Ramanujan wins again!!

Ramanujan would be proud!!

We calculate the integral from 0 to infinity of sin x^n using Ramanujan's master theorem. This involves by writing the function out as a Taylor or Maclaurin series and then converting it into the Gamma function, which is a generalization of the factorial function. We also use Euler's formula which relates the gamma function with sin and cos. This is a must see for anyone who appreciates Srinivasa Ramanujan's work and math and calculus in general

[jyotsanabenpanchal]

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!

[xinpingdonohoe]

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.

[joshuaiosevich]

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.

[Keithfert]

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

[YoungPhysicistsClub]

Actual ramanujan style answer

Step1: write answer

ends video

[dominicellis]

Mellin transform!!! Could you make video deriving it from the gamma function?

[alipourzand]

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.

[egoreremeev]

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.

[avosdelhevia-yf]

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.

[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?

[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)

[thomasjefferson]

I come for the fashion, the math comes secondary

[ianmii]

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)

[anestismoutafidis]

∫ ♾️ ->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

[ahmedlutfi]

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

[pedrocusinato]

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.

[ВикторПоплевко-ет]

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

[mintusaren]

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