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:
• How Ramanujan proved h...
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[@Player_is_I]

Bro tried to sneak in Feynman

[@cleojosei]

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

[@maths_505]

@@cleojosei well I've got a whole playlist so....

[@acelm8437]

More like reverse Feynman, since we're differentiating the simpler integral instead of integrating

[@parthasur6018]

The "Feynman Technique" is known in calculus as "differentiation under the integral sign" and was well known long before Feynman was born!

[@leroyzack265]

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

[@mcalkis5771]

That proof of convergence in the beginning is much appreciated.

[@stefanalecu9532]

You plugged your merch like a champ 🏆

[@Samir-zb3xk]

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

[@Flylearjet9]

That’s what I did as well. Doing gamma function later is better

[@Flylearjet9]

It’s not negative btw it’s positive e^(iax²)

[@Samir-zb3xk]

@@Flylearjet9 cosine is an even function so the real part of e^(ix) and real part of e^(-ix) is both cos(x)

[@Flylearjet9]

@@Samir-zb3xk oh crap I forgot, I used RMT, but for this technique we take the derivative of the gamma function of 1/2. Then what do we do. I have one solution, it I want to know the other solution. RMT is what I used though

[@Flylearjet9]

@@Samir-zb3xk also do you. Have insta, I can send you to RMT that I did

[@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

[@maths_505]

I'm not a genius but yeah I've been thinking of diversifying content for a while.

[@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ]

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.

[@MrWael1970]

Very cool. Thank you.

[@BikeArea]

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

[@maxvangulik1988]

11:49 "similarly" ahh cut

[@alielhajj7769]

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

[@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ]

How does convergence on an open interval prove differentiability? It doesn't even suffice to prove continuity.

[@barryzeeberg3672]

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

[@jakehobrath7721]

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

[@maths_505]

😂😂😂

[@anas-altaleb]

Why its ends with divergence when we use laplac tran

[@yoav613]

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?😂💯

[@maths_505]

I forgot 😂

[@TheIndianPrince_]

Destroying Integrals 😤😤

[@ericthegreat7805]

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

[@maths_505]

Aight

[@ericthegreat7805]

@@maths_505 ty bro 😊

[@CM63_France]

Hi, "ok, cool" : 0:58 , 8:04 , "terribly sorry about that" : 7:56 , 9:55 .

[@dharunpranay8581]

A nice promotion of your products of Gamma Function

[@empanador2748]

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

[@kingzenoiii]

LEIBNIZ REPRESENT!!

[@michakupczyk9560]

Response to title : By taking a nap

[@maths_505]

@@michakupczyk9560 😂😂😂

[@الْمَذْهَبُالْحَنْبَلِيُّ-ت9ذ]

Brilliant marketing techniques however I have to say!

[@chancia8990]

utter sick

[@zemm9003]

This is Feynman's trick.

[@Khushal435]

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

[@modern_genghis_khan0393]

Isn't it Feynman technique ?

[@maths_505]

More of a Collab between the two. I have more dedicated videos that I attribute to Feynman's trick.

[@wolfcanyon]

Thumbnail is cursed

[@maths_505]

How so?😂

[@mutenfuyael3461]

What? I find it fine

[@Pratyush641]

First

[@borhenbouchniba]

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

[@maths_505]

Instagram

[@diogeneslaertius3365]

Probably the same way he destroyed his underage wife.