Man, I remember when I watched Peyam solve this integral, you did it on 3x speed, this has been just beautiful, we should call you Magician Kamaal because you're pulling tricks left and right
15:09 i just finished calculus I and did not understand most of the things of the video, but i think it is pretty cool watching you solving these integrals
The same denominator of the integrand (x² +2xcosα + 1) appears in the result of the sum ∑ Rⁿcos(nθ), for n ≥ 0. (That is, if we let R=-x and α=θ) I wonder if this has any significance, of if it could be used to solve this integral in another way. edit: nevermind, after 4:00 you used a very similar argument
I couldn't help but notice that alpha=0 limit exsit in that gorgeous answer, so I did it as a homework and got 1/2ln(e^-γ * pi/2) which is a nice bonus result for a special case of first integral at alpha=0
On a serious note tho, most of your videos actually do in some way or another and it's really enjoyable to watch you layout the solution development for such problems. I really need to sleep now but I will give it another go in the morning am sure.
This is far harder than the coxeter's integral. What makes it insane is that it's linked to geometric series, Fourier series, Laplace transform. Almost all the craziest analysis stuff where required to solve it. It was really a worthy opponent.
Hello! Is there a "simple" function whose integral has a very "complex" closed form? By a simple function, I mean a function similar to the one in the video, consisting of a few elementary functions, at most rational. By a very complex closed form, I mean expressions that involve various constants, raficals, logarithms, sums and multiplications etc. (ellipitc integrals, hylergeometric, gamma, zeta functions are not allowed, solution must be close). Thanks for any insights!
1) you forgot a - sign at 14:30 in the second sum 2) If you replace cos(alpha) with cos(alpha * x) in the denominator, is this related to Coxeter's integral?
@@maddog5597 the parametrised results has 1/sin(alpha), which would suggest the integral diverges for multiples of alpha that is a whole multiple of pi.
@@Sugarman96 No. As alpha goes to zero, the Gamma functions cancel out leaving you with log(2*pi)^ (alpha/pi). You can bring the alpha/pi out in front of the log to give you a term that looks like alpha/sin(alpha). As alpha approaches zero, this approaches one.
@@maddog5597 why can you just take the limit when the result comes from just plugging in a value of alpha? And how does that work for values of alpha that aren't zero? The gamma functions _don't_ cancel out then
@@Sugarman96 I was just responding to your original post, where you claimed ln(ln(x))/(x+1)^2 diverged. The main problem is that Kamaal should have put limits on alpha. The integral diverges for alpha = pi. It probably diverges for other values of alpha that make the cos go negative.
At 10:26-ish, you need to divide the upstairs and downstairs by 2j, not just 2, to get the complex sine definition. Ultimately you're still multiplying by 1, but the point stands 😉
This one was from a patreon subscriber. Sorry I haven't been very responsive to requests, I've been extremely occupied these past few months. Even this integral was recommended to me almost a year ago.
Shoutout to one of the few world famous Swedish matematicians. We also have Ivar Fredholm and 50% of Harry Nyquist. The Norwegians of course have Marius Sophus Lie, who's on another level altogether. And the Danes have all the celebrated Physics/QM guys. The Finns have lots of great mathematicians too.
Me before watching this video: the polinomial in the denominator can be computed as (x+e^itheta)(x-e^itheta). now we define I(s) = int from 1 to infinity of log(x)^s/(x-a)(x-b). applying PFD. a = complement of b a = e^itheta. I(s) = int 1 to infinity of (1/b-a) * (log(x)^s/(x-b)) + (1/a-b) * (log(x)^s/(x-b)) . we split into I1(s) and I2(s). apply u = log(x) e^u du = dx e^u *u^s/(e^u-b) = u^s/(1-b*e^-u) so using some melliny mellin magic I1 evaluates to.. iuhun I'm just going to watch the vid now.
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
this integral is zeroeth degree murder
man I know you do crazy integrals but this is something else bro this is a murder
that's a horror movie
Man, I remember when I watched Peyam solve this integral, you did it on 3x speed, this has been just beautiful, we should call you Magician Kamaal because you're pulling tricks left and right
Why am I not surprised that peyam solved it before me😂
Peyam is probably the most underrated RUclipsr in the math space.
Finnaly legendery integral!
Next Coxeter integrals! 🖤
I do believe it's time to face your destiny and tackle the Borwein integrals.
I did not survive... Will try again tomorrow.
@@TheDhdk that's the spirit 🔥
I suffered the little death, in the Shakespearean sense.
The important thing is you survived....just as I did
@@maths_505 It means the big O and I'm not talking about notation
same here😅
“Thank You Math 505” we all said in Unison
so early there isn't even a thumbnail
Breathtaking!
@@trelosyiaellinika and you survived!
@@maths_505 Not only. I enjoyed it immensely!
Hi,
"ok, cool" : 2:23 , 6:24 , 14:54 , 20:55 , 22:28 , 22:57 ,
"terribly sorry about that" : 11:00 , 11:36 , 14:36 , 18:42 , 26:50 , 27:17 .
Extremely Insane. Thank you.
Early in the morning and I'm watching such an integral instead of going to work!!! 😅
Time well spent😂
15:09 i just finished calculus I and did not understand most of the things of the video, but i think it is pretty cool watching you solving these integrals
@@felps314 thanks mate
Is the integral related to the poisson kernel?
The same denominator of the integrand (x² +2xcosα + 1) appears in the result of the sum ∑ Rⁿcos(nθ), for n ≥ 0.
(That is, if we let R=-x and α=θ)
I wonder if this has any significance, of if it could be used to solve this integral in another way.
edit: nevermind, after 4:00 you used a very similar argument
I couldn't help but notice that alpha=0 limit exsit in that gorgeous answer, so I did it as a homework and got 1/2ln(e^-γ * pi/2) which is a nice bonus result for a special case of first integral at alpha=0
Coomer series coming in clutch! 🎉
You could say that the quadratic was a known kernel
Will come back to this video when my math skills are better
@@euler1 bro this video will literally improve your math skills in one go
On a serious note tho, most of your videos actually do in some way or another and it's really enjoyable to watch you layout the solution development for such problems. I really need to sleep now but I will give it another go in the morning am sure.
Aight bro
Sleep well and keep making those math gains.
This is far harder than the coxeter's integral. What makes it insane is that it's linked to geometric series, Fourier series, Laplace transform. Almost all the craziest analysis stuff where required to solve it. It was really a worthy opponent.
Coxeter is honestly just alot of algebra. This however is an entirely different story.
Hello! Is there a "simple" function whose integral has a very "complex" closed form? By a simple function, I mean a function similar to the one in the video, consisting of a few elementary functions, at most rational. By a very complex closed form, I mean expressions that involve various constants, raficals, logarithms, sums and multiplications etc. (ellipitc integrals, hylergeometric, gamma, zeta functions are not allowed, solution must be close). Thanks for any insights!
Great integral man
If I were a prof I'd put that in a test just for pranks
at 10:30 it should be divided by 2i, shouldn't it? ik it doesn't matter but still
1) you forgot a - sign at 14:30 in the second sum
2) If you replace cos(alpha) with cos(alpha * x) in the denominator, is this related to Coxeter's integral?
The negative was factored out. And I don't think this is related to any one of coxeter's integrals.
Wicked awesome.
@@jkid1134 another survivor!
Also, a weirdly specific conclusion is that the integral of ln(lnx)/(x±1)^2 over (1,inf) diverges
No! - ln(ln(x)/(x+1)^2 converges
@@maddog5597 the parametrised results has 1/sin(alpha), which would suggest the integral diverges for multiples of alpha that is a whole multiple of pi.
@@Sugarman96 No. As alpha goes to zero, the Gamma functions cancel out leaving you with log(2*pi)^ (alpha/pi). You can bring the alpha/pi out in front of the log to give you a term that looks like alpha/sin(alpha). As alpha approaches zero, this approaches one.
@@maddog5597 why can you just take the limit when the result comes from just plugging in a value of alpha? And how does that work for values of alpha that aren't zero? The gamma functions _don't_ cancel out then
@@Sugarman96 I was just responding to your original post, where you claimed ln(ln(x))/(x+1)^2 diverged. The main problem is that Kamaal should have put limits on alpha. The integral diverges for alpha = pi. It probably diverges for other values of alpha that make the cos go negative.
Wow that was just insane
I like Kummer series. Exciting integral!
Interestingly, I(a) where a = pi/2 is the result for the Vardi integral, correct?
The universe is now speaking to me
surviving is an overstatement 💀
At 10:26-ish, you need to divide the upstairs and downstairs by 2j, not just 2, to get the complex sine definition. Ultimately you're still multiplying by 1, but the point stands 😉
I survived, but now I carry with me the weight of the experience.
Como se llama el resultado en 15:36?
You must devide by 2i to get sin, not 2.
@@willemesterhuyse2547 yes that's 2 errors cancelling out 😂
Shredder! Minor quibble. At ten minutes thirty, divide by 2i to get sine.
Thankfully we are dividing in the numerator and denominator so the constants cancel.
You ruined my lazy Sunday. Will I survive? Greetings from Germany!
I really enjoyed this. But I have one question I dont understand why you wrote (2*pi)^(alpha/(2*pi)) shouldnt it be just (2*pi)^(alpha/2).
Where do you find all these integrals? Are they sent by the subscribers? I sent you two in the last few months.
This one was from a patreon subscriber. Sorry I haven't been very responsive to requests, I've been extremely occupied these past few months. Even this integral was recommended to me almost a year ago.
10:42 part is beautiful
I think I've survived
Can u cover Green's therom and scattering therom
Shoutout to one of the few world famous Swedish matematicians. We also have Ivar Fredholm and 50% of Harry Nyquist. The Norwegians of course have Marius Sophus Lie, who's on another level altogether. And the Danes have all the celebrated Physics/QM guys. The Finns have lots of great mathematicians too.
Not to mention Niels Henrik Abel.
I survived it! Dr Peyam did this integral on RUclips a while ago.
Me before watching this video:
the polinomial in the denominator can be computed as (x+e^itheta)(x-e^itheta).
now we define I(s) = int from 1 to infinity of log(x)^s/(x-a)(x-b).
applying PFD.
a = complement of b
a = e^itheta.
I(s) = int 1 to infinity of
(1/b-a) * (log(x)^s/(x-b)) + (1/a-b) * (log(x)^s/(x-b)) .
we split into I1(s) and I2(s).
apply u = log(x)
e^u du = dx
e^u *u^s/(e^u-b) = u^s/(1-b*e^-u)
so using some melliny mellin magic I1 evaluates to.. iuhun I'm just going to watch the vid now.
barely survived this one
Oh wow, didn’t realize this integral would invoke the Laplace transform and the Fourier series… wow
I died, but I turned out fine as well
Survived!
What am I missing: As alpha goes to zero, your answer approaches (1/2)log(2*pi). This is a positive number, but the integral should be negative. Huh?
13:31 dont mind me, this is just my personal timestamp
awesome
Now it’s time to prove that identity to ln(gamma)
@@alielhajj7769 it's that time alright....
15:05 Im... still breathing...
Love it
My question: WHERE would you encounter such a monstrosity???
Obviously in a near death experience 😂
why do you do this to yourself
Sorry Kamal but I died. However upon seeing the raw power of what killed me, Eru intervened and had me reborn
Somehow I survived and honestly… I don’t know how
IDK if it was integral or the notation that killed me. :)
Lordy. 🏆
Bro you survived too!
@@maths_505 Dazed, but not confused. 👍
I'm okay
Yoooo I’m alive, but my god man
Again poor maskeroni canseling out😂
RIP my boy kamaal(I hope I am writing your name right)
He will be remembered especially for the reverse cowgirl formulation of integral calculus. Oh wait here I am false alarm guys 😂
I didn't know Yngwie had his own integral. (This is a guitarist joke.)
YOU SURVIVED!
HELL YEAH!!!
Wow
OK COOL❤
Another survivor!
Wow.
You survived too!
i died
Sadly yes...but the good news is you're alive!
Anything
Unfortunately, I did not survive 😮💨
It's cool, try again tomorrow
OK COOL
anything
Er, I got the wrong channel
😂😂😂
Nice and easy😂😂😂😂😂
😂😂😂
I think it was you who recommended this to me a while back.
@@maths_505 no,but i'm glad someone else did,it is beautiful!
I stopped following you at 9:00 , cause I couldn't follow.