Ramanujan wins again!!

[@weo9473]

Why is Ramanujan so intelligent 😭

[@sajuvasu]

Coz he is ramanujan... 🗿

[@Eevee8858]

💀 bro I don't think intelligent is the right word for him, blud could spew random gibberish only to later realise "oh wow a new result alright"

[@FundamSrijan]

Cuz he *_was_*

[@agytjax]

Coz his diet was primarily rice with curd 😅😅

[@FundamSrijan]

@@agytjax and he was veg

[@ajiwibowo8736]

Interesting as always!

[@drpeyam]

Love that you're enjoying the channel! 😄

[@Calcprof]

Long live the Melin transform!

[@sajuvasu]

Merlin's beard😂 IYKYK

[@jyotsanabenpanchal7271]

Very long time seeing you doing these type of integrals. Your old videos are awesome (I have never seen explaining that good in those monster integrals.) Love you!❤ Dr. Peyam You're really helping me a lot in my prep days!

[@YoungPhysicistsClub1729]

Actual ramanujan style answer Step1: write answer ends video

[@Samir-zb3xk]

"Where did you get the answer from?" Ramanujan: "I saw the answer in my dream"

[@ianmi4i727]

Also "a goddess of my family showed me the formula in dreams"

[@xinpingdonohoe3978]

Ramanujan's master theorem is so powerful and underused. Now we need a video using Glasser's master theorem. Though for the end bit, Γ(1/n)→∞ because Γ(0)=(-1)! if you will. Still, x=1/n, xΓ(x)=Γ(x+1)→Γ(1)=1, and sin(πx/2)→0, so it's still 1×0=0.

[@mastershooter64]

KZbin should support LaTeX in its comment sections like MSE

[@Keithfert490]

I don't understand how the formula can be correct for all choices of phi. Like, the only condition we had on phi was that it was 0, +1, 0, -1, 0, ... at the positive integers. How can we then use it at real numbers less than 1? What if we had chosen [sin(pi×k/2)]×cos(2k×pi)? That would change the value at k=1/N but not at any of the integers

[@xinpingdonohoe3978]

@@Keithfert490 there will be conditions on φ. Perhaps real, analytic, etc.

[@Keithfert490]

@xinpingdonohoe3978 but the example in my comment is certainly analytic. I don't understand what condition is satisfied by sin(pi×k/2) that would not satisfied by sin(pi×k/2)×cos(2×k×pi)

[@xinpingdonohoe3978]

​@@Keithfert490 your φ(z) is analytic on Re(z)≤0, but exp(-|t|π/2)φ(it) does not have exponential decay as t→∞. They're both assumptions made when proving it.

[@Keithfert490]

@@xinpingdonohoe3978 this is super insightful! Thank you

[@Milan_Openfeint]

Not a single comment about how the sin(x^N) to infinity doesn't make sense? It oscillates wildly (for N>=1). Is this whole thing legitimate?

[@shaharjoselevich7169]

For N+1, you're right that the integral just oscillates. But, for N>1, the indefinite integral oscillates at a higher rate as x increases but the amplitude gets smaller. So it does approach one value

[@alipourzand6499]

Great video. I googled and I saw that there are collection of integrals using this theorem. I can't help but note the similarity with Fourier and Laplace transforms. I'm wondering if there is a physical meaning behind this theorem.

[@joshuaiosevich3727]

The sun from n=2 to infinity of integral from 0 to infinity sin(x^n) also converges. That’s a fun exercise for someone to prove. Finding what is actually converges to is a real trip as well.

[@thomasjefferson6225]

I come for the fashion, the math comes secondary

[@avosdelhevia-y7f]

While doing integration, we use LIATE or ILATE. But i dont understand why it makes sense, like what is the logic behind this rule? Hope you make a video on this.

[@xinpingdonohoe3978]

@avosdelhevia-y7f the logarithmic terms differentiate to 1/polynomial stuff. That can be used to cancel out polynomial stuff. The polynomials terms may eventually differentiate to 0. If we can't speed up the process by having them get cancelled or whatever, we may as well try to make them disappear. But also integrating them is something we know how to do, even though it will make the power higher. So unless there's a logarithmic term to get rid of some of the higher powers, we don't really want to integrate these, as it will just make the integral last longer, possibly never ending. The exponentials don't change too much when integrated. They still behave nicely. It's safest to integrate them whilst we try to simplify the other bits.

[@prateekmourya9567]

Nice shirt!!

[@drpeyam]

Thank you 😊

[@egoreremeev9969]

The moment I saw \int from 0 to inf x^(1/N-1) sin(x) dx i thought of expressing sin(x) = Im(e^ix) and changing the variable to ix=-z. This gives I = 1/N * Im[ (-i)^(1/N-1) \int z=0 to -i*infty z^(s-1) e^(-z) dz] And then using Jordans lemma to move the contour.

[@Samir-zb3xk]

What I usually do is make sin(x) into -Im(e^(-ix)) and then the integral is solved by either Laplace transform or a quick change of variable to make it into gamma function

[@agytjax]

Why can't we use the mod(2) function instead of sin(k*pi/2) ? For the negative sin we can always use (-1) ^ (2k+1)

[@drpeyam]

I guess it needs to be continuous

[@agytjax]

@@drpeyam - Even I thought so, but even sin(k*pi/2) is not continuous when k is an integer

[@drpeyam]

sin(kx) is

[@agytjax]

@@drpeyam - Agreed, but not when sin(k*pi/2) when k E Z

[@HPTopoG]

@@agytjaxThis is probably pedantic, but Z is closed discrete as a subspace of R. So every function on it is continuous. Also, is continuity really enough? I don’t know the master theorem well, but I find that really hard to believe. Continuous extensions to R of functions on Z are not unique without stronger conditions. Perhaps something like analyticity or log convexity is required as with the Gamma function?

[@Premkumar-nv8xf]

Why this integral is true for n=1?

[@rufusjasko]

The integral diverges for n=1, even though if you plug in n=1 to the final result then you get an answer of 1. I'm not entirely sure why this is the case but I presume it's due to one of the following reasons. 1) Ramanujan's Master Theorem assumes the integral converges, or 2) Ramanujan's Master Theorem only applies when s!=1. All the other steps seem to follow even when n=1 (although the u-sub achieves nothing) so I think it has to be something to do with some condition necessary to use the theorem.

[@gowipe-grandcross]

​@rufusjaskothe Ramanujan Master Theorem assume the convergence of the integral.

[@xinpingdonohoe3978]

@@Premkumar-nv8xf perhaps if we graph the integral as a function of n, that would be the analytic continuation's value at 1.

[@mattmolewski7475]

That's the thing that was bothering me, too. Why are we talking about the case of when N is infinity when it's not even clear for what values of N this can be said to be true?

[@gowipe-grandcross]

​@@mattmolewski7475 Well for that, we would have to first study the convergence of the integral depend of n. And I think this integral converge for n > 1

[@ianmi4i727]

Mathematicians ranking: 1- Ramanujan and Euler (absolute Kings) 2- You and the others 3- Me (yes, I'm a mathematician too, I have a Bachelor; also electrical engineer and a Master's Degree in Education)

[@Greninja352]

Where's gauss!

[@slavinojunepri7648]

Fantastic

[@drpeyam]

Thank you!!

[@ahmedlutfi4894]

i wonder if he trying to tone his skin color like Ramanujan

[@ianmi4i727]

lol

[@GUTY1729]

No entendí nada pero debe ser verdad

[@anestismoutafidis4575]

∫ ♾️ ->0 -cos [x^(n+1)/n]•dx ∫ -cosx^(n+1)/n •dx [-cos ♾️^(n+1)/n -(-cos 0^n+1/n)•dx] ♾️->0 [- ♾️/n -(-1/n)•dx]♾️ ->0 [(-♾️ +1)/n •dx] [- ♾️ /n•dx] ♾️ ->0 If n=ℝ\{ ±♾️ , 0}, then ∫ [-♾️ •dx]♾️->0

[@_Grative]

شما ایرانی هستین؟

[@drpeyam]

Bale!!

[@pedrocusinato1743]

I think your solution is wrong for the following reason: You had the sequence x_k = 0, -1, 0, 1, 0, -1, 0, 1, ... and said x_k=-sin(πk/2). That is correct. But you used n=1/k and you can't. There is a non enumerable amount of smooth real functions F such that the restriction of F to non-negative integers is x_k. And the values of different functions F may be different in 1/k. Your pick of -sin(πk/2) is arbitrary.

[@drpeyam]

But it’s continuous though! In any case, the solution is correct

[@jillerkore]

But Doctor Ramanujan is dead!!

[@ВикторПоплевко-е2т]

Please, don't rage like that! The mic is literally like 1 metre to you!🤣

[@ВикторПоплевко-е2т]

Btw, Wolframalfa can't solve this

[@mintusaren895]

Banglte bole kon thasa. Unlogo ki bichar me aya ,lekin buddhe ko keya karange ,LOCAL.