How Ramanujan would DESTROY this tough integral

A fascinating integral solved using Feynman's trick and ramanujan's master theorem.

Proof of the master theorem:

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Advanced MathWear:

[cleojosei]

I never get tired of the Feynman technique, so cool!

[mcalkis]

That proof of convergence in the beginning is much appreciated.

[leroyzack]

It was an absolute beast and the Euler Mascheroni constant did not disappear and the master theorem was of good help.

[GiornoYoshikage]

I guess one of solutions is substitution and Feynman's trick applied to the integral computed via Laplace transform (haven't watched the video yet)

[marcelobrenha]

hey Kamal, i have been following your videos for some 3 months or so and I think (know) that you are a fu%^$# genious. I want you to have as much success in the networks as you can get and for that reason I would suggest once in a while to go down on the difficulty level of the integral or any type of problem your are dealing with to talk to a much wider range of people. Of course what makes you fenomenal and this is what many of your current followers are after is that you are a fuck**ing genious that can solve unbelieveble problems like it is nothing, but just here and there, try to solve something that normal people would like to try and solve. I am pretty sure you will have a much broader audince by just piking some easier problems and solving them online once in a while. Cheers and keep going. Your are a monster

[BikeArea]

This must be one of the most hilarious product placements I've ever stumbled upon thus easy to forgive. 😊

[Samir-zbxk]

I solved it by placing the parameter in the argument of the sine function:
I(a) = (0 to ∞) ∫ sin(ax²) ln(x) / x² dx
I'(a) = (0 to ∞) ∫ cos(ax²) ln(x) dx
I'(a) = Re (0 to ∞) ∫ e^(-iax²) ln(x) dx
Which (after substitution) can by solved via the derivative of the gamma function using Γ'(1/2) = -√π (γ+ln(4))

And then after taking the real part, integrating to recover I(1) is fairly straightfoward

[الْمَذْهَبُالْحَنْبَلِيُّ-تذ]

The function isn't bounded on the interval (0, 1); at least not for negative values of α. In fact the integral only ever converges on that interval if α > -3.

[alielhajj]

The integrand is not bounded between 0 and 1, it is bounded for values of alpha greater than pr equal to -2 however this is not enough for differentiation, you need boundedness pr convergence on an open interval containing the value -2, fortunately we have convergence for value of alpha strictly greater than -3 (using some p value test), combining with the convergence of the tail for alpha strictly less than -1 we have garanteed convergence for alpha on the interval (-3, -1) which indeed contains -2 and hence we can differentiate under the integral at -2, the lower bound of -3 is strict however the upper limit can be pushed further and need some more advanced convergence test

[barryzeeberg]

9:45 pi/8 is how you would slice up a large pizza

[jakehobrath]

Love your videos but every version of any letter you write is some form of a rotated or reflected letter P

[yoav]

Nice!! 8 monthes ago you said that you are going to make a video about Hardy's proof to RMT, so should we wait 8 monthes or more?😂💯

[dharunpranay]

A nice promotion of your products of Gamma Function

[anas-altaleb]

Why its ends with divergence when we use laplac tran

[ericthegreat]

Can you make a video on the pattern for integrals of the form

Sin(x^n)/x^n

Since I assume they are recursively related to

Sin(x)/x

[CM_France]

Hi,

"ok, cool" : 0:58, 8:04,

"terribly sorry about that" : 7:56, 9:55 .

[empanador]

No views in 20 seconds? Bro fell off (love the vids)

[الْمَذْهَبُالْحَنْبَلِيُّ-تذ]

Brilliant marketing techniques however I have to say!

[Khushal]

First of all Ramanujans master theorem is very difficult to prove...

[borhenbouchniba]

Hello @Maths 505 I need a help to solve an equation how can I contact you please