula) Never thought about it graphicly

[@Jacob.Peyser]

Liouville's Theorem would simply decimate this integral...

[@fuckayo-ti1fj]

What is it?

[@amritlohia8240]

@@fuckayo-ti1fj Liouville's theorem says that all bounded holomorphic functions are constant ... but that doesn't seem to be relevant here. Instead, the most relevant theorem is Gauss's mean value theorem, which says that int_{0}^{2*pi}f(a+e^(ix)) dx = 2*pi*f(a) for all holomorphic functions f.

[@DeanOfBloons]

lol nice script on this one b, had me smiling

[@lmaorofl3229]

that is really cool

[@mikeoffthebox]

Very enjoyable!

[@CommanderdMtllca]

@6:00 where did the extra z in the numerator come from?

[@NaradaFox]

Consider this formula for calculating residues: RES z->z_0(f(z)) = lim z->z_0 {(z - z_0)•f(z))}, where z_0 is the singularity of f(z). In this case, f(z) = 1/z • G(z)sin(πz) innit has a singularity(pole) at z = 0, substitute that into the limit; lim z->0 {(z - 0)•1/z • G(z)sin(πz)} = lim z->0 {z • 1/z • G(z)sin(πz)} = lim z->0 {G(z + 1)sin(πz)/z} = lim z->0 {G(z + 1)} • lim z->0 sin(πz)/z = 1 • π•lim z->0 {sin(πz)/πz} = 1 • π = π. So that's why there's a new z from the numerator, by the formula itself

[@nerden1549]

1:30 I already discovered that in desmos, but can someone explain me quickly how can you prove that equation is true?

[@maths_505]

I proved the reflection formula in a separate video: kzbin.info/www/bejne/a4eoZp6Aiq95e5I

[@nerden1549]

@@maths_505 ok thank you

[@yoav613]

Thanks for this nice integral!,you are doing a great job👌

[@uzielfridlich]

Very nice

[@angusclark6170]

How does the z = e^ix sub work? I've seen something similar before where I had to make a similar substitution but I just tried to transform the limits of the integral to give me an integral from 0 to 0 of a function (which was setting off alarms in my head but i didn't know what to do) so i just said the integral evaluated to 0. this thing of turning the integral into a contour integral does make sense because the interval of integration does become a circle, but is there anything to it? any specific justification of why we can turn it into a contour integral, is that something you can always do if you path of integration happens to form a closed loop?

[@maths_505]

In Phineas voice: Yes, yes it is. The function e^(ix) traces out a circle for x going from 0 to 2pi. So the substitution will always yield a contour integral over the unit circle. If you're still confused about it, check out my video on the complex exponential function in the complex analysis lectures playlist.

[@abdoshaat3304]

I don't know what do you mean by solving integrals for living you are a math teacher or professor Because i don't think it's a job Enlighte me if you can

[@maths_505]

Full time KZbinr who pretty much solves integrals for a living.

[@vadymshtabovenko3532]

well couldn't we just say that e^ix equals to -1? Then gamma function of 1 - (-1) is simply 1! -> integrating 1 with respect to x is x -> plug 2pi - 0 and here's our answer! Maybe my approach is not so fascinating, but still :)

[@Nipppppppppp]

x is the variable of integration, ur thinking of when x is pi

[@vadymshtabovenko3532]

oh you are right, i get what you are talking about. So my approach is both wrong and boring lol@@Nipppppppppp