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
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
@@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
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.
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
Bro tried to sneak in Feynman
I never get tired of the Feynman technique, so cool!
@@cleojosei well I've got a whole playlist so....
More like reverse Feynman, since we're differentiating the simpler integral instead of integrating
The "Feynman Technique" is known in calculus as "differentiation under the integral sign" and was well known long before Feynman was born!
It was an absolute beast and the Euler Mascheroni constant did not disappear and the master theorem was of good help.
You plugged your merch like a champ 🏆
That proof of convergence in the beginning is much appreciated.
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)
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
I'm not a genius but yeah I've been thinking of diversifying content for a while.
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
That’s what I did as well. Doing gamma function later is better
It’s not negative btw it’s positive e^(iax²)
@@Flylearjet9 cosine is an even function so the real part of e^(ix) and real part of e^(-ix) is both cos(x)
@@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
@@Samir-zb3xk also do you. Have insta, I can send you to RMT that I did
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.
11:49 "similarly" ahh cut
This must be one of the most hilarious product placements I've ever stumbled upon thus easy to forgive. 😊
9:45 pi/8 is how you would slice up a large pizza
Very cool. Thank you.
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
How does convergence on an open interval prove differentiability? It doesn't even suffice to prove continuity.
Destroying Integrals 😤😤
Love your videos but every version of any letter you write is some form of a rotated or reflected letter P
😂😂😂
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?😂💯
I forgot 😂
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
Aight
@@maths_505 ty bro 😊
Why its ends with divergence when we use laplac tran
Brilliant marketing techniques however I have to say!
A nice promotion of your products of Gamma Function
LEIBNIZ REPRESENT!!
utter sick
Response to title :
By taking a nap
@@michakupczyk9560 😂😂😂
Hi,
"ok, cool" : 0:58 , 8:04 ,
"terribly sorry about that" : 7:56 , 9:55 .
First of all Ramanujans master theorem is very difficult to prove...
This is Feynman's trick.
No views in 20 seconds? Bro fell off (love the vids)
Thumbnail is cursed
How so?😂
What? I find it fine
Isn't it Feynman technique ?
More of a Collab between the two. I have more dedicated videos that I attribute to Feynman's trick.
First
Hello @Maths 505 I need a help to solve an equation how can I contact you please
Instagram
Probably the same way he destroyed his underage wife.